PH507 Homework
Green Function 내용 정리하고 과제 계산 따라하
HW1
1.a
$\nabla(\phi\psi) = \phi\nabla\psi + \psi\nabla\phi$
Proof:
We start with the definition of the gradient operator $\nabla$ in Cartesian coordinates:
$$\nabla = \hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z}$$
$$\nabla(\phi\psi) = \hat{i}\frac{\partial(\phi\psi)}{\partial x} + \hat{j}\frac{\partial(\phi\psi)}{\partial y} + \hat{k}\frac{\partial(\phi\psi)}{\partial z}$$
$$\nabla(\phi\psi) = \hat{i}\left(\phi\frac{\partial\psi}{\partial x} + \psi\frac{\partial\phi}{\partial x}\right) + \hat{j}\left(\phi\frac{\partial\psi}{\partial y} + \psi\frac{\partial\phi}{\partial y}\right) + \hat{k}\left(\phi\frac{\partial\psi}{\partial z} + \psi\frac{\partial\phi}{\partial z}\right)$$
$$\nabla(\phi\psi) = \left(\phi\frac{\partial\psi}{\partial x}\hat{i} + \phi\frac{\partial\psi}{\partial y}\hat{j} + \phi\frac{\partial\psi}{\partial z}\hat{k}\right) + \left(\psi\frac{\partial\phi}{\partial x}\hat{i} + \psi\frac{\partial\phi}{\partial y}\hat{j} + \psi\frac{\partial\phi}{\partial z}\hat{k}\right)$$
$$\nabla(\phi\psi) = \phi\left(\frac{\partial\psi}{\partial x}\hat{i} + \frac{\partial\psi}{\partial y}\hat{j} + \frac{\partial\psi}{\partial z}\hat{k}\right) + \psi\left(\frac{\partial\phi}{\partial x}\hat{i} + \frac{\partial\phi}{\partial y}\hat{j} + \frac{\partial\phi}{\partial z}\hat{k}\right)$$
$$\nabla(\phi\psi) = \phi\nabla\psi + \psi\nabla\phi$$
1.b
$\nabla \cdot (\phi \vec{f}) = \phi \nabla \cdot \vec{f} + \vec{f} \cdot \nabla \phi$
Proof:
Let the vector field $\vec{f}$ be represented in Cartesian coordinates as $\vec{f} = f_x\hat{i} + f_y\hat{j} + f_z\hat{k}$.
$$\phi\vec{f} = \phi f_x\hat{i} + \phi f_y\hat{j} + \phi f_z\hat{k}$$
$$\nabla \cdot (\phi\vec{f}) = \frac{\partial(\phi f_x)}{\partial x} + \frac{\partial(\phi f_y)}{\partial y} + \frac{\partial(\phi f_z)}{\partial z}$$
$$\nabla \cdot (\phi\vec{f}) = \left(\phi\frac{\partial f_x}{\partial x} + f_x\frac{\partial \phi}{\partial x}\right) + \left(\phi\frac{\partial f_y}{\partial y} + f_y\frac{\partial \phi}{\partial y}\right) + \left(\phi\frac{\partial f_z}{\partial z} + f_z\frac{\partial \phi}{\partial z}\right)$$
$$\nabla \cdot (\phi\vec{f}) = \left(\phi\frac{\partial f_x}{\partial x} + \phi\frac{\partial f_y}{\partial y} + \phi\frac{\partial f_z}{\partial z}\right) + \left(f_x\frac{\partial \phi}{\partial x} + f_y\frac{\partial \phi}{\partial y} + f_z\frac{\partial \phi}{\partial z}\right)$$
$$\nabla \cdot (\phi\vec{f}) = \phi\left(\frac{\partial f_x}{\partial x} + \frac{\partial f_y}{\partial y} + \frac{\partial f_z}{\partial z}\right) + (f_x\hat{i} + f_y\hat{j} + f_z\hat{k}) \cdot \left(\frac{\partial \phi}{\partial x}\hat{i} + \frac{\partial \phi}{\partial y}\hat{j} + \frac{\partial \phi}{\partial z}\hat{k}\right)$$
$$\nabla \cdot (\phi \vec{f}) = \phi(\nabla \cdot \vec{f}) + \vec{f} \cdot (\nabla \phi)$$
1.c
$\nabla \times (\phi \vec{f}) = \phi \nabla \times \vec{f} - \vec{f} \times \nabla \phi$
Proof:
The curl is calculated using a determinant. Let $\vec{f} = f_x\hat{i} + f_y\hat{j} + f_z\hat{k}$.
$$\nabla \times (\phi\vec{f}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \phi f_x & \phi f_y & \phi f_z \end{vmatrix}$$
We look at the $\hat{i}$ component first:
$$\left(\frac{\partial(\phi f_z)}{\partial y} - \frac{\partial(\phi f_y)}{\partial z}\right)\hat{i}$$
$$\left( \phi\frac{\partial f_z}{\partial y} + f_z\frac{\partial \phi}{\partial y} - \phi\frac{\partial f_y}{\partial z} - f_y\frac{\partial \phi}{\partial z} \right)\hat{i}$$
$$\left[ \phi\left(\frac{\partial f_z}{\partial y} - \frac{\partial f_y}{\partial z}\right) - \left(f_y\frac{\partial \phi}{\partial z} - f_z\frac{\partial \phi}{\partial y}\right) \right]\hat{i}$$
The first part, $\phi(\frac{\partial f_z}{\partial y} - \frac{\partial f_y}{\partial z})$, is the $\hat{i}$ component of $\phi(\nabla \times \vec{f})$.
The second part, $-(f_y\frac{\partial \phi}{\partial z} - f_z\frac{\partial \phi}{\partial y})$, is the $\hat{i}$ component of $-\vec{f} \times \nabla \phi$.
$$-\vec{f} \times \nabla\phi = -\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ f_x & f_y & f_z \\ \frac{\partial\phi}{\partial x} & \frac{\partial\phi}{\partial y} & \frac{\partial\phi}{\partial z} \end{vmatrix}$$
The $\hat{i}$ component of this is $-\left(f_y\frac{\partial\phi}{\partial z} - f_z\frac{\partial\phi}{\partial y}\right)$, which matches our term.
Since the logic is identical for the $\hat{j}$ and $\hat{k}$ components,
$$\nabla \times (\phi \vec{f}) = \phi(\nabla \times \vec{f}) - \vec{f} \times (\nabla \phi)$$
1.d
$\nabla(\vec{f} \cdot \vec{g}) = \vec{f} \times (\nabla \times \vec{g}) + (\vec{f} \cdot \nabla)\vec{g} + \vec{g} \times (\nabla \times \vec{f}) + (\vec{g} \cdot \nabla)\vec{f}$
Proof:
(LHS)
$$[\nabla(\vec{f} \cdot \vec{g})]_i = \partial_i (f_j g_j)$$
$$[\nabla(\vec{f} \cdot \vec{g})]_i = g_j (\partial_i f_j) + f_j (\partial_i g_j)$$
(RHS)
$[\vec{f} \times (\nabla \times \vec{g})]_i$:
$$[\vec{f} \times (\nabla \times \vec{g})]_i = \epsilon_{ijk} f_j (\nabla \times \vec{g})_k = \epsilon_{ijk} f_j (\epsilon_{klm} \partial_l g_m)$$
Using the identity $\epsilon_{ijk} \epsilon_{klm} = \epsilon_{kij} \epsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}$:
$$= (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}) f_j \partial_l g_m = f_j \partial_i g_j - f_j \partial_j g_i$$
$[(\vec{f} \cdot \nabla)\vec{g}]_i$:
$$[(\vec{f} \cdot \nabla)\vec{g}]_i = (f_j \partial_j) g_i = f_j \partial_j g_i$$
$[\vec{g} \times (\nabla \times \vec{f})]_i$:
Switch f and g from first term.
$$[\vec{g} \times (\nabla \times \vec{f})]_i = g_j \partial_i f_j - g_j \partial_j f_i$$
$[(\vec{g} \cdot \nabla)\vec{f}]_i$
Switch f and g from second term.
$$[(\vec{g} \cdot \nabla)\vec{f}]_i = (g_j \partial_j) f_i = g_j \partial_j f_i$$
Now, we sum the components of the RHS:
$$[RHS]_i = (f_j \partial_i g_j - f_j \partial_j g_i) + (f_j \partial_j g_i) + (g_j \partial_i f_j - g_j \partial_j f_i) + (g_j \partial_j f_i)$$
Canceling out the appropriate terms,
$$[RHS]_i = f_j (\partial_i g_j) + g_j (\partial_i f_j)$$
1.e
$\nabla \cdot (\vec{f} \times \vec{g}) = \vec{g} \cdot (\nabla \times \vec{f}) - \vec{f} \cdot (\nabla \times \vec{g})$
Proof:
We'll prove this by expanding both sides in Cartesian coordinates.
$$\vec{f} = f_x\hat{i} + f_y\hat{j} + f_z\hat{k}$$
$$\vec{g} = g_x\hat{i} + g_y\hat{j} + g_z\hat{k}$$
(LHS)
$$\vec{f} \times \vec{g} = (f_y g_z - f_z g_y)\hat{i} + (f_z g_x - f_x g_z)\hat{j} + (f_x g_y - f_y g_x)\hat{k}$$
$$\nabla \cdot (\vec{f} \times \vec{g}) = \frac{\partial}{\partial x}(f_y g_z - f_z g_y) + \frac{\partial}{\partial y}(f_z g_x - f_x g_z) + \frac{\partial}{\partial z}(f_x g_y - f_y g_x)$$
$$\nabla \cdot (\vec{f} \times \vec{g}) = \left(\frac{\partial f_y}{\partial x}g_z + f_y\frac{\partial g_z}{\partial x} - \frac{\partial f_z}{\partial x}g_y - f_z\frac{\partial g_y}{\partial x}\right) + \left(\frac{\partial f_z}{\partial y}g_x + f_z\frac{\partial g_x}{\partial y} - \frac{\partial f_x}{\partial y}g_z - f_x\frac{\partial g_z}{\partial y}\right) + \left(\frac{\partial f_x}{\partial z}g_y + f_x\frac{\partial g_y}{\partial z} - \frac{\partial f_y}{\partial z}g_x - f_y\frac{\partial g_x}{\partial z}\right)$$
omg..
(RHS)
$\vec{g} \cdot (\nabla \times \vec{f})$:
$$\nabla \times \vec{f} = \left(\frac{\partial f_z}{\partial y} - \frac{\partial f_y}{\partial z}\right)\hat{i} + \left(\frac{\partial f_x}{\partial z} - \frac{\partial f_z}{\partial x}\right)\hat{j} + \left(\frac{\partial f_y}{\partial x} - \frac{\partial f_x}{\partial y}\right)\hat{k}$$
$$\vec{g} \cdot (\nabla \times \vec{f}) = g_x\left(\frac{\partial f_z}{\partial y} - \frac{\partial f_y}{\partial z}\right) + g_y\left(\frac{\partial f_x}{\partial z} - \frac{\partial f_z}{\partial x}\right) + g_z\left(\frac{\partial f_y}{\partial x} - \frac{\partial f_x}{\partial y}\right)$$
$\vec{f} \cdot (\nabla \times \vec{g})$:
$$\nabla \times \vec{g} = \left(\frac{\partial g_z}{\partial y} - \frac{\partial g_y}{\partial z}\right)\hat{i} + \left(\frac{\partial g_x}{\partial z} - \frac{\partial g_z}{\partial x}\right)\hat{j} + \left(\frac{\partial g_y}{\partial x} - \frac{\partial g_x}{\partial y}\right)\hat{k}$$
$$\vec{f} \cdot (\nabla \times \vec{g}) = f_x\left(\frac{\partial g_z}{\partial y} - \frac{\partial g_y}{\partial z}\right) + f_y\left(\frac{\partial g_x}{\partial z} - \frac{\partial g_z}{\partial x}\right) + f_z\left(\frac{\partial g_y}{\partial x} - \frac{\partial g_x}{\partial y}\right)$$
12 terms are same.
1.f
$\nabla \times (\vec{f} \times \vec{g}) = (\vec{g} \cdot \nabla)\vec{f} - (\vec{f} \cdot \nabla)\vec{g} + (\nabla \cdot \vec{g})\vec{f} - (\nabla \cdot \vec{f})\vec{g}$
Proof:
(LHS)
$\nabla \times (\vec{f} \times \vec{g})$:
$$[\nabla \times (\vec{f} \times \vec{g})]_i = \epsilon_{ijk} \partial_j (\vec{f} \times \vec{g})_k = \epsilon_{ijk} \partial_j (\epsilon_{klm} f_l g_m)$$
Using $\partial_j (f_l g_m) = (\partial_j f_l) g_m + f_l (\partial_j g_m)$:
$$= \epsilon_{ijk} \epsilon_{klm} [(\partial_j f_l) g_m + f_l (\partial_j g_m)]$$
Using $\epsilon_{ijk} \epsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}$:
$$= (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}) [(\partial_j f_l) g_m + f_l (\partial_j g_m)]$$
Then
$$= \underbrace{\delta_{il}\delta_{jm}(\partial_j f_l) g_m}_{\text{1}} - \underbrace{\delta_{im}\delta_{jl}(\partial_j f_l) g_m}_{\text{2}} + \underbrace{\delta_{il}\delta_{jm}f_l (\partial_j g_m)}_{\text{3}} - \underbrace{\delta_{im}\delta_{jl}f_l (\partial_j g_m)}_{\text{4}}$$
1: $(\partial_j f_i) g_j = g_j \partial_j f_i$
2: $(\partial_j f_j) g_i$
3: $f_i (\partial_j g_j)$
4: $f_j (\partial_j g_i)$
$$[LHS]_i = g_j \partial_j f_i - (\partial_j f_j) g_i + (\partial_j g_j) f_i - f_j \partial_j g_i$$
(RHS)
$[(\vec{g} \cdot \nabla)\vec{f}]_i = (g_j \partial_j) f_i = g_j \partial_j f_i$
$[(\vec{f} \cdot \nabla)\vec{g}]_i = (f_j \partial_j) g_i = f_j \partial_j g_i$
$[(\nabla \cdot \vec{g})\vec{f}]_i = (\partial_j g_j) f_i$
$[(\nabla \cdot \vec{f})\vec{g}]_i = (\partial_j f_j) g_i$
4 terms are same.
1.g
$\nabla \times \nabla \phi = 0$
Proof:
$$\nabla \phi = \frac{\partial \phi}{\partial x}\hat{i} + \frac{\partial \phi}{\partial y}\hat{j} + \frac{\partial \phi}{\partial z}\hat{k}$$
$$\nabla \times (\nabla \phi) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} & \frac{\partial \phi}{\partial z} \end{vmatrix}$$
$$\nabla \times (\nabla \phi) = \left( \frac{\partial}{\partial y}\left(\frac{\partial \phi}{\partial z}\right) - \frac{\partial}{\partial z}\left(\frac{\partial \phi}{\partial y}\right) \right)\hat{i} + \left( \frac{\partial}{\partial z}\left(\frac{\partial \phi}{\partial x}\right) - \frac{\partial}{\partial x}\left(\frac{\partial \phi}{\partial z}\right) \right)\hat{j} + \left( \frac{\partial}{\partial x}\left(\frac{\partial \phi}{\partial y}\right) - \frac{\partial}{\partial y}\left(\frac{\partial \phi}{\partial x}\right) \right)\hat{k}$$
$$\nabla \times (\nabla \phi) = \left( \frac{\partial^2 \phi}{\partial y \partial z} - \frac{\partial^2 \phi}{\partial z \partial y} \right)\hat{i} + \left( \frac{\partial^2 \phi}{\partial z \partial x} - \frac{\partial^2 \phi}{\partial x \partial z} \right)\hat{j} + \left( \frac{\partial^2 \phi}{\partial x \partial y} - \frac{\partial^2 \phi}{\partial y \partial x} \right)\hat{k}$$
Clairaut's theorem (e.g, $\frac{\partial^2 \phi}{\partial y \partial z} = \frac{\partial^2 \phi}{\partial z \partial y}$):
$$\nabla \times \nabla \phi = (0)\hat{i} + (0)\hat{j} + (0)\hat{k} = \vec{0}$$
1.h
$\nabla \cdot (\nabla \times \vec{f}) = 0$
Proof:
Curl of a vector field $\vec{f} = f_x\hat{i} + f_y\hat{j} + f_z\hat{k}$:
$$\nabla \times \vec{f} = \left(\frac{\partial f_z}{\partial y} - \frac{\partial f_y}{\partial z}\right)\hat{i} + \left(\frac{\partial f_x}{\partial z} - \frac{\partial f_z}{\partial x}\right)\hat{j} + \left(\frac{\partial f_y}{\partial x} - \frac{\partial f_x}{\partial y}\right)\hat{k}$$
$$\nabla \cdot (\nabla \times \vec{f}) = \frac{\partial}{\partial x}\left(\frac{\partial f_z}{\partial y} - \frac{\partial f_y}{\partial z}\right) + \frac{\partial}{\partial y}\left(\frac{\partial f_x}{\partial z} - \frac{\partial f_z}{\partial x}\right) + \frac{\partial}{\partial z}\left(\frac{\partial f_y}{\partial x} - \frac{\partial f_x}{\partial y}\right)$$
$$\nabla \cdot (\nabla \times \vec{f}) = \frac{\partial^2 f_z}{\partial x \partial y} - \frac{\partial^2 f_y}{\partial x \partial z} + \frac{\partial^2 f_x}{\partial y \partial z} - \frac{\partial^2 f_z}{\partial y \partial x} + \frac{\partial^2 f_y}{\partial z \partial x} - \frac{\partial^2 f_x}{\partial z \partial y}$$
Clairaut's theorem: $$\nabla \cdot (\nabla \times \vec{f}) = \left(\frac{\partial^2 f_z}{\partial x \partial y} - \frac{\partial^2 f_z}{\partial y \partial x}\right) + \left(\frac{\partial^2 f_y}{\partial z \partial x} - \frac{\partial^2 f_y}{\partial x \partial z}\right) + \left(\frac{\partial^2 f_x}{\partial y \partial z} - \frac{\partial^2 f_x}{\partial z \partial y}\right)$$
Each term is zero.
1.i
$\nabla \times (\nabla \times \vec{f}) = \nabla(\nabla \cdot \vec{f}) - \nabla^2 \vec{f}$
Proof:
(LHS) $\hat{i}$-component
$$\nabla \times \vec{f} = \left(\frac{\partial f_z}{\partial y} - \frac{\partial f_y}{\partial z}\right)\hat{i} + \left(\frac{\partial f_x}{\partial z} - \frac{\partial f_z}{\partial x}\right)\hat{j} + \left(\frac{\partial f_y}{\partial x} - \frac{\partial f_x}{\partial y}\right)\hat{k}$$
$$[\text{LHS}]_{\hat{i}} = \frac{\partial}{\partial y}\left(\frac{\partial f_y}{\partial x} - \frac{\partial f_x}{\partial y}\right) - \frac{\partial}{\partial z}\left(\frac{\partial f_x}{\partial z} - \frac{\partial f_z}{\partial x}\right)$$
$$[\text{LHS}]_{\hat{i}} = \frac{\partial^2 f_y}{\partial y \partial x} - \frac{\partial^2 f_x}{\partial y^2} - \frac{\partial^2 f_x}{\partial z^2} + \frac{\partial^2 f_z}{\partial z \partial x}$$
$$[\text{LHS}]_{\hat{i}} = \left(\frac{\partial^2 f_y}{\partial y \partial x} + \frac{\partial^2 f_z}{\partial z \partial x} + \frac{\partial^2 f_x}{\partial x^2}\right) - \left(\frac{\partial^2 f_x}{\partial x^2} + \frac{\partial^2 f_x}{\partial y^2} + \frac{\partial^2 f_x}{\partial z^2}\right)$$
(RHS)
$\nabla(\nabla \cdot \vec{f})$:
$$\nabla \cdot \vec{f} = \frac{\partial f_x}{\partial x} + \frac{\partial f_y}{\partial y} + \frac{\partial f_z}{\partial z}$$
$$[\nabla(\nabla \cdot \vec{f})]_{\hat{i}} = \frac{\partial}{\partial x}\left(\frac{\partial f_x}{\partial x} + \frac{\partial f_y}{\partial y} + \frac{\partial f_z}{\partial z}\right) = \frac{\partial^2 f_x}{\partial x^2} + \frac{\partial^2 f_y}{\partial x \partial y} + \frac{\partial^2 f_z}{\partial x \partial z}$$
$-\nabla^2 \vec{f}$:
$$[-\nabla^2 \vec{f}]_{\hat{i}} = -(\nabla^2 f_x) = -\left(\frac{\partial^2 f_x}{\partial x^2} + \frac{\partial^2 f_x}{\partial y^2} + \frac{\partial^2 f_x}{\partial z^2}\right)$$
6 terms are same.
2.a
$$\int_V \nabla f \, d\tau = \oint_S f \hat{n} \, dS$$
Proof:
Let's prove the x-component of the identity:
$$\int_V \frac{\partial f}{\partial x} \, d\tau = \oint_S f n_x \, dS$$
Divergence Theorem:
$$\int_V (\nabla \cdot \vec{F}) \, d\tau = \oint_S (\vec{F} \cdot \hat{n}) \, dS$$
Let $\vec{F} = f\hat{i}$, $\hat{n} = n_x\hat{i} + n_y\hat{j} + n_z\hat{k}$:
$$\nabla \cdot \vec{F} = \nabla \cdot (f\hat{i}) = \frac{\partial(f)}{\partial x} + \frac{\partial(0)}{\partial y} + \frac{\partial(0)}{\partial z} = \frac{\partial f}{\partial x}$$
$$\vec{F} \cdot \hat{n} = (f\hat{i}) \cdot (n_x\hat{i} + n_y\hat{j} + n_z\hat{k}) = f n_x$$
2.b
$$\oint_S \vec{A} (\vec{B} \cdot \hat{n}) \, dS = \int_V \vec{A}(\nabla \cdot \vec{B}) \, d\tau + \int_V (\vec{B} \cdot \nabla)\vec{A} \, d\tau$$
Proof:
Let's prove the x-component of the identity:
$$\oint_S A_x (\vec{B} \cdot \hat{n}) \, dS = \int_V A_x(\nabla \cdot \vec{B}) \, d\tau + \int_V (\vec{B} \cdot \nabla)A_x \, d\tau$$
Use identity:
$$\nabla \cdot (\phi \vec{F}) = \phi(\nabla \cdot \vec{F}) + \vec{F} \cdot \nabla\phi$$
Let $\phi = A_x$, $\vec{F} = \vec{B}$,
$$\nabla \cdot (A_x \vec{B}) = A_x(\nabla \cdot \vec{B}) + \vec{B} \cdot \nabla A_x$$
$$[\text{RHS}]_x = \int_V \nabla \cdot (A_x \vec{B}) \, d\tau$$
Divergence Theorem:
$$\int_V \nabla \cdot (A_x \vec{B}) \, d\tau = \oint_S (A_x \vec{B}) \cdot \hat{n} \, dS$$
3.
$$\int_S (\nabla\phi \times \nabla\psi) \cdot d\vec{s} = \oint_l \phi \, d\psi$$
Proof.
$$\nabla \times (\phi \vec{F}) = (\nabla\phi) \times \vec{F} + \phi(\nabla \times \vec{F})$$
Let $\vec{F} = \nabla\psi$: $$\nabla \times (\phi \nabla\psi) = (\nabla\phi) \times (\nabla\psi) + \phi(\nabla \times \nabla\psi)$$
Since $\nabla \times \nabla\psi = \vec{0}$: $$\nabla \times (\phi \nabla\psi) = (\nabla\phi) \times (\nabla\psi)$$
$$\text{LHS} = \int_S (\nabla\phi \times \nabla\psi) \cdot d\vec{s} = \int_S (\nabla \times (\phi \nabla\psi)) \cdot d\vec{s}$$
Stokes' Theorem:$$\int_S (\nabla \times \vec{F}) \cdot d\vec{s} = \oint_l \vec{F} \cdot d\vec{l}$$
Let $\vec{F} = \phi \nabla\psi$: $$\int_S (\nabla \times (\phi \nabla\psi)) \cdot d\vec{s} = \oint_l (\phi \nabla\psi) \cdot d\vec{l}$$
Since
$$d\psi = \frac{\partial \psi}{\partial x}dx + \frac{\partial \psi}{\partial y}dy + \frac{\partial \psi}{\partial z}dz$$
So $$\nabla\psi \cdot d\vec{l} = \left(\frac{\partial \psi}{\partial x}\hat{i} + \frac{\partial \psi}{\partial y}\hat{j} + \frac{\partial \psi}{\partial z}\hat{k}\right) \cdot (dx\hat{i} + dy\hat{j} + dz\hat{k})$$
$$\nabla\psi \cdot d\vec{l} = \frac{\partial \psi}{\partial x}dx + \frac{\partial \psi}{\partial y}dy + \frac{\partial \psi}{\partial z}dz = d\psi$$
Then: $$\oint_l (\phi \nabla\psi) \cdot d\vec{l} = \oint_l \phi \, d\psi$$
4.
$$\delta(\vec{r}_1 - \vec{r}_2) = \frac{1}{r_1^2} \delta(r_1 - r_2) \delta(\cos\theta_1 - \cos\theta_2) \delta(\phi_1 - \phi_2)$$
Proof.
Goal is
$$\int_{\text{all space}} f(\vec{r}_1) \delta(\vec{r}_1 - \vec{r}_2) \, dV_1 = f(\vec{r}_2)$$
Use $dV_1 = r_1^2 \sin\theta_1 \, dr_1 \, d\theta_1 \, d\phi_1$:
$$\int_0^{2\pi} \int_0^\pi \int_0^\infty f(r_1, \theta_1, \phi_1) \delta(\vec{r}_1 - \vec{r}_2) \, r_1^2 \sin\theta_1 \, dr_1 \, d\theta_1 \, d\phi_1$$
$$I = \int_0^{2\pi} \int_0^\pi \int_0^\infty f(r_1, \theta_1, \phi_1) \left[ \frac{1}{r_1^2} \delta(r_1 - r_2) \delta(\cos\theta_1 - \cos\theta_2) \delta(\phi_1 - \phi_2) \right] r_1^2 \sin\theta_1 \, dr_1 \, d\theta_1 \, d\phi_1$$
$$I = \int_0^{2\pi} \int_0^\pi \int_0^\infty f(r_1, \theta_1, \phi_1) \delta(r_1 - r_2) \delta(\cos\theta_1 - \cos\theta_2) \delta(\phi_1 - \phi_2) \sin\theta_1 \, dr_1 \, d\theta_1 \, d\phi_1$$
$\phi_1$:
$$I = \int_0^\pi \int_0^\infty f(r_1, \theta_1, \phi_2) \delta(r_1 - r_2) \delta(\cos\theta_1 - \cos\theta_2) \sin\theta_1 \, dr_1 \, d\theta_1$$
$r_1$:
$$I = \int_0^\pi f(r_2, \theta_1, \phi_2) \delta(\cos\theta_1 - \cos\theta_2) \sin\theta_1 \, d\theta_1$$
$\theta_1$:
Let $u = \cos\theta_1$, $du = -\sin\theta_1 \, d\theta_1$.
$$I = \int_{u=1}^{u=-1} f(r_2, \arccos(u), \phi_2) \delta(u - \cos\theta_2) \, (-du)$$
$$I = \int_{-1}^{1} f(r_2, \arccos(u), \phi_2) \delta(u - \cos\theta_2) \, du$$
$$I = f(r_2, \arccos(\cos\theta_2), \phi_2)$$
$$I = f(r_2, \theta_2, \phi_2)$$
Finally,
$$\int_{\text{all space}} f(\vec{r}_1) \left[ \frac{1}{r_1^2} \delta(r_1 - r_2) \delta(\cos\theta_1 - \cos\theta_2) \delta(\phi_1 - \phi_2) \right] dV_1 = f(\vec{r}_2)$$
5.
$\vec{B} = \nabla \times \left( \frac{\vec{m} \times \vec{r}}{r^3} \right)$.
Proof.
Use
$$\nabla \times (\phi \vec{f}) = \phi(\nabla \times \vec{f}) + (\nabla\phi) \times \vec{f}$$
Let $\phi = \frac{1}{r^3}$, $\vec{f} = \vec{m} \times \vec{r}$
$\phi(\nabla \times \vec{f})$
Use
$$\nabla \times (\vec{u} \times \vec{v}) = \vec{u}(\nabla \cdot \vec{v}) - \vec{v}(\nabla \cdot \vec{u}) + (\vec{v} \cdot \nabla)\vec{u} - (\vec{u} \cdot \nabla)\vec{v}$$
Let $\vec{u} = \vec{m}$ (Constant), $\vec{v} = \vec{r}$:
Since $\nabla \cdot \vec{r} = 3$, $\nabla \cdot \vec{m} = 0$, $(\vec{r} \cdot \nabla)\vec{m} = 0$, $(\vec{m} \cdot \nabla)\vec{r} = \vec{m}$
$$\nabla \times (\vec{m} \times \vec{r}) = \vec{m}(3) - \vec{r}(0) + 0 - \vec{m} = 2\vec{m}$$
$$\phi(\nabla \times \vec{f}) = \frac{2\vec{m}}{r^3}$$
$(\nabla\phi) \times \vec{f}$:
$$\nabla\phi = \nabla(r^{-3}) = -3r^{-4} \nabla r = -3r^{-4} \hat{r} = -\frac{3\vec{r}}{r^5}$$
$$(\nabla\phi) \times \vec{f} = \left(-\frac{3\vec{r}}{r^5}\right) \times (\vec{m} \times \vec{r})$$
Use $\vec{u} \times (\vec{v} \times \vec{w}) = \vec{v}(\vec{u} \cdot \vec{w}) - \vec{w}(\vec{u} \cdot \vec{v})$: $$(\nabla\phi) \times \vec{f} = -\frac{3}{r^5} \left[ \vec{m}(\vec{r} \cdot \vec{r}) - \vec{r}(\vec{r} \cdot \vec{m}) \right]$$
$$(\nabla\phi) \times \vec{f} = -\frac{3}{r^5} [ \vec{m}r^2 - \vec{r}(\vec{r} \cdot \vec{m}) ] = -\frac{3\vec{m}}{r^3} + \frac{3\vec{r}(\vec{r} \cdot \vec{m})}{r^5}$$
Finally:
$$\vec{B} = \left(\frac{2\vec{m}}{r^3}\right) + \left(-\frac{3\vec{m}}{r^3} + \frac{3\vec{r}(\vec{r} \cdot \vec{m})}{r^5}\right)$$
$$\vec{B} = -\frac{\vec{m}}{r^3} + \frac{3\vec{r}(\vec{r} \cdot \vec{m})}{r^5}$$
$$\vec{B} = -\frac{\vec{m}}{r^3} + \frac{3(r\hat{r})(\hat{r} \cdot \vec{m})}{r^3}$$
$$\vec{B} = \frac{3\hat{r}(\hat{r} \cdot \vec{m}) - \vec{m}}{r^3}$$
6.
(a) Let $r = |\vec{r}| = \sqrt{x^2+y^2+z^2}$.
$$\nabla f(r) = \frac{df}{dr} \nabla r$$
$$\nabla r = \nabla\sqrt{x^2+y^2+z^2} = \frac{\partial r}{\partial x}\hat{i} + \frac{\partial r}{\partial y}\hat{j} + \frac{\partial r}{\partial z}\hat{k}$$
$$\frac{\partial r}{\partial x} = \frac{\partial}{\partial x}(x^2+y^2+z^2)^{1/2} = \frac{1}{2}(x^2+y^2+z^2)^{-1/2}(2x) = \frac{x}{r}$$
$$\nabla r = \frac{x\hat{i} + y\hat{j} + z\hat{k}}{r} = \frac{\vec{r}}{r} = \hat{r}$$
$$\nabla f(r) = \frac{df}{dr}\hat{r} = \frac{df}{dr} \frac{\vec{r}}{r}$$
Since $\frac{df}{dr}\frac{1}{r}$ is a scalar function, $\nabla f(r)$ is proportional to the position vector $\vec{r}$.
$\nabla r$: $f(r)=r$, so $\frac{df}{dr} = 1$.
$$\nabla r = 1 \cdot \hat{r} = \hat{r}$$
$\nabla(\frac{1}{r})$: $f(r)=1/r = r^{-1}$, so $\frac{df}{dr} = -r^{-2} = -1/r^2$.
$$\nabla\left(\frac{1}{r}\right) = \left(-\frac{1}{r^2}\right)\hat{r} = -\frac{\hat{r}}{r^2} = -\frac{\vec{r}}{r^3}$$
(b) Laplacian of $1/r$
For $r \neq 0$:
$$\nabla^2\left(\frac{1}{r}\right) = \nabla \cdot \left( \nabla \frac{1}{r} \right) = \nabla \cdot \left( -\frac{\vec{r}}{r^3} \right)$$
Use $\nabla \cdot (\phi \vec{F}) = \phi(\nabla \cdot \vec{F}) + (\nabla\phi)\cdot\vec{F}$, let $\phi = -1/r^3$ and $\vec{F}=\vec{r}$:
$$\nabla \cdot \left(-\frac{\vec{r}}{r^3}\right) = -\frac{1}{r^3}(\nabla \cdot \vec{r}) + \left(\nabla(-\frac{1}{r^3})\right) \cdot \vec{r}$$
Then $\nabla \cdot \vec{r} = 3$ and $\nabla(-r^{-3}) = 3r^{-4}\hat{r} = 3\vec{r}/r^5$.
$$\nabla^2\left(\frac{1}{r}\right) = -\frac{3}{r^3} + \left(\frac{3\vec{r}}{r^5}\right)\cdot\vec{r} = -\frac{3}{r^3} + \frac{3(\vec{r}\cdot\vec{r})}{r^5} = -\frac{3}{r^3} + \frac{3r^2}{r^5} = 0$$
The singularity at $r=0$:
$$\int_V \nabla^2\left(\frac{1}{r}\right) dV = \int_V \nabla \cdot \left(\nabla\frac{1}{r}\right) dV = \oint_S \left(\nabla\frac{1}{r}\right) \cdot d\vec{S}$$
On the spherical surface $S$, $d\vec{S} = \hat{r} dS$ and $\nabla(1/r) = -\hat{r}/r^2$. At $r=R$:$$\oint_S \left(-\frac{\hat{r}}{R^2}\right) \cdot (\hat{r} dS) = \oint_S -\frac{1}{R^2} dS = -\frac{1}{R^2} (4\pi R^2) = -4\pi$$
$$\nabla^2\left(\frac{1}{r}\right) = -4\pi\delta(\vec{r})$$
(c)
$\nabla \cdot \vec{r}$:
$$\nabla \cdot \vec{r} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 1 + 1 + 1 = 3$$
$\nabla \times \vec{r}$:
$$\nabla \times \vec{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x & y & z \end{vmatrix}$$ $$ = \left(\frac{\partial z}{\partial y} - \frac{\partial y}{\partial z}\right)\hat{i} + \left(\frac{\partial x}{\partial z} - \frac{\partial z}{\partial x}\right)\hat{j} + \left(\frac{\partial y}{\partial x} - \frac{\partial x}{\partial y}\right)\hat{k}$$ $$ = (0-0)\hat{i} + (0-0)\hat{j} + (0-0)\hat{k} = \vec{0}$$
$\nabla(\vec{a} \cdot \vec{r})$:
Let $\vec{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}$ be a constant vector.
$$\vec{a} \cdot \vec{r} = a_x x + a_y y + a_z z$$
$$\nabla(\vec{a} \cdot \vec{r}) = \frac{\partial}{\partial x}(a_x x + \dots)\hat{i} + \frac{\partial}{\partial y}(a_y y + \dots)\hat{j} + \frac{\partial}{\partial z}(a_z z + \dots)\hat{k}$$ $$\nabla(\vec{a} \cdot \vec{r}) = a_x\hat{i} + a_y\hat{j} + a_z\hat{k} = \vec{a}$$
HW2
1
(a) Since charge $q$ at origin is screened, $Q=-q$.
(b)
$$\nabla^2 \phi(r) = \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{d\phi}{dr} \right)$$
We use Poisson's equation($\nabla^2 \phi = -4\pi\rho$), also separate charge into two.
$$\rho(\vec{r}) = q\delta^{(3)}(\vec{r}) + \rho_{screen}(r)$$
$$\frac{d\phi}{dr} = q \frac{d}{dr} \left(\frac{e^{-\alpha r}}{r}\right) = q \left( \frac{- \alpha r e^{-\alpha r} - e^{-\alpha r}}{r^2} \right) = -q \frac{e^{-\alpha r}}{r^2} (1 + \alpha r)$$
$$r^2 \frac{d\phi}{dr} = -q e^{-\alpha r} (1 + \alpha r)$$
$$\frac{d}{dr} \left( r^2 \frac{d\phi}{dr} \right) = -q \left[ (-\alpha e^{-\alpha r})(1+\alpha r) + (e^{-\alpha r})(\alpha) \right] = -q e^{-\alpha r} [-\alpha - \alpha^2 r + \alpha] = q\alpha^2 r e^{-\alpha r}$$
$$\nabla^2 \phi = \frac{1}{r^2} (q\alpha^2 r e^{-\alpha r}) = q\alpha^2 \frac{e^{-\alpha r}}{r}$$
Finally, screen charge is
$$\rho_{screen}(r) = -\frac{1}{4\pi} \nabla^2 \phi = -\frac{q\alpha^2}{4\pi} \frac{e^{-\alpha r}}{r}$$
(c)
The volume element in spherical coordinates is $d\tau = r^2 \sin\theta \, dr \, d\theta \, d\phi$.
$$Q = \iiint_V \rho(\vec{r}) \, d\tau = \int \left( q\delta^{(3)}(\vec{r}) - \frac{q\alpha^2}{4\pi} \frac{e^{-\alpha r}}{r} \right) d\tau$$
$$Q_{screen} = -\frac{q\alpha^2}{4\pi} \int_0^{2\pi} d\phi \int_0^{\pi} \sin\theta \, d\theta \int_0^{\infty} \left(\frac{e^{-\alpha r}}{r}\right) r^2 \, dr$$
$$\int_0^{2\pi} d\phi = 2\pi \quad \text{and} \quad \int_0^{\pi} \sin\theta \, d\theta = 2$$
$$Q_{screen} = -\frac{q\alpha^2}{4\pi} (4\pi) \int_0^{\infty} r e^{-\alpha r} \, dr = -q\alpha^2 \int_0^{\infty} r e^{-\alpha r} \, dr$$
$$\int_0^{\infty} r e^{-\alpha r} \, dr = \left[ -\frac{r}{\alpha}e^{-\alpha r} \right]_0^{\infty} - \int_0^{\infty} \left(-\frac{1}{\alpha}e^{-\alpha r}\right) dr = 0 - \left[ \frac{1}{\alpha^2}e^{-\alpha r} \right]_0^{\infty} = \frac{1}{\alpha^2}$$
Finally,
$$Q_{screen} = -q\alpha^2 \left(\frac{1}{\alpha^2}\right) = -q$$
The total charge is the sum of the two parts:
$$Q = Q_{origin} + Q_{screen} = q + (-q) = 0$$
2
Green's second identity:
$$\int_V (\Phi \nabla^2 \Phi' - \Phi' \nabla^2 \Phi) d^3x = \oint_S \left(\Phi \frac{\partial \Phi'}{\partial n} - \Phi' \frac{\partial \Phi}{\partial n}\right) da$$
($\frac{\partial}{\partial n}$ represents the normal derivative at the surface $S$, directed outward from the volume $V$)
Poisson's Equation:
$$\nabla^2 \Phi = -\frac{\rho}{\epsilon_0} \quad \text{and} \quad \nabla^2 \Phi' = -\frac{\rho'}{\epsilon_0}$$
Conductor Boundary Condition:
$$\frac{\partial \Phi}{\partial n} = \frac{\sigma}{\epsilon_0} \quad \text{and} \quad \frac{\partial \Phi'}{\partial n} = \frac{\sigma'}{\epsilon_0}$$
Substitute these relationships into Green's identity.
Volume integral:
$$\int_V \left( \Phi \left(-\frac{\rho'}{\epsilon_0}\right) - \Phi' \left(-\frac{\rho}{\epsilon_0}\right) \right) d^3x = \frac{1}{\epsilon_0} \int_V (\rho \Phi' - \rho' \Phi) d^3x$$
Surface integral:
$$\oint_S \left( \Phi \left(\frac{\sigma'}{\epsilon_0}\right) - \Phi' \left(\frac{\sigma}{\epsilon_0}\right) \right) da = \frac{1}{\epsilon_0} \oint_S (\sigma' \Phi - \sigma \Phi') da$$
Then
$$\int_V (\rho \Phi' - \rho' \Phi) d^3x = \oint_S (\sigma' \Phi - \sigma \Phi') da$$
$$\int_V \rho \Phi' d^3x + \oint_S \sigma \Phi' da = \int_V \rho' \Phi d^3x + \oint_S \sigma' \Phi da$$
3
Green's reciprocation theorem provides a relationship between two different electrostatic situation that share the same volume $V$ and bounding surface $S$.
$$\int_V \rho_1 \Phi_2 d^3x + \oint_S \sigma_1 \Phi_2 da = \int_V \rho_2 \Phi_1 d^3x + \oint_S \sigma_2 \Phi_1 da$$
$\int_V \rho_1 \Phi_2 d^3x$:
$$\int_V q\delta(z-z_0) \left(V_0 \frac{z}{d}\right) d^3x = q \cdot \Phi_2(z_0) = qV_0\frac{z_0}{d}$$
$\oint_S \sigma_1 \Phi_2 da$:
Bottom plate ($z=0$), $\Phi_2 = 0$, 0.
Top plate ($z=d$), $\Phi_2 = V_0$, $\int_{z=d} \sigma_1 V_0 da = V_0 \int_{z=d} \sigma_1 da = V_0 Q_{top}$.
$\int_V \rho_2 \Phi_1 d^3x$:$\rho_2=0$, 0.
$\oint_S \sigma_2 \Phi_1 da$: $\Phi_1=0$, 0.
Then
$$qV_0\frac{z_0}{d} + V_0 Q_{top} = 0 + 0$$
$$Q_{top} = -q \frac{z_0}{d}$$
The induced charge on the top plate ($Q_{top}$) is equal to $(-q)$ times the fractional perpendicular distance of the point charge from the other plane (the one at $z=0$), which is $z_0/d$.
4
For a system of conductors, the charge on the $i$-th conductor ($q_i$) is related to the potentials of all the conductors ($V_j$) by the linear relationship
$$q_i = \sum_{j} C_{ij} V_j$$
Then goal is
$$C_{13} = \left. \frac{q_1}{V_3} \right|_{V_1=0, V_2=0}$$
Since $\vec{E}(r) = \frac{1}{4\pi\epsilon_0} \frac{q_1}{r^2} \hat{r}$,
$$V_2 - V_1 = -\frac{q_1}{4\pi\epsilon_0} \left[ -\frac{1}{r} \right]_{r_1}^{r_2} = \frac{q_1}{4\pi\epsilon_0} \left( \frac{1}{r_2} - \frac{1}{r_1} \right)$$
Apply the conditions required to find $C_{13}$: $V_1 = 0$ and $V_2 = 0$.
$$0 - 0 = \frac{q_1}{4\pi\epsilon_0} \left( \frac{1}{r_2} - \frac{1}{r_1} \right)$$
Then
$$q_1 = 0$$
$$C_{13} = \frac{0}{V_3} = 0$$
The result $C_{13}=0$ is a mathematical statement of electrostatic shielding, also known as the Faraday cage effect.
It means that changing the potential of conductor 3 ($V_3$) has no effect on the charge of conductor 1 ($q_1$), as long as the intermediate conductor (conductor 2) is held at a constant potential (in this case, grounded at $V_2=0$).
5
By sphere's volume $V = \frac{4}{3}\pi R^3$,
$$\rho = \frac{Q}{V} = \frac{Q}{\frac{4}{3}\pi R^3} = \frac{3Q}{4\pi R^3}$$
Using Gauss's Law for a spherical surface of radius $r < R$ inside the sphere,
$$\vec{E}_{total}(r) = \frac{Q'}{4\pi\epsilon_0 r^2} \hat{r}$$
$Q' = \rho \cdot V_{r} = \left(\frac{3Q}{4\pi R^3}\right) \left(\frac{4}{3}\pi r^3\right) = Q\frac{r^3}{R^3}$.
$$\vec{E}_{total}(r) = \frac{1}{4\pi\epsilon_0 r^2} \left(Q\frac{r^3}{R^3}\right) \hat{r} = \frac{Q}{4\pi\epsilon_0} \frac{r}{R^3} \hat{r}$$
By symmetry, the net force will be purely vertical. $dF_z = \rho E_z d\tau$.
The vertical component of the electric field is $E_z = \vec{E}_{total} \cdot \hat{k}$. In spherical coordinates, $z = r\cos\theta$ and $\hat{r} \cdot \hat{k} = \cos\theta$.
$$E_z = \left(\frac{Qr}{4\pi\epsilon_0 R^3}\right) \cos\theta$$
$$F_z = \iiint_{V_{upper}} \rho E_z \, d\tau$$
$$F_z = \int_0^{2\pi} d\phi \int_0^{\pi/2} d\theta \int_0^R dr \, \left(\frac{3Q}{4\pi R^3}\right) \left(\frac{Qr\cos\theta}{4\pi\epsilon_0 R^3}\right) (r^2 \sin\theta)$$
$$F_z = \frac{3Q^2}{16\pi^2\epsilon_0 R^6} (2\pi) \left(\frac{R^4}{4}\right) \left(\frac{1}{2}\right) = \frac{3Q^2 (2\pi R^4)}{128\pi^2\epsilon_0 R^6}$$
$$\vec{F} = \frac{3Q^2}{64\pi\epsilon_0 R^2} \hat{k}$$
HW3
1
We place an image charge $q'$ outside the sphere, also on the z-axis, at a distance $b$ from the origin. The values for $q'$ and $b$ that satisfy $V(r=a)=0$ are:
$q' = -q \frac{a}{d}$
$b = \frac{a^2}{d}$
(a)
Let the real charge $q$ be at position $\vec{d}$ and the image charge $q'$ be at position $\vec{b}$. The potential at a point $\vec{r}$ is:
$$V(\vec{r}) = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{|\vec{r} - \vec{d}|} + \frac{q'}{|\vec{r} - \vec{b}|} \right)$$
$$V(\vec{r}) = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{|\vec{r} - d\hat{z}|} - \frac{qa/d}{|\vec{r} - (a^2/d)\hat{z}|} \right)$$
(b)
The induced surface-charge density, $\sigma$, on the inner surface of the conductor is related to the electric field normal to the surface.
$$\sigma = \epsilon_0 E_n$$
$$E_n = -\vec{E} \cdot \hat{r} = -E_r = \frac{\partial V}{\partial r}$$
$$\sigma = \epsilon_0 \left( \frac{\partial V}{\partial r} \right)_{r=a}$$
$$\sigma(\theta) = -\frac{q(a^2 - d^2)}{4\pi a (a^2 + d^2 - 2ad\cos\theta)^{3/2}}$$
where $\theta$ is the angle with respect to the z-axis.
(c)
The distance between $q$ and $q'$ is $b-d = \frac{a^2}{d} - d = \frac{a^2-d^2}{d}$.
$$\vec{F} = \frac{1}{4\pi\epsilon_0} \frac{q q'}{(b-d)^2} \hat{z}$$Substituting the values for $q'$ and $(b-d)$:$$\vec{F} = \frac{1}{4\pi\epsilon_0} \frac{q(-qa/d)}{\left(\frac{a^2-d^2}{d}\right)^2} \hat{z} = -\frac{1}{4\pi\epsilon_0} \frac{q^2 a d}{(a^2-d^2)^2} \hat{z}$$
(d)
Sphere at a fixed potential V:
We can place a second image charge, $q''$, at the center of the sphere.
The potential at the surface ($r=a$) from this new charge is $V_{q''} = \frac{q''}{4\pi\epsilon_0 a}$.
We require the total potential to be $V$, so $0 + V_{q''} = V$, which gives $q'' = 4\pi\epsilon_0 a V$.
(a) Potential: $V_{new}(\vec{r}) = V_{grounded}(\vec{r}) + \frac{q''}{4\pi\epsilon_0 r} = V_{grounded}(\vec{r}) + \frac{aV}{r}$.
(b) Surface Density: The charge $q''$ adds a uniform charge density $\sigma'' = \frac{q''}{4\pi a^2} = \frac{\epsilon_0 V}{a}$. So, $\sigma_{new}(\theta) = \sigma_{grounded}(\theta) + \frac{\epsilon_0 V}{a}$.
(c) Force: The new charge $q''$ at the origin exerts a force on $q$. $\vec{F}_{q''} = \frac{1}{4\pi\epsilon_0}\frac{q q''}{d^2}\hat{z} = \frac{q a V}{d^2}\hat{z}$. The total force is $\vec{F}_{new} = \vec{F}_{grounded} + \frac{q a V}{d^2}\hat{z}$.
Sphere with a total charge Q:
If the outer radius is $R$, the potential of the sphere is $V_{sphere} = \frac{Q+q}{4\pi\epsilon_0 R}$.
(a) Potential: The potential inside is shifted by this constant value: $V_{new}(\vec{r}) = V_{grounded}(\vec{r}) + V_{sphere}$.
(b) Surface Density: The density on the inner surface is identical to the grounded case, $\sigma(\theta)$. There is an additional uniform density of $\sigma_{outer} = \frac{Q+q}{4\pi R^2}$ on the outer surface.
(c) Force: The force on $q$ is unchanged from the grounded case, as it is only affected by the charge distribution on the inner surface.
2
The total force acting on the conducting shell is zero.
The shell is hanging in a static position, which means it is in equilibrium.
Gravitational Force:
$$F_g = -mg$$
where $m$ is the mass of the shell and $g$ is the acceleration due to gravity.
Spring Force:
$$F_s = k(L_{eq} - d)$$
where $k$ is the spring constant.
Electrostatic Force:
$$F_e = \frac{1}{4\pi\epsilon_0} \left( \frac{q^2 R L_{eq}}{(L_{eq}^2 - R^2)^2} - \frac{qQ}{L_{eq}^2} \right)$$
The first term, $\frac{q^2 R L_{eq}}{(L_{eq}^2 - R^2)^2}$, represents the always attractive force from the induced charges.
The second term, $-\frac{qQ}{L_{eq}^2}$, represents the force from the net charge `Q`. It is repulsive if $q$ and $Q$ have the same sign.
$$\sum F = F_g + F_s + F_e = 0$$
$$-mg + k(L_{eq} - d) + \frac{1}{4\pi\epsilon_0} \left( \frac{q^2 R L_{eq}}{(L_{eq}^2 - R^2)^2} - \frac{qQ}{L_{eq}^2} \right) = 0$$
3
(a)
The potential at $\mathbf{x}$ due to a unit point charge at $\mathbf{x'} = (x', y', z')$ is $\frac{1}{|\mathbf{x}-\mathbf{x'}|}$. To make the potential zero on the $z'=0$ plane, we place an image charge of -1 at the mirror-image point $\mathbf{x''} = (x', y', -z')$.
$$G(\mathbf{x}, \mathbf{x'}) = \frac{1}{|\mathbf{x}-\mathbf{x'}|} - \frac{1}{|\mathbf{x}-\mathbf{x''}|}$$
(b)
For a Dirichlet problem with no charges in the volume, the potential $\Phi(\mathbf{x})$ is given by Green's theorem:
$$\Phi(\mathbf{x}) = -\frac{1}{4\pi} \oint_S \Phi(\mathbf{x'}) \frac{\partial G}{\partial n'} dS'$$
Here, the surface $S$ is the plane $z'=0$, and the outward normal from the volume is $\hat{n}' = -\hat{z}$. Therefore, $\frac{\partial}{\partial n'} = -\frac{\partial}{\partial z'}$.
Normal derivative of $G$ and evaluate it at $z'=0$:
$$\frac{\partial G}{\partial z'} = \frac{z-z'}{|\mathbf{x}-\mathbf{x'}|^3} + \frac{z+z'}{|\mathbf{x}-\mathbf{x''}|^3}$$
$$\left. \frac{\partial G}{\partial z'} \right|_{z'=0} = \frac{z}{( (x-x')^2 + (y-y')^2 + z^2 )^{3/2}} + \frac{z}{( (x-x')^2 + (y-y')^2 + z^2 )^{3/2}} = \frac{2z}{( (x-x')^2 + (y-y')^2 + z^2 )^{3/2}}$$
So, $\left. \frac{\partial G}{\partial n'} \right|_{z'=0} = -\left. \frac{\partial G}{\partial z'} \right|_{z'=0} = -\frac{2z}{( (x-x')^2 + (y-y')^2 + z^2 )^{3/2}}$.
$$\Phi(\mathbf{x}) = -\frac{1}{4\pi} \int_{z'=0} \Phi(\mathbf{x'}) \left( -\frac{2z}{( (x-x')^2 + (y-y')^2 + z^2 )^{3/2}} \right) dS'$$
$$\Phi(\mathbf{x}) = \frac{z}{2\pi} \int_{z'=0} \frac{\Phi(\mathbf{x'})}{( (x-x')^2 + (y-y')^2 + z^2 )^{3/2}} dS'$$
The denominator becomes $(\rho^2 + \rho'^2 - 2\rho\rho'\cos(\phi-\phi') + z^2)^{3/2}$, and $dS' = \rho'd\rho'd\phi'$.
The potential on the plane is $\Phi(\rho') = V$ for $\rho' \le a$ and 0 otherwise.
$$\Phi(\rho, \phi, z) = \frac{zV}{2\pi} \int_0^{2\pi} d\phi' \int_0^a \frac{\rho' d\rho'}{(\rho^2 + \rho'^2 - 2\rho\rho'\cos(\phi-\phi') + z^2)^{3/2}}$$
(c)
We set $\rho=0$ in the integral from part (b):
$$\Phi(0, z) = \frac{zV}{2\pi} \int_0^{2\pi} d\phi' \int_0^a \frac{\rho' d\rho'}{(\rho'^2 + z^2)^{3/2}}$$
$$\Phi(0, z) = zV \int_0^a (\rho'^2 + z^2)^{-3/2} \rho' d\rho'$$
This integral can be solved with a u-substitution: let $u = \rho'^2 + z^2$, so $du = 2\rho'd\rho'$.
$$\Phi(0, z) = zV \int_{z^2}^{a^2+z^2} \frac{1}{2} u^{-3/2} du = \frac{zV}{2} \left[ -2u^{-1/2} \right]_{z^2}^{a^2+z^2}$$
$$\Phi(0, z) = -zV \left( \frac{1}{\sqrt{a^2+z^2}} - \frac{1}{\sqrt{z^2}} \right) = -zV \left( \frac{1}{\sqrt{a^2+z^2}} - \frac{1}{z} \right)$$
$$\Phi(0, z) = V \left( 1 - \frac{z}{\sqrt{a^2+z^2}} \right)$$
(d)
For large distances, where $R^2 = \rho^2+z^2 \gg a^2$, we can expand
$$[R^2 + \rho'^2 - 2\rho\rho'\cos(\phi-\phi')]^{-3/2} = R^{-3} \left[ 1 + \frac{\rho'^2 - 2\rho\rho'\cos(\phi-\phi')}{R^2} \right]^{-3/2}$$
Using the binomial expansion $(1+\epsilon)^{-3/2} \approx 1 - \frac{3}{2}\epsilon + \frac{15}{8}\epsilon^2 - \dots$,
$$\Phi \approx \frac{Va^2 z}{2 (\rho^2+z^2)^{3/2}} \left[ 1 - \frac{3a^2}{4(\rho^2+z^2)} + \frac{5(3\rho^2a^2+a^4)}{8(\rho^2+z^2)^2} + \dots \right]$$
Verification:
(d) with $\rho=0$
$$\Phi(0,z) \approx \frac{Va^2 z}{2(z^2)^{3/2}} \left[ 1 - \frac{3a^2}{4z^2} + \frac{5a^4}{8z^4} + \dots \right] = \frac{Va^2}{2z^2} \left[ 1 - \frac{3a^2}{4z^2} + \dots \right]$$ $$\Phi(0,z) \approx \frac{Va^2}{2z^2} - \frac{3Va^4}{8z^4} + \dots$$
(c) expanded for $z \gg a$:
$$\Phi(0, z) = V \left( 1 - \frac{z}{z\sqrt{1+a^2/z^2}} \right) = V \left( 1 - (1+a^2/z^2)^{-1/2} \right)$$
Using the binomial expansion $(1+x)^{-1/2} \approx 1 - \frac{1}{2}x + \frac{3}{8}x^2 - \dots$ with $x = a^2/z^2$:
$$\Phi(0, z) \approx V \left( 1 - \left[1 - \frac{1}{2}\frac{a^2}{z^2} + \frac{3}{8}\frac{a^4}{z^4} - \dots\right] \right) = V \left( \frac{a^2}{2z^2} - \frac{3a^4}{8z^4} + \dots \right)$$ $$\Phi(0,z) \approx \frac{Va^2}{2z^2} - \frac{3Va^4}{8z^4} + \dots$$
4
(a)
Grounded conducting plane ($V=0$) with a hemispherical boss in a uniform external electric field $\vec{E} = -E_0\hat{z}$. This system is equivalent to the upper half of a full conducting sphere in the same field.
The potential $\Phi$ outside the conductor ($r \ge a$) is:
$$\Phi(r, \theta) = -E_0 \left( 1 - \frac{a^3}{r^3} \right) r \cos\theta$$
Hemispherical Boss:
The charge density is $\sigma = \epsilon_0 E_r$ evaluated at the surface $r=a$.
$$\sigma_{boss}(\theta) = 3\epsilon_0 E_0 \cos\theta \quad (0 \le \theta \le \pi/2)$$
Plane:
The charge density is $\sigma = \epsilon_0 E_z$ evaluated on the plane $z=0$ (where $r=\rho > a$).
$$\sigma_{plane}(\rho) = \epsilon_0 E_0 \left(1 - \frac{a^3}{\rho^3}\right) \quad (\rho > a)$$
(b)
The total charge is found by integrating the surface charge density over the hemisphere's area ($dS = a^2 \sin\theta d\theta d\phi$).
$$Q_{boss} = \int_{hemisphere} \sigma_{boss} dS = \int_0^{2\pi} d\phi \int_0^{\pi/2} (3\epsilon_0 E_0 \cos\theta)(a^2 \sin\theta d\theta)$$
$$Q_{boss} = 3\pi\epsilon_0 E_0 a^2$$
(c)
$$\Phi(\mathbf{x}) = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{|\mathbf{x}-\mathbf{d}|} + \frac{q'}{|\mathbf{x}-\mathbf{b}|} \right)$$
$$\Phi(r, \theta) = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{\sqrt{r^2 + d^2 - 2rd\cos\theta}} + \frac{-qa/d}{\sqrt{r^2 + (a^4/d^2) - 2r(a^2/d)\cos\theta}} \right)$$
$$\frac{\partial \Phi_q}{\partial r} = -\frac{q}{4\pi\epsilon_0} \frac{r - d\cos\theta}{(r^2 + d^2 - 2rd\cos\theta)^{3/2}}$$
$$\frac{\partial \Phi_{q'}}{\partial r} = \frac{qa/d}{4\pi\epsilon_0} \frac{r - (a^2/d)\cos\theta}{(r^2 + a^4/d^2 - 2ra^2/d\cos\theta)^{3/2}}$$
$$\sigma(\theta) = \epsilon_0 E_r \Big|_{r=a} = -\epsilon_0 \frac{\partial \Phi}{\partial r} \Bigg|_{r=a} = -\frac{q}{4\pi a} \frac{d^2 - a^2}{(a^2 + d^2 - 2ad\cos\theta)^{3/2}}$$
$$E_r(a) = \frac{1}{4\pi\epsilon_0 (a^2+d^2-2ad\cos\theta)^{3/2}} \left[ q(a - d\cos\theta) - \frac{qd(d-a\cos\theta)}{a} \right]$$
$$E_r(a) = \frac{1}{4\pi\epsilon_0 (\dots)^{3/2}} \left[ \frac{q(a^2 - d^2)}{a} \right] = -\frac{q(d^2-a^2)}{4\pi\epsilon_0 a (a^2+d^2-2ad\cos\theta)^{3/2}}$$
$$\sigma(\theta) = -\frac{q}{4\pi a} \frac{d^2 - a^2}{(a^2 + d^2 - 2ad\cos\theta)^{3/2}}$$
We integrate this over the hemisphere to find the induced charge, $q'_{hemi}$:
$$q'_{hemi} = \int_0^{2\pi} \int_0^{\pi/2} \left( -\frac{q(d^2 - a^2)}{4\pi a (a^2 + d^2 - 2ad\cos\theta)^{3/2}} \right) (a^2 \sin\theta \, d\theta) \, d\phi$$
$$q'_{hemi} = (2\pi) \left( -\frac{q a^2 (d^2-a^2)}{4\pi a} \right) \int_0^{\pi/2} \frac{\sin\theta}{(a^2 + d^2 - 2ad\cos\theta)^{3/2}} \, d\theta$$ $$q'_{hemi} = -\frac{q a (d^2-a^2)}{2} \int_0^{\pi/2} (a^2 + d^2 - 2ad\cos\theta)^{-3/2} \sin\theta \, d\theta$$
Let $u = a^2 + d^2 - 2ad\cos\theta$, Then $du = 2ad\sin\theta \, d\theta$, which means $\sin\theta \, d\theta = \frac{du}{2ad}$
When $\theta=0$, $u = a^2+d^2-2ad = (d-a)^2$.
When $\theta=\pi/2$, $u = a^2+d^2$.
$$\int_{(d-a)^2}^{a^2+d^2} u^{-3/2} \, \frac{du}{2ad} = -\frac{1}{ad} \left[ \frac{1}{\sqrt{u}} \right]_{(d-a)^2}^{a^2+d^2}$$ $$= -\frac{1}{ad} \left( \frac{1}{\sqrt{a^2+d^2}} - \frac{1}{\sqrt{(d-a)^2}} \right) = \frac{1}{ad} \left( \frac{1}{d-a} - \frac{1}{\sqrt{a^2+d^2}} \right)$$
$$q'_{hemi} = -\frac{q a (d^2-a^2)}{2} \times \left[ \frac{1}{ad} \left( \frac{1}{d-a} - \frac{1}{\sqrt{a^2+d^2}} \right) \right]$$
$$q' = -q \left[ 1 - \frac{d^2 - a^2}{d\sqrt{d^2 + a^2}} \right]$$
HW4
1
2
(a)
$\Phi=0$ on the four side walls ($x=0, a$ and $y=0, a$),
$$\Phi(x, y, z) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \sin\left(\frac{n\pi x}{a}\right) \sin\left(\frac{m\pi y}{a}\right) Z_{nm}(z)$$
$\frac{d^2Z}{dz^2} = k_{nm}^2 Z$, where $k_{nm} = \frac{\pi}{a}\sqrt{n^2+m^2}$
$$Z_{nm}(z) = A \cosh(k_{nm}z) + B \sinh(k_{nm}z)$$
Symmetric about the plane $z=a/2$,
$$Z_{nm}(z) = C_{nm} \cosh\left(k_{nm} (z - a/2)\right)$$
$$\Phi(x, y, z) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} C_{nm} \sin\left(\frac{n\pi x}{a}\right) \sin\left(\frac{m\pi y}{a}\right) \cosh\left(k_{nm} (z - a/2)\right)$$
At $z=0$,
$$\Phi(x, y, 0) = V = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} C_{nm} \sin\left(\frac{n\pi x}{a}\right) \sin\left(\frac{m\pi y}{a}\right) \cosh\left(\frac{k_{nm}a}{2}\right)$$
$$C_{nm} \cosh\left(\frac{k_{nm}a}{2}\right) = \left(\frac{2}{a}\right)^2 \int_0^a \int_0^a V \sin\left(\frac{n\pi x}{a}\right) \sin\left(\frac{m\pi y}{a}\right) dx \, dy$$
$$= \frac{4V}{a^2} \left[ \int_0^a \sin\left(\frac{n\pi x}{a}\right) dx \right] \left[ \int_0^a \sin\left(\frac{m\pi y}{a}\right) dy \right]$$
Use
$$\int_0^a \sin\left(\frac{k\pi u}{a}\right) du = \begin{cases} \frac{2a}{k\pi} & \text{if } k \text{ is odd} \\ 0 & \text{if } k \text{ is even} \end{cases}$$
For odd n and m,
$$C_{nm} \cosh\left(\frac{k_{nm}a}{2}\right) = \frac{4V}{a^2} \left(\frac{2a}{n\pi}\right) \left(\frac{2a}{m\pi}\right) = \frac{16V}{nm\pi^2}$$
$$C_{nm} = \frac{16V}{nm\pi^2 \cosh\left(\frac{k_{nm}a}{2}\right)}$$
$$\Phi(x, y, z) = \frac{16V}{\pi^2} \sum_{n, m \text{ odd}} \frac{\sin\left(\frac{n\pi x}{a}\right) \sin\left(\frac{m\pi y}{a}\right)}{nm \cosh\left(\frac{\pi}{2}\sqrt{n^2+m^2}\right)} \cosh\left(\frac{\pi}{a}\sqrt{n^2+m^2}\left(z-\frac{a}{2}\right)\right)$$
(b)
At $(a/2, a/2, a/2)$,
$\sin\left(\frac{n\pi (a/2)}{a}\right) = \sin(n\pi/2) = (-1)^{(n-1)/2}$ for odd n.
$\sin\left(\frac{m\pi (a/2)}{a}\right) = \sin(m\pi/2) = (-1)^{(m-1)/2}$ for odd m.
$\cosh\left(\frac{\pi}{a}\sqrt{n^2+m^2}\left(\frac{a}{2}-\frac{a}{2}\right)\right) = \cosh(0) = 1$.
$$\Phi\left(\frac{a}{2}, \frac{a}{2}, \frac{a}{2}\right) = \frac{16V}{\pi^2} \sum_{n, m \text{ odd}} \frac{(-1)^{(n-1)/2}(-1)^{(m-1)/2}}{nm \cosh\left(\frac{\pi}{2}\sqrt{n^2+m^2}\right)}$$
(n=1, m=1): $\frac{(-1)^0(-1)^0}{(1)(1)\cosh(\frac{\pi}{2}\sqrt{2})} \approx +0.21442$
(n=1, m=3) & (n=3, m=1): $2 \times \frac{(-1)^0(-1)^1}{(1)(3)\cosh(\frac{\pi}{2}\sqrt{10})} \approx -0.00928$
(n=3, m=3): $\frac{(-1)^1(-1)^1}{(3)(3)\cosh(\frac{\pi}{2}\sqrt{18})} \approx +0.00028$
$$\Phi_{center} \approx \frac{16V}{\pi^2} (0.20542) \approx 0.3330V$$
$$\langle\Phi\rangle_{walls} = \frac{\sum_i \Phi_i A_i}{\sum_i A_i} = \frac{V(a^2) + V(a^2) + 0(4a^2)}{6a^2} = \frac{2Va^2}{6a^2} = \frac{V}{3}$$
Similar in 3 terms.
(c)
$$\sigma = -\epsilon_0 \frac{\partial\Phi}{\partial n}$$
For the surface at $z=a$
$$\sigma = \epsilon_0 (\vec{E}_{metal} - \vec{E}_{cavity}) \cdot \hat{n}_{cavity \to metal} = \epsilon_0 (0 - (-\nabla\Phi)) \cdot \hat{z} = \epsilon_0 \left.\frac{\partial\Phi}{\partial z}\right|_{z=a}$$
$$\frac{\partial\Phi}{\partial z} = \frac{16V}{\pi^2} \sum_{n, m \text{ odd}} \frac{\sin(\frac{n\pi x}{a}) \sin(\frac{m\pi y}{a})}{nm \cosh(\frac{k_{nm}a}{2})} \left[ k_{nm} \sinh\left(k_{nm} \left(z-\frac{a}{2}\right)\right) \right]$$
$$\left.\frac{\partial\Phi}{\partial z}\right|_{z=a} = \frac{16V}{\pi^2} \sum_{n, m \text{ odd}} \frac{k_{nm} \sinh(k_{nm}a/2)}{nm \cosh(k_{nm}a/2)} \sin\left(\frac{n\pi x}{a}\right) \sin\left(\frac{m\pi y}{a}\right)$$
$$= \frac{16V}{\pi^2} \sum_{n, m \text{ odd}} \frac{k_{nm} \tanh(k_{nm}a/2)}{nm} \sin\left(\frac{n\pi x}{a}\right) \sin\left(\frac{m\pi y}{a}\right)$$
Substituting $k_{nm} = \frac{\pi}{a}\sqrt{n^2+m^2}$,
$$\sigma(x, y) = \frac{16V\epsilon_0}{\pi a} \sum_{n,m \text{ odd}} \frac{\sqrt{n^2+m^2}}{nm} \tanh\left(\frac{\pi}{2}\sqrt{n^2+m^2}\right) \sin\left(\frac{n\pi x}{a}\right) \sin\left(\frac{m\pi y}{a}\right)$$
3
(a)
Superposition of terms of the form $(\rho^k, \rho^{-k})$ and $(\cos(k\phi), \sin(k\phi))$.
The potential must be zero at $\phi=0$ and $\phi=\beta$.
$\Phi(\rho, 0) = 0$: Leaving only sine terms.
$\Phi(\rho, \beta) = 0$: $\sin(k\beta) = 0$, $k\beta = n\pi$, $k = n\pi/\beta$.
$$\Phi(\rho, \phi) = \sum_{n=1}^{\infty} \left( A_n \rho^{n\pi/\beta} + B_n \rho^{-n\pi/\beta} \right) \sin\left(\frac{n\pi\phi}{\beta}\right)$$
$\Phi(a, \phi) = 0$,
$$A_n a^{n\pi/\beta} + B_n a^{-n\pi/\beta} = 0 \implies B_n = -A_n a^{2n\pi/\beta}$$
$$\Phi(\rho, \phi) = \sum_{n=1}^{\infty} A_n \left( \rho^{n\pi/\beta} - a^{2n\pi/\beta} \rho^{-n\pi/\beta} \right) \sin\left(\frac{n\pi\phi}{\beta}\right)$$
$$\Phi(\rho, \phi) = \sum_{n=1}^{\infty} C_n \left[ \left(\frac{\rho}{a}\right)^{n\pi/\beta} - \left(\frac{a}{\rho}\right)^{n\pi/\beta} \right] \sin\left(\frac{n\pi\phi}{\beta}\right)$$
(b)
n=1, let $k_1 = \pi/\beta$.
$$\Phi(\rho, \phi) \approx C_1 \left[ \left(\frac{\rho}{a}\right)^{k_1} - \left(\frac{a}{\rho}\right)^{k_1} \right] \sin(k_1\phi)$$
$E_\rho = -\frac{\partial\Phi}{\partial\rho} = -C_1 \frac{k_1}{a} \left[ \left(\frac{\rho}{a}\right)^{k_1-1} + \left(\frac{a}{\rho}\right)^{k_1+1} \right] \sin(k_1\phi)$
$E_\phi = -\frac{1}{\rho}\frac{\partial\Phi}{\partial\phi} = -\frac{C_1 k_1}{\rho} \left[ \left(\frac{\rho}{a}\right)^{k_1} - \left(\frac{a}{\rho}\right)^{k_1} \right] \cos(k_1\phi)$
$\sigma = \epsilon_0 E_{\perp}$
$\sigma(\rho, 0) = \epsilon_0 E_{\phi}|_{\phi=0} = -\frac{\epsilon_0 C_1 k_1}{\rho} \left[ \left(\frac{\rho}{a}\right)^{k_1} - \left(\frac{a}{\rho}\right)^{k_1} \right]$
$\sigma(\rho, \beta) = -\epsilon_0 E_{\phi}|_{\phi=\beta} = \epsilon_0 \frac{C_1 k_1}{\rho} \left[ \left(\frac{\rho}{a}\right)^{k_1} - \left(\frac{a}{\rho}\right)^{k_1} \right] \cos(k_1\beta = \pi) = -\frac{\epsilon_0 C_1 k_1}{\rho} \left[ \left(\frac{\rho}{a}\right)^{k_1} - \left(\frac{a}{\rho}\right)^{k_1} \right]$
$\sigma(a, \phi) = \epsilon_0 E_{\rho}|_{\rho=a} = -\epsilon_0 C_1 \frac{k_1}{a} [1+1] \sin(k_1\phi) = -\frac{2\epsilon_0 C_1 k_1}{a} \sin(k_1\phi)$
(c)
Set $\beta=\pi$, so $k_1 = \pi/\pi = 1$.
$$\Phi(\rho, \phi) \approx C_1 \left( \frac{\rho}{a} - \frac{a}{\rho} \right) \sin(\phi)$$
As $\rho \to \infty$, the $a/\rho$ term vanishes, and $\Phi \to \frac{C_1}{a} \rho\sin\phi=\Phi \to \frac{C_1}{a} y$, $\vec{E} = E_0 \hat{y}$.
$\frac{C_1}{a} = -E_0$, so $C_1 = -aE_0$,
$$\Phi(\rho, \phi) = -E_0 (\rho - a^2/\rho) \sin(\phi)$$
Substituting $C_1 = -aE_0$ and $k_1=1$ into (b),
$\sigma(a, \phi) = -\frac{2\epsilon_0 (-aE_0) (1)}{a} \sin(\phi) = 2\epsilon_0 E_0 \sin(\phi)$
$\sigma(\rho, 0) = -\frac{\epsilon_0 (-aE_0) (1)}{\rho} \left[ \frac{\rho}{a} - \frac{a}{\rho} \right] = \epsilon_0 E_0 (1 - a^2/\rho^2)$
$$\lambda_{cyl} = \int_0^{\pi} \sigma(a, \phi) (a\,d\phi) = \int_0^{\pi} (2\epsilon_0 E_0 \sin\phi) a\,d\phi = 2a\epsilon_0 E_0 [-\cos\phi]_0^{\pi} = 4a\epsilon_0 E_0$$
$$\lambda_{strip} = \sigma_{flat} \times (\text{width}) = (\epsilon_0 E_0)(2a) = 2a\epsilon_0 E_0$$
C$\lambda_{cyl} = 2\lambda_{strip}$
Flat Plane: $\lambda_{total, flat} = \sigma_{flat} \times (2L) = 2L\epsilon_0 E_0$.
Bumped Plane:
$$\lambda_{total, bump} = \lambda_{cyl} + 2 \int_a^L \sigma(\rho, 0) d\rho$$ $$= 4a\epsilon_0 E_0 + 2 \int_a^L \epsilon_0 E_0(1-a^2/\rho^2)d\rho$$ $$= 4a\epsilon_0 E_0 + 2\epsilon_0 E_0 [\rho + a^2/\rho]_a^L$$ $$= 4a\epsilon_0 E_0 + 2\epsilon_0 E_0 [(L + a^2/L) - (a+a)] = 4a\epsilon_0 E_0 + 2\epsilon_0 E_0 [L - 2a + a^2/L]$$ $$= 2L\epsilon_0 E_0 + 2\epsilon_0 E_0 a^2/L$$
As $L \to \infty$, the term $2\epsilon_0 E_0 a^2/L \to 0$,
$$\lim_{L \to \infty} \lambda_{total, bump} = 2L\epsilon_0 E_0 = \lambda_{total, flat}$$
4
By superposition principle, problem is sum of $n$ cases, each are one face have potential $V_k$, others are $0$. Then $\Phi_k(0)=C_k\cdot V_k$.
Polyhedron is regular, by symmetry, $\Phi_k(0)=C\cdot V_k$, $C$ is the same constant for all faces.
So, for $V_k=V_0$ case, we can get $C=1/n$, then $$\Phi(0)=\frac{1}{n}\sum^n_{i=1}V_i$$
HW5
1
The general solution for the potential $\Phi(r, \theta)$ in the region between the spheres ($a \le r \le b$) is a series of Legendre polynomials:
$$\Phi(r, \theta) = \sum_{l=0}^{\infty} \left( A_l r^l + \frac{B_l}{r^{l+1}} \right) P_l(\cos\theta)$$
coefficients $A_l$ and $B_l$ that satisfy the given boundary conditions.
$$V_a(\theta) = \Phi(a, \theta) = \begin{cases} V & \text{for } 0 \le \theta < \pi/2 \quad \text{(upper hemisphere)} \\ 0 & \text{for } \pi/2 < \theta \le \pi \quad \text{(lower hemisphere)} \end{cases}$$
$$V_b(\theta) = \Phi(b, \theta) = \begin{cases} 0 & \text{for } 0 \le \theta < \pi/2 \quad \text{(upper hemisphere)} \\ V & \text{for } \pi/2 < \theta \le \pi \quad \text{(lower hemisphere)} \end{cases}$$
Let $x = \cos\theta$.
The coefficient for each $l$ in equation (1) is:
$$C_l = A_l a^l + \frac{B_l}{a^{l+1}} = \frac{2l+1}{2} \int_0^\pi V_a(\theta) P_l(\cos\theta) \sin\theta \,d\theta = \frac{V(2l+1)}{2} \int_0^{\pi/2} P_l(\cos\theta) \sin\theta \,d\theta$$
$$C_l = \frac{V(2l+1)}{2} \int_0^1 P_l(x) \,dx$$
The coefficient for each $l$ in equation (2) is:
$$D_l = A_l b^l + \frac{B_l}{b^{l+1}} = \frac{2l+1}{2} \int_0^\pi V_b(\theta) P_l(\cos\theta) \sin\theta \,d\theta = \frac{V(2l+1)}{2} \int_{\pi/2}^\pi P_l(\cos\theta) \sin\theta \,d\theta$$
$$D_l = \frac{V(2l+1)}{2} \int_{-1}^0 P_l(x) \,dx$$
Using the property $P_l(-x) = (-1)^l P_l(x)$, we find that $\int_{-1}^0 P_l(x) dx = (-1)^l \int_0^1 P_l(x) dx$. Therefore, $D_l = (-1)^l C_l$.
Now we evaluate the integral $\int_0^1 P_l(x) dx$ for the first few values of $l$:
l=0: $C_0 = \frac{V}{2} \int_0^1 (1) \,dx = \frac{V}{2}$. So $D_0 = V/2$.
l=1: $C_1 = \frac{3V}{2} \int_0^1 x \,dx = \frac{3V}{2} \left[\frac{x^2}{2}\right]_0^1 = \frac{3V}{4}$. So $D_1 = -3V/4$.
l=2: $C_2 = \frac{5V}{2} \int_0^1 \frac{1}{2}(3x^2-1) \,dx = \frac{5V}{4} [x^3-x]_0^1 = 0$. So $D_2 = 0$.
l=3: $C_3 = \frac{7V}{2} \int_0^1 \frac{1}{2}(5x^3-3x) \,dx = \frac{7V}{4} \left[\frac{5}{4}x^4-\frac{3}{2}x^2\right]_0^1 = \frac{7V}{4}\left(\frac{5}{4}-\frac{3}{2}\right) = -\frac{7V}{16}$. So $D_3 = 7V/16$.
l=4: The integral is zero, so $C_4=0$ and $D_4=0$.
In general, $C_l = D_l = 0$ for all even $l > 0$.
For each $l$, we solve the system of two linear equations:
$$A_l a^l + B_l a^{-(l+1)} = C_l$$
$$A_l b^l + B_l b^{-(l+1)} = D_l = (-1)^l C_l$$
Solving this system yields:
$$A_l = \frac{C_l ((-1)^l b^{l+1} - a^{l+1})}{b^{2l+1} - a^{2l+1}}$$
$$B_l = \frac{C_l a^{l+1}b^{l+1}(b^l - (-1)^l a^l)}{b^{2l+1} - a^{2l+1}}$$
For l=0: $C_0=D_0=V/2$. This gives $A_0 = V/2$ and $B_0=0$.
For odd l: $(-1)^l = -1$.
$$A_l = -C_l \frac{a^{l+1} + b^{l+1}}{b^{2l+1} - a^{2l+1}}$$
$$B_l = C_l \frac{a^{l+1}b^{l+1}(a^l + b^l)}{b^{2l+1} - a^{2l+1}}$$
Combining the results, the potential $\Phi(r, \theta)$ is:
$$\Phi(r, \theta) = \frac{V}{2} + \sum_{l=1,3,5,...}^{\infty} \left( A_l r^l + \frac{B_l}{r^{l+1}} \right) P_l(\cos\theta)$$
Including terms up to $l=4$, we only need to write the terms for $l=1$ and $l=3$.
The potential in the region $a \le r \le b$ is:
$$\Phi(r, \theta) = \frac{V}{2} + \left( A_1 r + \frac{B_1}{r^2} \right) P_1(\cos\theta) + \left( A_3 r^3 + \frac{B_3}{r^4} \right) P_3(\cos\theta) + \dots$$
where:
$P_1(\cos\theta) = \cos\theta$
$P_3(\cos\theta) = \frac{1}{2}(5\cos^3\theta - 3\cos\theta)$
and the coefficients are:
$A_1 = -\frac{3V}{4} \frac{a^2+b^2}{b^3-a^3}$
$B_1 = \frac{3V}{4} \frac{a^2 b^2 (a+b)}{b^3-a^3}$
$A_3 = \frac{7V}{16} \frac{a^4+b^4}{b^7-a^7}$
$B_3 = -\frac{7V}{16} \frac{a^4 b^4 (a^3+b^3)}{b^7-a^7}$
Case 1: $b \to \infty$
This corresponds to the potential outside a single sphere of radius $a$. For the potential to be well-behaved at infinity ($\Phi \to 0$ as $r \to \infty$), all coefficients $A_l$ must be zero. Let's check our solution. As $b \to \infty$:
$A_l \approx -C_l \frac{b^{l+1}}{b^{2l+1}} = -C_l b^{-l} \to 0$ for $l \ge 1$.
$B_l \approx C_l \frac{a^{l+1}b^{l+1}b^l}{b^{2l+1}} = C_l a^{l+1}$.
The term $A_0 = V/2$ implies a constant potential at infinity. If we set the potential at infinity to be zero (a more standard choice), then $A_0=0$. Assuming this reference, the potential becomes:
$$\Phi(r, \theta) = \sum_{l=0}^{\infty} \frac{B_l}{r^{l+1}} P_l(\cos\theta) = \sum_{l=0}^{\infty} C_l \left(\frac{a}{r}\right)^{l+1} P_l(\cos\theta)$$
$$\Phi(r, \theta) = V \left[ \frac{1}{2}\frac{a}{r} + \frac{3}{4}\left(\frac{a}{r}\right)^2 P_1(\cos\theta) - \frac{7}{16}\left(\frac{a}{r}\right)^4 P_3(\cos\theta) + \dots \right]$$
Case 2: $a \to 0$}
This corresponds to the potential inside a single sphere of radius $b$. For the potential to be finite at the origin ($r=0$), all coefficients $B_l$ must be zero. Let's check our solution. As $a \to 0$:
$B_l \propto a^{l+1} \to 0$ for $l \ge 0$.
$A_l = \frac{C_l ((-1)^l b^{l+1} - a^{l+1})}{b^{2l+1} - a^{2l+1}} \to \frac{C_l (-1)^l b^{l+1}}{b^{2l+1}} = \frac{(-1)^l C_l}{b^l} = \frac{D_l}{b^l}$.
The potential becomes:
$$\Phi(r, \theta) = \sum_{l=0}^{\infty} A_l r^l P_l(\cos\theta) = \sum_{l=0}^{\infty} D_l \left(\frac{r}{b}\right)^l P_l(\cos\theta)$$
$$\Phi(r, \theta) = V \left[ \frac{1}{2} - \frac{3}{4}\frac{r}{b}P_1(\cos\theta) + \frac{7}{16}\left(\frac{r}{b}\right)^3 P_3(\cos\theta) + \dots \right]$$
2
The problem states that the surface charge density $\sigma(\rho)$ is proportional to $(R^2 - \rho^2)^{-1/2}$. Let's write this as:
$$ \sigma(\rho) = \frac{K}{\sqrt{R^2 - \rho^2}} $$
where $K$ is a proportionality constant. We can find $K$ by relating it to the given potential $V$. The potential of the disc at its center ($\rho=0, z=0$) is $V$:
$$ V = \Phi(0) = \frac{1}{4\pi\epsilon_0} \int_S \frac{\sigma(\rho')}{|\mathbf{r}-\mathbf{r'}|} dS' = \frac{1}{4\pi\epsilon_0} \int_0^{2\pi} d\phi' \int_0^R \frac{\sigma(\rho')}{\rho'} \rho' d\rho' $$
$$ V = \frac{2\pi}{4\pi\epsilon_0} \int_0^R \frac{K}{\sqrt{R^2 - \rho'^2}} d\rho' = \frac{K}{2\epsilon_0} \left[ \arcsin\left(\frac{\rho'}{R}\right) \right]_0^R = \frac{K}{2\epsilon_0} \left(\frac{\pi}{2}\right) = \frac{\pi K}{4\epsilon_0} $$
Solving for $K$, we get:
$$ K = \frac{4\epsilon_0 V}{\pi} $$
So, the charge density is:
$$ \sigma(\rho) = \frac{4\epsilon_0 V}{\pi \sqrt{R^2 - \rho^2}} $$
For $r > R$, we expand the on-axis potential in powers of $1/r$ (or $1/z$). The Taylor series for $\arctan(x)$ is $\sum_{l=0}^\infty \frac{(-1)^l x^{2l+1}}{2l+1}$.
$$ \Phi(r, 0) = \frac{2V}{\pi} \sum_{l=0}^{\infty} \frac{(-1)^l}{2l+1} \left(\frac{R}{r}\right)^{2l+1} $$
The general solution to Laplace's equation for $r>R$ with azimuthal symmetry is $\Phi(r, \theta) = \sum_{l=0}^\infty B_l r^{-l-1} P_l(\cos\theta)$. On the axis ($\theta=0$), $P_l(1)=1$, so $\Phi(r, 0) = \sum B_l r^{-l-1}$.
Comparing the two series for $\Phi(r, 0)$:
$$ \sum_{l=0}^\infty B_l r^{-l-1} = \frac{2V}{\pi} \sum_{l=0}^{\infty} \frac{(-1)^l R^{2l+1}}{2l+1} r^{-2l-1} $$
By matching the powers of $r$, we see that only terms where the exponent is odd survive. The coefficients $B_{2l}$ are non-zero, while $B_{2l+1}=0$.
$$ B_{2l} = \frac{2V}{\pi} \frac{(-1)^l R^{2l+1}}{2l+1} $$
The general off-axis potential is found by reintroducing the Legendre polynomials:
$$ \Phi(r, \theta) = \sum_{l=0}^{\infty} B_{2l} r^{-2l-1} P_{2l}(\cos\theta) = \sum_{l=0}^{\infty} \frac{2V}{\pi} \frac{(-1)^l R^{2l+1}}{2l+1} r^{-2l-1} P_{2l}(\cos\theta) $$
Rearranging this to match the desired form:
$$ \Phi(r, \theta) = \frac{2VR}{\pi r} \sum_{l=0}^{\infty} \frac{(-1)^l}{2l+1} \left(\frac{R}{r}\right)^{2l} P_{2l}(\cos\theta) $$
This successfully shows the potential for $r > R$.
(b)
We use the same expression $\Phi(z, 0) = \frac{2V}{\pi} \arctan(R/z)$, but now $z < R$. It's more convenient to expand in powers of $z/R$. We use the identity $\arctan(x) = \frac{\pi}{2} - \arctan(1/x)$ for $x>0$.
$$ \Phi(z, 0) = \frac{2V}{\pi} \left( \frac{\pi}{2} - \arctan\left(\frac{z}{R}\right) \right) = V - \frac{2V}{\pi} \arctan\left(\frac{z}{R}\right) $$
Expanding $\arctan(z/R)$ for $z/R < 1$:
$$ \Phi(r, 0) = V - \frac{2V}{\pi} \sum_{l=0}^{\infty} \frac{(-1)^l}{2l+1} \left(\frac{r}{R}\right)^{2l+1} $$
The general solution regular at the origin is $\Phi(r, \theta) = \sum_{l=0}^\infty A_l r^l P_l(\cos\theta)$. On the axis, this is $\Phi(r, 0) = \sum A_l r^l$.
By comparing coefficients, we find:
- $A_0 = V$
- $A_{2l} = 0$ for $l > 0$
- $A_{2l+1} = - \frac{2V}{\pi} \frac{(-1)^l}{(2l+1) R^{2l+1}}$
The general potential for $r < R$ is therefore:
$$ \Phi(r, \theta) = A_0 P_0(\cos\theta) + \sum_{l=0}^{\infty} A_{2l+1} r^{2l+1} P_{2l+1}(\cos\theta) $$
$$ \Phi(r, \theta) = V - \frac{2V}{\pi} \sum_{l=0}^{\infty} \frac{(-1)^l}{2l+1} \left(\frac{r}{R}\right)^{2l+1} P_{2l+1}(\cos\theta) $$
(c)
$$ Q = \int_S \sigma(\rho) dS = \int_0^{2\pi} d\phi \int_0^R \frac{4\epsilon_0 V}{\pi \sqrt{R^2 - \rho^2}} \rho\, d\rho $$
$$ Q = \frac{8\pi\epsilon_0 V}{\pi} \int_0^R \frac{\rho}{\sqrt{R^2 - \rho^2}} d\rho = 8\epsilon_0 V \int_0^R \frac{\rho}{\sqrt{R^2 - \rho^2}} d\rho $$
Let $u = R^2 - \rho^2$, so $du = -2\rho\,d\rho$.
$$ Q = 8\epsilon_0 V \int_{R^2}^0 \frac{-du/2}{\sqrt{u}} = 4\epsilon_0 V \int_0^{R^2} u^{-1/2} du $$
$$ Q = 4\epsilon_0 V \left[ 2u^{1/2} \right]_0^{R^2} = 4\epsilon_0 V (2\sqrt{R^2}) = 8\epsilon_0 V R $$
Capacitance:
$$ C_{\text{cap}} = \frac{Q}{V} = \frac{8\epsilon_0 V R}{V} $$
$$ C_{\text{cap}} = 8\epsilon_0 R $$
3
(a)
$$\Phi(r, \theta, \phi) = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} \left( A_{lm} r^l + B_{lm} r^{-(l+1)} \right) Y_{lm}(\theta, \phi)$$
For the region inside the sphere ($r<a$), $B_{lm} = 0$.
$$\Phi(r, \theta, \phi) = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} A_{lm} r^l Y_{lm}(\theta, \phi)$$
$A_{lm}$ are determined by the boundary condition at $r=a$, where $\Phi(a, \theta, \phi) = V(\theta, \phi)$.
$$V(\theta, \phi) = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} A_{lm} a^l Y_{lm}(\theta, \phi)$$
Using the orthogonality property of spherical harmonics, $\int Y_{l'm'}^* Y_{lm} d\Omega = \delta_{ll'} \delta_{mm'}$,
$$A_{lm} a^l = \int_{0}^{2\pi} \int_{0}^{\pi} V(\theta, \phi) Y_{lm}^*(\theta, \phi) \sin\theta \,d\theta \,d\phi$$
$$A_{lm} = \frac{1}{a^l} \int_{0}^{2\pi} \int_{0}^{\pi} V(\theta, \phi) Y_{lm}^*(\theta, \phi) \sin\theta \,d\theta \,d\phi$$
$$V(\phi) = \begin{cases} +V & 0 < \phi < \pi/2 \\ -V & \pi/2 < \phi < \pi \\ +V & \pi < \phi < 3\pi/2 \\ -V & 3\pi/2 < \phi < 2\pi \end{cases}$$
(b)
$Y_{lm}^*(\theta, \phi) = \sqrt{\frac{(2l+1)}{4\pi} \frac{(l-m)!}{(l+m)!}} P_l^m(\cos\theta) e^{-im\phi}$, into the integral for $A_{lm}$:
$$A_{lm} = \frac{1}{a^l} \sqrt{\frac{(2l+1)}{4\pi} \frac{(l-m)!}{(l+m)!}} \left( \int_{0}^{2\pi} V(\phi) e^{-im\phi} d\phi \right) \left( \int_{0}^{\pi} P_l^m(\cos\theta) \sin\theta d\theta \right)$$
$\phi$ Integral:
Let $I_\phi(m) = \int_{0}^{2\pi} V(\phi) e^{-im\phi} d\phi$.
$$I_\phi(m) = V \left[ \int_{0}^{\pi/2} e^{-im\phi} d\phi - \int_{\pi/2}^{\pi} e^{-im\phi} d\phi + \int_{\pi}^{3\pi/2} e^{-im\phi} d\phi - \int_{3\pi/2}^{2\pi} e^{-im\phi} d\phi \right]$$
For $m=0$, the integral evaluates to $V[\pi/2 - \pi/2 + \pi/2 - \pi/2] = 0$.
For $m \neq 0$:
$$I_\phi(m) = \frac{V}{-im} \left[ (e^{-im\pi/2} - 1) - (e^{-im\pi} - e^{-im\pi/2}) + (e^{-im3\pi/2} - e^{-im\pi}) - (e^{-im2\pi} - e^{-im3\pi/2}) \right]$$
This simplifies to:
$$I_\phi(m) = \frac{2V}{-im} (e^{-im\pi/2} - 1)(1 - e^{-im\pi})$$
This expression is non-zero only for $m = \pm 2, \pm 6, \pm 10, \ldots$, which can be written as $m = 2(2n+1)$ for any integer $n$. For the required range ($l \le 3$), the only possible non-zero $|m|$ value is 2 (which requires $l \ge 2$).
For $m = 2$:
$$I_\phi(2) = \frac{V}{-2i}[ (e^{-i\pi}-1) - (e^{-i2\pi}-e^{-i\pi}) + (e^{-i3\pi}-e^{-i2\pi}) - (e^{-i4\pi}-e^{-i3\pi}) ] = \frac{V}{-2i}[(-2) - (1 - (-1)) + (-1 - 1) - (1 - (-1))] = \frac{V}{-2i}[-2 - 2 - 2 - 2] = \frac{-8V}{-2i} = -4iV$$
For $m = -2$: $I_\phi(-2) = (I_\phi(2))^* = (-4iV)^* = 4iV$.
Therefore, for $l \le 3$, all coefficients $A_{lm}$ are zero except possibly for those with $m = \pm 2$. This implies that we only need to check $A_{2, \pm 2}$ and $A_{3, \pm 2}$.
$\theta$ Integral:
Let $I_\theta(l,m) = \int_{0}^{\pi} P_l^m(\cos\theta) \sin\theta d\theta = \int_{-1}^{1} P_l^m(x) dx$.
For $(l, m) = (3, \pm 2)$: The associated Legendre polynomial $P_3^2(x) = 15x(1-x^2)$ is an odd function. The integral of an odd function over a symmetric interval $[-1, 1]$ is zero. Thus, $I_\theta(3, \pm 2) = 0$, which means $A_{3, \pm 2} = 0$.
For $(l, m) = (2, 2)$: $P_2^2(x) = 3(1-x^2)$.
$$I_\theta(2, 2) = \int_{-1}^{1} 3(1-x^2)dx = 3\left[x - \frac{x^3}{3}\right]_{-1}^{1} = 3\left( \left(1-\frac{1}{3}\right) - \left(-1+\frac{1}{3}\right) \right) = 3\left(\frac{2}{3} + \frac{2}{3}\right) = 4$$
For $(l, m) = (2, -2)$: $P_2^{-2}(x) = \frac{(2-2)!}{(2+2)!}P_2^2(x) = \frac{1}{24}P_2^2(x)$.
$I_\theta(2, -2) = \frac{1}{24} \int_{-1}^{1} 3(1-x^2)dx = \frac{1}{24}(4) = \frac{1}{6}$.
We now assemble the non-zero coefficients for $l \le 3$.
$A_{2,2}$:
$$A_{2,2} = \frac{1}{a^2} \sqrt{\frac{5}{4\pi} \frac{0!}{4!}} \cdot I_\phi(2) \cdot I_\theta(2,2) = \frac{1}{a^2} \sqrt{\frac{5}{96\pi}} \cdot (-4iV) \cdot (4)$$
$$A_{2,2} = -\frac{16iV}{a^2} \frac{\sqrt{5}}{4\sqrt{6\pi}} = -\frac{4iV}{a^2}\sqrt{\frac{5}{6\pi}}$$
$A_{2,-2}$:
The potential is a real function, so the coefficients must satisfy $A_{l,-m} = (-1)^m A_{lm}^*$.
$$A_{2,-2} = (-1)^2 A_{2,2}^* = A_{2,2}^* = \left( -\frac{4iV}{a^2}\sqrt{\frac{5}{6\pi}} \right)^* = \frac{4iV}{a^2}\sqrt{\frac{5}{6\pi}}$$
4
(a)
The potential $\Phi$ at a point $\vec{r}$ is found by the principle of \textbf{superposition}:
$$\Phi(\vec{r}) = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{|\vec{r} - a\hat{z}|} + \frac{-2q}{|\vec{r}|} + \frac{q}{|\vec{r} + a\hat{z}|} \right)$$
$$\frac{1}{\sqrt{r^2 \mp 2ar\cos\theta + a^2}} = \frac{1}{r} \left( 1 \mp \frac{2a}{r}\cos\theta + \frac{a^2}{r^2} \right)^{-1/2} \approx \frac{1}{r} \left( 1 \pm \frac{a}{r}\cos\theta + \frac{a^2}{2r^2}(3\cos^2\theta - 1) + \dots \right)$$
$$\frac{1}{\sqrt{r^2 \mp 2ar\cos\theta + a^2}} = \frac{1}{r} \left( 1 \mp \frac{2a}{r}\cos\theta + \frac{a^2}{r^2} \right)^{-1/2} \approx \frac{1}{r} \left( 1 \pm \frac{a}{r}\cos\theta + \frac{a^2}{2r^2}(3\cos^2\theta - 1) + \dots \right)$$
Using $P_1(\cos\theta) = \cos\theta$ and $P_2(\cos\theta) = \frac{1}{2}(3\cos^2\theta - 1)$,
$$\frac{1}{|\vec{r} \mp a\hat{z}|} \approx \frac{1}{r} \left( 1 \pm \frac{a}{r}P_1(\cos\theta) + \frac{a^2}{r^2}P_2(\cos\theta) \right)$$
$$\Phi \approx \frac{q}{4\pi\epsilon_0} \left[ \frac{1}{r}\left(1 - \frac{a}{r}P_1 + \frac{a^2}{r^2}P_2\right) - \frac{2}{r} + \frac{1}{r}\left(1 + \frac{a}{r}P_1 + \frac{a^2}{r^2}P_2\right) \right]$$
Monopole term ($a^0$): $\frac{q}{4\pi\epsilon_0 r} (1 - 2 + 1) = 0$. (The total charge is zero).
Dipole term ($a^1$): $\frac{qa}{4\pi\epsilon_0 r^2} (-P_1 + P_1) = 0$. (The dipole moment is zero).
Quadrupole term ($a^2$): $\frac{qa^2}{4\pi\epsilon_0 r^3} (P_2 + P_2) = \frac{2qa^2}{4\pi\epsilon_0 r^3} P_2(\cos\theta)$.
$$\Phi(r, \theta) = \frac{2Q}{4\pi\epsilon_0 r^3} P_2(\cos\theta) = \frac{Q}{2\pi\epsilon_0 r^3} P_2(\cos\theta)$$
(b)
The {boundary condition is that the potential is zero on the surface of the sphere, $\Phi(r=b, \theta) = 0$.
For a charge $q_{\text{orig}}$ at a distance $d$ from the origin, its image charge $q_{\text{img}}$ in a grounded sphere of radius $b$ is located at $d_{\text{img}} = b^2/d$ with magnitude $q_{\text{img}} = -q_{\text{orig}} \frac{b}{d}$.
For the charge $q$ at $z=a$: $q'_1 = -q(b/a)$, $z'_1 = b^2/a$
For the charge $q$ at $z=-a$: $q'_2 = -q(b/a)$, $z'_2 = -b^2/a$
For the charge $-2q$ at the origin ($z=0$): $\Phi_{\text{central}}(r) = \frac{1}{4\pi\epsilon_0} \left( \frac{-2q}{r} + \frac{2q}{b} \right)$.
$$\Phi_{\text{in}}(\vec{r}) = \underbrace{\frac{1}{4\pi\epsilon_0} \left( \frac{q}{|\vec{r} - a\hat{z}|} - \frac{2q}{r} + \frac{q}{|\vec{r} + a\hat{z}|} \right)}_{\Phi_{\text{orig}}} + \underbrace{\frac{1}{4\pi\epsilon_0} \left( \frac{-q(b/a)}{|\vec{r} - (b^2/a)\hat{z}|} + \frac{-q(b/a)}{|\vec{r} + (b^2/a)\hat{z}|} + \frac{2q}{b} \right)}_{\Phi_{\text{image}}}$$
HW6
1
(a)
The general solution to Laplace's equation in cylindrical coordinates:
$$\Phi(\rho, \phi, z) = \sum_{m,n} \left( A_{mn} I_n(k_m \rho) + B_{mn} K_n(k_m \rho) \right) \left( C_{mn} \cos(n\phi) + D_{mn} \sin(n\phi) \right) \left( E_{mn} \sin(k_m z) + F_{mn} \cos(k_m z) \right)$$
Boundary Conditions in z:
$\Phi(\rho, \phi, 0) = 0$ implies $F_{mn} = 0$.
$\Phi(\rho, \phi, L) = 0$ implies $\sin(k_m L) = 0$, which means $k_m = \frac{m\pi}{L}$ for integers $m = 1, 2, 3, \ldots$.
So the z-dependence must be of the form $\sin\left(\frac{m\pi z}{L}\right)$.
Boundary Condition at $\rho=0$:
The modified Bessel function of the second kind, $K_n(x)$, diverges as $x \to 0$. Therefore, $B_{mn} = 0$.
Symmetry in $\phi$:
Even in $\phi$, $V(-\phi) = V(\phi)$. Therefore, the coefficients of the $\sin(n\phi)$ terms must be zero.
Combining these results,
$$\Phi(\rho, \phi, z) = \sum_{m=1}^{\infty} \sum_{n=0}^{\infty} A_{mn} I_n\left(\frac{m\pi \rho}{L}\right) \cos(n\phi) \sin\left(\frac{m\pi z}{L}\right)$$
At the cylindrical surface $\rho = b$, the potential is given by $V(\phi)$.
$$\Phi(b, \phi, z) = V(\phi) = \sum_{m=1}^{\infty} \left[ \sum_{n=0}^{\infty} A_{mn} I_n\left(\frac{m\pi b}{L}\right) \cos(n\phi) \right] \sin\left(\frac{m\pi z}{L}\right)$$
This is a double Fourier series. We can find the coefficients $A_{mn}$ using orthogonality.
Let $C_m(\phi)$ be the term in the square brackets.
$$C_m(\phi) = \frac{2}{L} \int_0^L V(\phi) \sin\left(\frac{m\pi z}{L}\right) dz$$
$$C_m(\phi) = \frac{2V(\phi)}{L} \left[ -\frac{L}{m\pi} \cos\left(\frac{m\pi z}{L}\right) \right]_0^L = \frac{2V(\phi)}{m\pi} (1 - \cos(m\pi))$$
This term is zero if $m$ is even and $\frac{4V(\phi)}{m\pi}$ if $m$ is odd.
So, for $m$ odd:
$$\sum_{n=0}^{\infty} A_{mn} I_n\left(\frac{m\pi b}{L}\right) \cos(n\phi) = \frac{4V(\phi)}{m\pi}$$
Now we find the coefficients $A_{mn}$ for the cosine series in $\phi$.
The coefficient for $n=0$ is:
$$A_{m0} I_0\left(\frac{m\pi b}{L}\right) = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{4V(\phi)}{m\pi} d\phi = \frac{2}{m\pi^2} \left( \int_{-\pi/2}^{\pi/2} V d\phi + \int_{\pi/2}^{3\pi/2} -V d\phi \right) = 0$$
So, $A_{m0} = 0$.
For $n > 0$:
Since the boundary condition is an even function,
$$A_{mn} I_n\left(\frac{m\pi b}{L}\right) = \frac{2}{\pi} \int_0^\pi \frac{4V(\phi)}{m\pi} \cos(n\phi)d\phi = \frac{8V}{m\pi^2} \left( \int_{0}^{\pi/2} \cos(n\phi)d\phi - \int_{\pi/2}^{\pi} \cos(n\phi)d\phi \right)$$
$$= \frac{8V}{m\pi^2} \left( \left[\frac{\sin(n\phi)}{n}\right]_0^{\pi/2} - \left[\frac{\sin(n\phi)}{n}\right]_{\pi/2}^{\pi} \right) = \frac{8V}{mn\pi^2} \left( \sin\left(\frac{n\pi}{2}\right) - (\sin(n\pi) - \sin\left(\frac{n\pi}{2}\right)) \right)$$
$$= \frac{16V}{mn\pi^2} \sin\left(\frac{n\pi}{2}\right)$$
This is non-zero only for $n$ odd. If n is odd, $\sin(n\pi/2) = (-1)^{(n-1)/2}$.
Solving for $A_{mn}$:
$$A_{mn} = \frac{16V}{mn\pi^2 I_n(m\pi b/L)} \sin\left(\frac{n\pi}{2}\right) \quad (\text{for m odd, n odd})$$
Substituting the coefficients back into the series solution gives the final potential inside the cylinder:
$$\Phi(\rho, \phi, z) = \frac{16V}{\pi^2} \sum_{m=1,3,...}^{\infty} \sum_{n=1,3,...}^{\infty} \frac{1}{mn} \frac{I_n(m\pi \rho/L)}{I_n(m\pi b/L)} \sin\left(\frac{n\pi}{2}\right) \cos(n\phi) \sin\left(\frac{m\pi z}{L}\right)$$
(b)
At $z=L/2$, the term $\sin\left(\frac{m\pi z}{L}\right)$ becomes $\sin\left(\frac{m\pi}{2}\right)$.
For the condition $L \gg b$, the argument of the Bessel functions, $x = m\pi\rho/L$, is very small.
$$I_n(x) \approx \frac{1}{n!} \left(\frac{x}{2}\right)^n$$
Using this approximation,
$$\frac{I_n(m\pi \rho/L)}{I_n(m\pi b/L)} \approx \frac{\frac{1}{n!} (m\pi \rho/2L)^n}{\frac{1}{n!} (m\pi b/2L)^n} = \left(\frac{\rho}{b}\right)^n$$
Substituting this approximation and $z=L/2$ into the solution from part (a):
$$\Phi(\rho, \phi, L/2) \approx \frac{16V}{\pi^2} \sum_{m=1,3,...}^{\infty} \sum_{n=1,3,...}^{\infty} \frac{1}{mn} \left(\frac{\rho}{b}\right)^n \sin\left(\frac{n\pi}{2}\right) \cos(n\phi) \sin\left(\frac{m\pi}{2}\right)$$
Separate the sums over $m$ and $n$:
$$\Phi(\rho, \phi, L/2) \approx \left[ \frac{4V}{\pi} \sum_{n=1,3,...}^{\infty} \frac{1}{n} \left(\frac{\rho}{b}\right)^n \sin\left(\frac{n\pi}{2}\right) \cos(n\phi) \right] \times \left[ \frac{4}{\pi} \sum_{m=1,3,...}^{\infty} \frac{1}{m} \sin\left(\frac{m\pi}{2}\right) \right]$$
The first bracket is precisely the solution to the 2D version of this problem. The 2D solution given (from Problem 2.13) is:
$$\Phi_{2D}(\vec{r}) = \frac{V_1+V_2}{2} + \frac{2(V_1-V_2)}{\pi} \sum_{n=1}^{\infty} \left(\frac{\rho}{b}\right)^n \frac{1}{n} \sin\left(\frac{n\pi}{2}\right) \cos(n\phi)$$
Setting $V_1 = V$ and $V_2 = -V$, we get:
$$\Phi_{2D}(\vec{r}) = \frac{V-V}{2} + \frac{2(V-(-V))}{\pi} \sum \dots = \frac{4V}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{\rho}{b}\right)^n \sin\left(\frac{n\pi}{2}\right) \cos(n\phi)$$
Since $\sin(n\pi/2)$ is zero for even $n$, the sum is only over odd $n$. This perfectly matches the first bracket.
The second bracket is a numerical series:
$$S = \frac{4}{\pi} \sum_{m=1,3,...}^{\infty} \frac{1}{m} \sin\left(\frac{m\pi}{2}\right) = \frac{4}{\pi} \left( \frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots \right)$$
This is the well-known Gregory-Leibniz series for $\pi/4$.
$$S = \frac{4}{\pi} \left( \frac{\pi}{4} \right) = 1$$
Combining these results,
$$\Phi(\rho, \phi, L/2) \approx \Phi_{2D}(\rho, \phi)$$
2
(a)
Laplace's equation in cylindrical coordinates:
$$\nabla^2 \Phi = \frac{1}{\rho} \frac{\partial}{\partial \rho} \left( \rho \frac{\partial \Phi}{\partial \rho} \right) + \frac{\partial^2 \Phi}{\partial z^2} = 0$$
Solution of the form $\Phi(\rho, z) = R(\rho)Z(z)$,
$$\frac{1}{\rho R} \frac{d}{d\rho} \left( \rho \frac{dR}{d\rho} \right) = -\frac{1}{Z} \frac{d^2 Z}{dz^2} = -k^2$$
where $-k^2$ is the separation constant.
Z-dependence: $\frac{d^2 Z}{dz^2} - k^2 Z = 0$. The general solution is $Z(z) = A e^{kz} + B e^{-kz}$. Since the potential must vanish as $z \to \infty$, we must have $A=0$. Thus, $Z(z) = B e^{-kz}$.
$\rho$-dependence: $\rho^2 \frac{d^2 R}{d\rho^2} + \rho \frac{dR}{d\rho} + k^2 \rho^2 R = 0$. This is Bessel's differential equation of order zero. The general solution is $R(\rho) = C J_0(k\rho) + D Y_0(k\rho)$. Since the potential must be finite at the axis ($\rho=0$) and the Neumann function $Y_0(k\rho)$ diverges there, we must have $D=0$. Thus, $R(\rho) = C J_0(k\rho)$.
The elementary solution for a given $k$ is of the form $\Phi_k(\rho, z) = J_0(k\rho)e^{-kz}$. Since $k$ is a continuous variable, the most general solution is a superposition (integral) over all possible values of $k$:
$$\Phi(\rho, z) = \int_0^\infty A(k) J_0(k\rho) e^{-kz} dk$$
The function $A(k)$ is determined by the boundary conditions on the plane $z=0$.
The boundary conditions at $z=0$ are:
$$\Phi(\rho, 0) = \begin{cases} V & \text{for } \rho < a \\ 0 & \text{for } \rho > a \end{cases}$$
Setting $z=0$ in our general solution, we get:
$$\Phi(\rho, 0) = \int_0^\infty A(k) J_0(k\rho) dk$$
This is a Fourier-Bessel transform. We can find $A(k)$ using the inverse transform:
$$A(k) = k \int_0^\infty \rho' \Phi(\rho', 0) J_0(k\rho') d\rho'$$
Substituting the given potential at $z=0$:
$$A(k) = k \int_0^a \rho' (V) J_0(k\rho') d\rho'$$
To solve this integral, we use the standard identity $\int x J_0(x) dx = x J_1(x)$. Let $u = k\rho'$, so $d\rho' = du/k$.
$$A(k) = V k \int_0^{ka} \frac{u}{k} J_0(u) \frac{du}{k} = \frac{V}{k} \int_0^{ka} u J_0(u) du = \frac{V}{k} [u J_1(u)]_0^{ka}$$
$$A(k) = \frac{V}{k} (ka J_1(ka)) = V a J_1(ka)$$
Substituting this expression for $A(k)$ back into our general solution:
$$\Phi(\rho, z) = V a \int_0^\infty e^{-kz} J_0(k\rho) J_1(ka) dk$$
(b)
We use the property that $J_0(0) = 1$. The potential along the z-axis, $\Phi_0(z) = \Phi(0, z)$, is:
$$\Phi_0(z) = V a \int_0^\infty e^{-kz} J_0(0) J_1(ka) dk = V a \int_0^\infty e^{-kz} J_1(ka) dk$$
Laplace transform of a Bessel function:
$$\mathcal{L} \{J_n(at)\} \equiv \int_0^\infty e^{-st} J_n(at) dt = \frac{a^{-n}(\sqrt{s^2 + a^2} - s)^n}{\sqrt{s^2 + a^2}}$$
- Integration variable: $t \to k$
- Exponential parameter: $s \to z$
- Bessel function order: $n=1$
- Bessel function argument parameter: $a \to a$
With these substitutions:
$$\int_0^\infty e^{-zk} J_1(ak) dk = \frac{a^{-1}(\sqrt{z^2 + a^2} - z)^1}{\sqrt{z^2 + a^2}} = \frac{\sqrt{z^2 + a^2} - z}{a \sqrt{z^2 + a^2}}$$
Substitute this result back into the expression for $\Phi_0(z)$:
$$\Phi_0(z) = V \left( 1 - \frac{z}{\sqrt{a^2 + z^2}} \right)$$
(c)
To find the potential at the edge of the disc, we set $\rho = a$:
$$\Phi_a(z) = \Phi(a, z) = V a \int_0^\infty e^{-kz} J_0(ka) J_1(ka) dk$$
Weber-Schafheitlin type integral
$$\int_0^\infty e^{-st} J_0(at) J_1(at) dt = \frac{1}{2a} \left( 1 - \frac{s}{\pi} \int_0^\pi \frac{d\theta}{\sqrt{s^2 + 4a^2 \cos^2\theta}} \right)$$
Match the variables in our potential expression with this formula: $t \to k$, $s \to z$, and $a \to a$.
$$\Phi_a(z) = \frac{V}{2} \left( 1 - \frac{z}{\pi} \int_0^\pi \frac{d\theta}{\sqrt{z^2 + 4a^2 \cos^2\theta}} \right)$$
Elliptic integral of the first kind, $K(k)$.
$$K(k) = \int_0^{\pi/2} \frac{d\phi}{\sqrt{1 - k^2 \sin^2\phi}}$$
Let's work on the integral from our expression for $\Phi_a(z)$:
$$I = \int_0^\pi \frac{d\theta}{\sqrt{z^2 + 4a^2 \cos^2\theta}}$$
$$I = \frac{1}{z} \int_0^\pi \frac{d\theta}{\sqrt{1 + (4a^2/z^2) \cos^2\theta}}$$
$$I = \frac{1}{z} \int_0^\pi \frac{d\theta}{\sqrt{1 + (4a^2/z^2)(1 - \sin^2\theta)}} = \frac{1}{z} \int_0^\pi \frac{d\theta}{\sqrt{1 + 4a^2/z^2 - (4a^2/z^2)\sin^2\theta}}$$
Factor out the term $\sqrt{1 + 4a^2/z^2} = \frac{\sqrt{z^2+4a^2}}{z}$:
$$I = \frac{1}{\sqrt{z^2+4a^2}} \int_0^\pi \frac{d\theta}{\sqrt{1 - \frac{4a^2}{z^2+4a^2}\sin^2\theta}}$$
The integrand is symmetric about $\theta = \pi/2$,
$$I = \frac{2}{\sqrt{z^2+4a^2}} \int_0^{\pi/2} \frac{d\theta}{\sqrt{1 - \frac{4a^2}{z^2+4a^2}\sin^2\theta}}$$
Define the parameter $k$ as given in the problem statement:
$k^2 = \left( \frac{2a}{\sqrt{z^2 + 4a^2}} \right)^2 = \frac{4a^2}{z^2 + 4a^2}$
$$I = \frac{2}{\sqrt{z^2+4a^2}} \int_0^{\pi/2} \frac{d\theta}{\sqrt{1 - k^2\sin^2\theta}} = \frac{2}{\sqrt{z^2+4a^2}} K(k)$$
$$\Phi_a(z) = \frac{V}{2} \left( 1 - \frac{z}{\pi} \cdot \frac{2K(k)}{\sqrt{z^2+4a^2}} \right)$$
Use the definition of $k$:
$k = \frac{2a}{\sqrt{z^2+4a^2}} \implies \frac{1}{\sqrt{z^2+4a^2}} = \frac{k}{2a}$
$$\Phi_a(z) = \frac{V}{2} \left( 1 - \frac{z}{\pi} \cdot 2 K(k) \cdot \frac{k}{2a} \right) = \frac{V}{2} \left( 1 - \frac{kz}{\pi a} K(k) \right)$$
3
(a)
For $r \neq r'$, the equation is homogeneous:
$$r^2 \frac{d^2y}{dr^2} + 2r \frac{dy}{dr} - l(l+1)y = 0$$
This is Euler's differential equation. Assuming a solution of the form $y = r^k$, we get the characteristic equation $k(k-1) + 2k - l(l+1) = 0$, which simplifies to $k^2 + k - l(l+1) = 0$. Factoring gives $(k-l)(k+l+1) = 0$, so the roots are $k=l$ and $k=-(l+1)$. The two linearly independent solutions are $r^l$ and $r^{-(l+1)}$.
Two solutions
$u(r) = r^l - a^{2l+1}r^{-(l+1)}$, $u(a)=0$
$v(r) = r^{-(l+1)} - b^{-2l-1}r^l$, $v(b)=0$
The Green function is given by:
$$g_l(r, r') = C_l \cdot u(r_<) \cdot v(r_>)$$
where $r_< = \min(r, r')$ and $r_> = \max(r, r')$. The constant $C_l$ is determined by the jump condition derived from the delta function.
$$\left[ r^2 \frac{d g_l}{dr} \right]_{r=r'^+} - \left[ r^2 \frac{d g_l}{dr} \right]_{r=r'^-} = -4\pi$$
This leads to the constant $C_l$ being related to the Wronskian $W(u, v)$ of the solutions:
$$C_l = \frac{-4\pi}{r'^2 W(u,v)(r')}$$
The Wronskian is $W(u,v) = u(r)v'(r) - u'(r)v(r)$.
$u'(r) = l r^{l-1} + (l+1) a^{2l+1} r^{-(l+2)}$
$v'(r) = -(l+1) r^{-(l+2)} - l b^{-2l-1} r^{l-1}$
$r^2 W(u,v) = -(2l+1)(1 - (a/b)^{2l+1})$
$$C_l = \frac{-4\pi}{-(2l+1)(1 - (a/b)^{2l+1})} = \frac{4\pi}{(2l+1)(1 - (a/b)^{2l+1})}$$
The Green function for the radial equation is:
$$g_l(r, r') = \frac{4\pi}{(2l+1)(1 - (a/b)^{2l+1})} \left(r_<^l - \frac{a^{2l+1}}{r_<^ {l+1}}\right) \left(\frac{1}{r_>^ {l+1}} - \frac{r_>^l}{b^{2l+1}}\right)$$
(b)
Use Green's integral theorem for a source-free volume $(\rho=0)$:
$$\Phi(\mathbf{r}) = \frac{1}{4\pi} \oint_S \left( G \frac{\partial \Phi}{\partial n'} - \Phi \frac{\partial G}{\partial n'} \right) dS'$$
Using a Dirichlet Green function, which is zero on the boundary $S$, the formula simplifies to:
$$\Phi(\mathbf{r}) = - \frac{1}{4\pi} \oint_S \Phi(\mathbf{r'}) \frac{\partial G(\mathbf{r}, \mathbf{r'})}{\partial n'} dS'$$
The full 3D Green function $G(\mathbf{r}, \mathbf{r'})$ is built from the radial Green function $g_l(r,r')$:
$$G(\mathbf{r}, \mathbf{r'}) = \sum_{l=0}^\infty \frac{2l+1}{4\pi} g_l(r, r') P_l(\cos\gamma)$$
where $\gamma$ is the angle between $\mathbf{r}$ and $\mathbf{r'}$. Since the problem has azimuthal symmetry, the potential will only depend on $r$ and $\theta$, and we can write the solution as a series of Legendre polynomials:
$$\Phi(r, \theta) = \sum_{l=0}^\infty \phi_l(r) P_l(\cos\theta)$$
The boundary conditions from Problem 3.1 are:
- $\Phi(a, \theta) = V$ for $0 \le \theta < \pi/2$ and $0$ for $\pi/2 < \theta \le \pi$.
- $\Phi(b, \theta) = 0$ for $0 \le \theta < \pi/2$ and $V$ for $\pi/2 < \theta \le \pi$.
We expand these boundary potentials in Legendre polynomials, $\Phi(r_0, \theta) = \sum_l c_l(r_0) P_l(\cos\theta)$:
$c_l(a) = V \frac{2l+1}{2} \int_0^1 P_l(x) dx $$ c_l(b) = V \frac{2l+1}{2} \int_{-1}^0 P_l(x) dx = (-1)^l c_l(a) $
The radial function $\phi_l(r)$ that matches these boundary conditions is:
$$\phi_l(r) = \frac{1}{b^{2l+1} - a^{2l+1}} \left[ (c_l(b)b^{l+1} - c_l(a)a^{l+1}) r^l + (ab)^{l+1}(c_l(a)b^l - c_l(b)a^l) r^{-l-1} \right]$$
The total potential is $\Phi(r, \theta) = \sum_{l=0}^\infty \phi_l(r) P_l(\cos\theta)$.
(c)
$$\phi_l(r) = \underbrace{\left( \frac{c_l(b)b^{l+1} - c_l(a)a^{l+1}}{b^{2l+1} - a^{2l+1}} \right)}_{A_l} r^l + \underbrace{\left( \frac{(ab)^{l+1}(c_l(a)b^l - c_l(b)a^l)}{b^{2l+1} - a^{2l+1}} \right)}_{B_l} r^{-l-1}$$
The expressions for the coefficients $A_l$ and $B_l$ are identical.
Comparison for l=0, 1, 2:
l = 0:
$c_0(a) = V/2$, $c_0(b) = V/2$.
$A_0 = \frac{V/2 \cdot b - V/2 \cdot a}{b-a} = V/2$.
$B_0 = \frac{ab(V/2 \cdot 1 - V/2 \cdot 1)}{b-a} = 0$.
Both methods give $\phi_0(r) = V/2$. The coefficients match.
l = 1:
$c_1(a) = 3V/4$, $c_1(b) = -3V/4$.
$A_1 = \frac{(-3V/4)b^2 - (3V/4)a^2}{b^3-a^3} = -\frac{3V}{4}\frac{a^2+b^2}{b^3-a^3}$.
$B_1 = \frac{(ab)^2((3V/4)b - (-3V/4)a)}{b^3-a^3} = \frac{3V}{4}\frac{a^2b^2(a+b)}{b^3-a^3}$.
l = 2:
$c_2(a) = 0$, $c_2(b) = 0$.
$A_2 = \frac{0 - 0}{b^5-a^5} = 0$.
$B_2 = \frac{(ab)^3(0 - 0)}{b^5-a^5} = 0$.
4
(a)
The Green function is expanded as:
$$G(\mathbf{x}, \mathbf{x'}) = \sum_{n=1}^{\infty} \sum_{m=-\infty}^{\infty} g_{nm}(\rho, \rho') e^{im(\phi-\phi')} \sin\left(\frac{n\pi z}{L}\right) \sin\left(\frac{n\pi z'}{L}\right)$$
The delta function in cylindrical coordinates is $\delta(\mathbf{x} - \mathbf{x'}) = \frac{1}{\rho}\delta(\rho-\rho')\delta(\phi-\phi')\delta(z-z')$.
We expand the $\phi$ and $z$ parts:
$$\delta(\phi-\phi') = \frac{1}{2\pi} \sum_{m=-\infty}^{\infty} e^{im(\phi - \phi')}$$
$$\delta(z-z') = \frac{2}{L} \sum_{n=1}^{\infty} \sin\left(\frac{n\pi z}{L}\right) \sin\left(\frac{n\pi z'}{L}\right)$$
We substitute these expansions into $\nabla^2 G = -4\pi \delta(\mathbf{x} - \mathbf{x'})$. The Laplacian in cylindrical coordinates is $\nabla^2 = \frac{1}{\rho}\frac{\partial}{\partial\rho}(\rho\frac{\partial}{\partial\rho}) + \frac{1}{\rho^2}\frac{\partial^2}{\partial\phi^2} + \frac{\partial^2}{\partial z^2}$. Applying the operators gives:
$$\frac{\partial^2}{\partial z^2} \sin\left(\frac{n\pi z}{L}\right) = -\left(\frac{n\pi}{L}\right)^2 \sin\left(\frac{n\pi z}{L}\right)$$
$$\frac{\partial^2}{\partial \phi^2} e^{im(\phi-\phi')} = -m^2 e^{im(\phi-\phi')}$$
Matching the coefficients for each $(n,m)$ mode, we obtain a radial differential equation for $g_{nm}(\rho, \rho')$:
$$\left[ \frac{1}{\rho}\frac{d}{d\rho}\left(\rho\frac{d}{d\rho}\right) - \frac{m^2}{\rho^2} - \left(\frac{n\pi}{L}\right)^2 \right] g_{nm}(\rho, \rho') = -\frac{4\pi}{\rho} \left(\frac{1}{2\pi}\right) \left(\frac{2}{L}\right) \delta(\rho-\rho') \sin\left(\frac{n\pi z'}{L}\right)$$
Let $k_n = \frac{n\pi}{L}$. The equation simplifies to:
$$\left[ \frac{1}{\rho}\frac{d}{d\rho}\left(\rho\frac{d}{d\rho}\right) - \left(k_n^2 + \frac{m^2}{\rho^2}\right) \right] g_{nm}(\rho) = -\frac{4}{L\rho} \delta(\rho-\rho') \sin\left(\frac{n\pi z'}{L}\right)$$
This is the equation for a 1D Green function in the radial coordinate, and the homogeneous equation is the modified Bessel equation.
The two linearly independent solutions are $I_m(k_n \rho)$ and $K_m(k_n \rho)$.
It must be regular at $\rho=0$ and vanish as $\rho \to \infty$. Thus, we choose $I_m(k_n \rho)$ for $\rho < \rho'$ and $K_m(k_n \rho)$ for $\rho > \rho'$.
The solution takes the form: $g_{nm}(\rho) = C \cdot I_m(k_n \rho_<) K_m(k_n \rho_>)$, where $\rho_<$ is the smaller of $(\rho, \rho')$ and $\rho_>$ is the larger.
By integrating the differential equation across the singularity at $\rho=\rho'$, we find the constant $C$. The jump in the derivative gives:
$$C = \frac{4}{L} \sin\left(\frac{n\pi z'}{L}\right)$$
Substituting the expression for $g_{nm}$ back into the expansion for $G(\mathbf{x},
$$G(\mathbf{x}, \mathbf{x'}) = \frac{4}{L} \sum_{n=1}^{\infty} \sum_{m=-\infty}^{\infty} e^{im(\phi-\phi')} \sin\left(\frac{n\pi z}{L}\right) \sin\left(\frac{n\pi z'}{L}\right) I_m\left(\frac{n\pi}{L} \rho_<\right) K_m\left(\frac{n\pi}{L} \rho_>\right)$$
(b)
Green function in a Fourier series in $\phi$ and a Fourier-Bessel (or Hankel) transform in $\rho$.
$$G(\mathbf{x}, \mathbf{x'}) = \sum_{m=-\infty}^{\infty} \int_0^\infty k\,dk \, g_m(z, z'; k) J_m(k\rho) e^{im(\phi-\phi')}$$
The delta function is:
$$\frac{\delta(\rho-\rho')}{\rho}\delta(\phi-\phi') = \left(\frac{1}{2\pi} \sum_{m=-\infty}^{\infty} e^{im(\phi-\phi')}\right) \left(\int_0^\infty k'dk' J_m(k'\rho) J_m(k'\rho')\right)$$
We again substitute into $\nabla^2 G = -4\pi \delta(\mathbf{x} - \mathbf{x'})$. The radial part of the Laplacian acting on the Bessel function $J_m(k\rho)$ gives $-k^2 J_m(k\rho)$. By matching coefficients, we obtain a 1D differential equation for $g_m(z, z'; k)$:
$$\left( \frac{d^2}{dz^2} - k^2 \right) g_m(z, z'; k) J_m(k\rho') = - \frac{4\pi}{2\pi} \delta(z-z') J_m(k\rho')$$
$$\left( \frac{d^2}{dz^2} - k^2 \right) g_m(z, z'; k) = -2 \delta(z-z')$$
We now need to solve this 1D equation subject to the boundary conditions $g_m(0, z'; k) = 0$ and $g_m(L, z'; k) = 0$.
The homogeneous equation $(g'' - k^2g = 0)$ has solutions $\sinh(kz)$ and $\cosh(kz)$.
For $z < z'$, the solution must be zero at $z=0$, so we must use $g_<(z) = A \sinh(kz)$.
For $z > z'$, the solution must be zero at $z=L$. A suitable solution is $g_>(z) = B \sinh[k(L-z)]$.
At $z=z'$, the function must be continuous ($g_<(z') = g_>(z')$), and the derivative must have a jump discontinuity determined by integrating the ODE across the delta function: $g'_>(z') - g'_<(z') = -2$.
Solving these two conditions gives the coefficients $A$ and $B$, leading to the solution:
$$g_m(z, z'; k) = \frac{2 \sinh(kz_<) \sinh[k(L-z_>)]}{k \sinh(kL)}$$
where $z_<$ is the smaller of $(z, z')$ and $z_>$ is the larger.
We substitute this 1D Green function back into our integral expansion for $G(\mathbf{x}, \mathbf{x'})$:
$$G(\mathbf{x}, \mathbf{x'}) = \sum_{m=-\infty}^{\infty} \int_0^\infty k\,dk \left( \frac{2 \sinh(kz_<) \sinh[k(L-z_>)]}{k \sinh(kL)} \right) J_m(k\rho) J_m(k\rho') e^{im(\phi-\phi')}$$
$$G(\mathbf{x}, \mathbf{x'}) = 2 \sum_{m=-\infty}^{\infty} \int_0^\infty dk \, e^{im(\phi-\phi')} J_m(k\rho) J_m(k\rho') \frac{\sinh(kz_<) \sinh[k(L-z_>)]}{\sinh(kL)}$$
HW7
1
a)
(a)
l = 0 (Monopole):
The total charge is $Q = q - q + q - q = 0$.
$$q_{00} = \sum q_i Y_{00}^* = Q \frac{1}{\sqrt{4\pi}} = 0$$
l = 1 (Dipole):
The dipole moment is non-zero, as we will see in part (b). The spherical moments are:
$q_{10} = \sum q_i r_i Y_{10}^* \propto \sum q_i a \cos(\theta_i)$. Since $\theta_i = \pi/2$ for all charges, $\cos(\theta_i)=0$, so $q_{10} = 0$.
$q_{1, \pm 1} = \sum q_i a Y_{1, \pm 1}^* = \mp a \sqrt{\frac{3}{8\pi}} \sum q_i \sin(\theta_i) e^{\mp i \phi_i}$. Since $\sin(\pi/2)=1$:
$\sum q_i e^{-i\phi_i} = q e^0 - q e^{-i\pi} + q e^{-i\pi/2} - q e^{-i3\pi/2} = q(1) - q(-1) + q(-i) - q(i) = 2q - 2iq = 2q(1-i)$.
$\sum q_i e^{i\phi_i} = q e^0 - q e^{i\pi} + q e^{i\pi/2} - q e^{i3\pi/2} = q(1) - q(-1) + q(i) - q(-i) = 2q + 2iq = 2q(1+i)$.
Therefore:
$$q_{1,1} = -a \sqrt{\frac{3}{8\pi}} [2q(1-i)] = -qa(1-i) \sqrt{\frac{3}{2\pi}}$$
$$q_{1,-1} = a \sqrt{\frac{3}{8\pi}} [2q(1+i)] = qa(1+i) \sqrt{\frac{3}{2\pi}}$$
l = 2 (Quadrupole):
Since the $q_{2m}$ moments are linear combinations of the $Q_{ij}$ components, all spherical quadrupole moments are also zero.
$$q_{2m} = 0 \quad \text{for } m = -2, -1, 0, 1, 2$$
(b)
l = 0 (Monopole):
The total charge is $Q = q - 2q + q = 0$.
$$q_{00} = 0$$
l = 1 (Dipole):
The dipole moment is $\vec{p} = q(a\hat{z}) + q(-a\hat{z}) = 0$. Therefore, all dipole moments are zero.
$$q_{1m} = 0 \quad \text{for } m = -1, 0, 1$$
l = 2 (Quadrupole):
For charges on the z-axis, spherical harmonics $Y_{lm}(\theta, \phi)$ are zero unless $m=0$. So, only $q_{20}$ can be non-zero. The charge at the origin does not contribute for $l>0$.
$$q_{20} = \sum_i q_i r_i^2 Y_{20}^*(\theta_i, \phi_i) = qa^2 Y_{20}^*(\theta=0) + qa^2 Y_{20}^*(\theta=\pi)$$
Using $Y_{20}(\theta, \phi) = \sqrt{\frac{5}{16\pi}}(3\cos^2\theta - 1)$:
$$q_{20} = qa^2 \sqrt{\frac{5}{16\pi}} [(3\cos^2(0)-1) + (3\cos^2(\pi)-1)]$$
$$q_{20} = qa^2 \sqrt{\frac{5}{16\pi}} [(2) + (2)] = 4qa^2 \sqrt{\frac{5}{16\pi}} = qa^2\sqrt{\frac{5}{\pi}}$$
The other quadrupole moments are zero: $q_{2, \pm 1} = q_{2, \pm 2} = 0$.
b)
The Cartesian moments are defined as:
Dipole: $\vec{p} = \sum_i q_i \vec{r}_i$
Quadrupole: $Q_{ij} = \sum_k q_k (3x_{ki}x_{kj} - r_k^2 \delta_{ij})$
(a)
Dipole Moment:
$$\vec{p} = q(a\hat{x}) - q(-a\hat{x}) + q(a\hat{y}) - q(-a\hat{y}) = 2qa\hat{x} + 2qa\hat{y}$$
Quadrupole Moment:
$$Q_{ij} = 0$$
(b)
Dipole Moment:
$$\vec{p} = q(a\hat{z}) - 2q(0) + q(-a\hat{z}) = 0$$
Quadrupole Moment:
The charges are on the z-axis, so $x_k=y_k=0$ for all charges.
$Q_{xx} = \sum q_k(3x_k^2 - r_k^2) = q(0-a^2) + q(0-a^2) = -2qa^2$
$Q_{yy} = \sum q_k(3y_k^2 - r_k^2) = q(0-a^2) + q(0-a^2) = -2qa^2$
$Q_{zz} = \sum q_k(3z_k^2 - r_k^2) = q(3a^2-a^2) + q(3(-a)^2-a^2) = 4qa^2$
Off-diagonal terms are zero.
The quadrupole tensor is:
$$Q = \begin{pmatrix} -2qa^2 & 0 & 0 \\ 0 & -2qa^2 & 0 \\ 0 & 0 & 4qa^2 \end{pmatrix}$$
c)
(a)
The potential is given by the dipole term:
$$V(\mathbf{r}) \approx \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \mathbf{r}}{r^3}$$
Using $\vec{p} = 2qa(\hat{x}+\hat{y})$ and $\mathbf{r} = r\sin\theta\cos\phi\hat{x} + r\sin\theta\sin\phi\hat{y} + r\cos\theta\hat{z}$:
$$\vec{p} \cdot \mathbf{r} = 2qa(r\sin\theta\cos\phi + r\sin\theta\sin\phi) = 2qar\sin\theta(\cos\phi + \sin\phi)$$
$$V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \frac{2qar\sin\theta(\cos\phi + \sin\phi)}{r^3} = \frac{qa}{2\pi\epsilon_0 r^2} \sin\theta(\cos\phi + \sin\phi)$$
(b)
The potential is given by the quadrupole term, where only $q_{20}$ is non-zero.
$$V(\mathbf{r}) \approx \frac{1}{4\pi\epsilon_0} \frac{4\pi}{5} \frac{q_{20}Y_{20}(\theta, \phi)}{r^3}$$
Substituting $q_{20}$ and $Y_{20}$:
$$V(\mathbf{r}) = \frac{1}{5\epsilon_0 r^3} \left( qa^2\sqrt{\frac{5}{\pi}} \right) \left( \sqrt{\frac{5}{16\pi}}(3\cos^2\theta - 1) \right)$$
$$V(\mathbf{r}) = \frac{qa^2}{5\epsilon_0 r^3} \frac{5}{4\pi} (3\cos^2\theta - 1) = \frac{qa^2}{4\pi\epsilon_0 r^3} (3\cos^2\theta - 1)$$
d)
(a)
The electric field of a dipole $\vec{p}$ is $\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{3(\vec{p}\cdot\hat{r})\hat{r} - \vec{p}}{r^3}$. With $\vec{p}=2qa(\hat{x}+\hat{y})$:
$$\vec{E}(\vec{r}) = \frac{1}{4\pi\epsilon_0 r^3} \left[ 3 \frac{2qa(x+y)}{r} \frac{x\hat{x}+y\hat{y}+z\hat{z}}{r} - 2qa(\hat{x}+\hat{y}) \right]$$
$$\vec{E}(\vec{r}) = \frac{qa}{2\pi\epsilon_0 r^5} \left[ 3(x+y)(x\hat{x}+y\hat{y}+z\hat{z}) - r^2(\hat{x}+\hat{y}) \right]$$
$$\vec{E}(\vec{r}) = \frac{qa}{2\pi\epsilon_0 r^5} \left( [3x(x+y)-r^2]\hat{x} + [3y(x+y)-r^2]\hat{y} + [3z(x+y)]\hat{z} \right)$$
(b)
The potential is $V(\vec{r}) = \frac{qa^2}{4\pi\epsilon_0} \frac{2z^2-x^2-y^2}{r^5}$. The field is $\vec{E}=-\nabla V$.
$$\vec{E}(\vec{r}) = \frac{3qa^2}{4\pi\epsilon_0 r^7} \left( x(4z^2-x^2-y^2)\hat{x} + y(4z^2-x^2-y^2)\hat{y} + z(2z^2-3x^2-3y^2)\hat{z} \right)$$
e)
$$\vec{E}_{quad}(\vec{r}) = \frac{3qa^2}{4\pi\epsilon_0 r^4} \left[ (5(\hat{r}\cdot\hat{z})^2-1)\hat{r} - 2(\hat{r}\cdot\hat{z})\hat{z} \right]$$
where $\hat{z}$ is the axis of the quadrupole and $\hat{r}$ is the direction vector to the point of interest.
The interaction energy is then:
$$U = -\vec{P} \cdot \vec{E}_{quad}$$
$$U = -\frac{3qa^2}{4\pi\epsilon_0 r^4} \left[ (5(\hat{r}\cdot\hat{z})^2-1)(\vec{P}\cdot\hat{r}) - 2(\hat{r}\cdot\hat{z})(\vec{P}\cdot\hat{z}) \right]$$
This expression gives the interaction energy between the linear quadrupole of system (b) and an external dipole $\vec{P}$ located at position $\vec{r}$.
2
(a)
Inside the sphere ($r < a$):
The potential $\Phi_{in}$ must be a solution to Laplace's equation, $\nabla^2\Phi_{in} = 0$, and must be finite at the origin ($r=0$). The general solution with azimuthal symmetry is:
$$\Phi_{in}(r, \theta) = \sum_{l=0}^{\infty} A_l r^l P_l(\cos\theta)$$
Outside the sphere ($r > a$):
The potential $\Phi_{out}$ is the sum of the potential from the point charge $q$ and the potential due to the polarization of the sphere, $\Phi_{ind}$.
$$\Phi_{out}(r, \theta) = \frac{q}{4\pi\epsilon_0 |\mathbf{r} - \mathbf{d}|} + \Phi_{ind}(r, \theta)$$
The potential of the point charge can be expanded in Legendre polynomials. For a point on the z-axis, this is:
$$\frac{q}{4\pi\epsilon_0 |\mathbf{r} - \mathbf{d}|} = \frac{q}{4\pi\epsilon_0} \sum_{l=0}^{\infty} \frac{r_<^l}{r_>^{l+1}} P_l(\cos\theta)$$
where $r_< = \min(r,d)$ and $r_> = \max(r,d)$.
The induced potential $\Phi_{ind}$ must satisfy Laplace's equation and vanish at infinity ($r \to \infty$). Its general form is:
$$\Phi_{ind}(r, \theta) = \sum_{l=0}^{\infty} \frac{B_l}{r^{l+1}} P_l(\cos\theta)$$
So, for the region outside the sphere but with $r<d$ (which is necessary for applying boundary conditions at $r=a$), the total potential is:
$$\Phi_{out}(r, \theta) = \sum_{l=0}^{\infty} \left[ \frac{q}{4\pi\epsilon_0} \frac{r^l}{d^{l+1}} + \frac{B_l}{r^{l+1}} \right] P_l(\cos\theta)$$
Boundary Conditions:
1. The potential is continuous: $\Phi_{in}(a, \theta) = \Phi_{out}(a, \theta)$.
$$A_l a^l = \frac{q}{4\pi\epsilon_0} \frac{a^l}{d^{l+1}} + \frac{B_l}{a^{l+1}}$$
2. The normal component of the electric displacement vector $\mathbf{D}$ is continuous: $\epsilon \frac{\partial\Phi_{in}}{\partial r}\bigg|_{r=a} = \epsilon_0 \frac{\partial\Phi_{out}}{\partial r}\bigg|_{r=a}$.
$$\epsilon (l A_l a^{l-1}) = \epsilon_0 \left( \frac{q}{4\pi\epsilon_0} \frac{l a^{l-1}}{d^{l+1}} - \frac{(l+1)B_l}{a^{l+2}} \right)$$
Solving this system yields:
$$A_l = \frac{q}{4\pi\epsilon_0} \frac{2l+1}{l(\kappa+1)+1} \frac{1}{d^{l+1}}$$
$$B_l = \frac{q}{4\pi\epsilon_0} \frac{l(1-\kappa)}{l(\kappa+1)+1} \frac{a^{2l+1}}{d^{l+1}}$$
where $\kappa = \epsilon/\epsilon_0$ is the dielectric constant. Note that for $l=0$, $B_0=0$.
Inside the sphere ($r < a$):
$$ \Phi_{in}(r, \theta) = \frac{q}{4\pi\epsilon_0} \sum_{l=0}^{\infty} \frac{2l+1}{l(\kappa+1)+1} \frac{r^l}{d^{l+1}} P_l(\cos\theta)$$
Outside the sphere ($r > a$):
$$\Phi_{out}(r, \theta) = \frac{q}{4\pi\epsilon_0 |\mathbf{r} - \mathbf{d}|} + \frac{q}{4\pi\epsilon_0} \sum_{l=0}^{\infty} \frac{l(1-\kappa)}{l(\kappa+1)+1} \frac{a^{2l+1}}{d^{l+1}r^{l+1}} P_l(\cos\theta)$$
(b)
$$\Phi_{in}(r, \theta) = A_0 + A_1 r P_1(\cos\theta) + A_2 r^2 P_2(\cos\theta) + \dots$$
The $l=0$ term is a constant potential, $A_0 = \frac{q}{4\pi\epsilon_0 d}$, which produces no electric field. The dominant term for the field is the $l=1$ term.
For $l=1$:
$$A_1 = \frac{q}{4\pi\epsilon_0} \frac{3}{1(\kappa+1)+1} \frac{1}{d^2} = \frac{q}{4\pi\epsilon_0} \frac{3}{(\kappa+2)d^2}$$
The potential from this term is:
$$\Phi_1 = A_1 r P_1(\cos\theta) = A_1 r \cos\theta$$
In Cartesian coordinates, $z = r\cos\theta$, so $\Phi_1 = A_1 z$. The electric field is uniform near the origin:
$$\mathbf{E} = -\nabla \Phi_1 = -\nabla(A_1 z) = -A_1 \mathbf{\hat{z}}$$
The rectangular components of the electric field near the center are:
$E_x = 0$
$E_y = 0$
$E_z = -A_1 = - \frac{q}{4\pi\epsilon_0} \frac{3}{(\kappa+2)d^2}$
(c)
Potential Inside ($r<a$)
Let's examine the coefficients $A_l$ as $\kappa \to \infty$:
$$A_l = \frac{q}{4\pi\epsilon_0 d^{l+1}} \frac{2l+1}{l(\kappa+1)+1}$$
For $l=0$: $A_0 = \frac{q}{4\pi\epsilon_0 d}$.
For $l \ge 1$: $\lim_{\kappa \to \infty} A_l = 0$.
Therefore, the potential inside becomes constant:
$$\lim_{\kappa \to \infty} \Phi_{in}(r, \theta) = A_0 = \frac{q}{4\pi\epsilon_0 d}$$
This matches the potential inside a neutral conducting sphere, which is constant and equal to the potential at its surface.
Potential Outside ($r>a$)
Let's examine the coefficients $B_l$ as $\kappa \to \infty$:
$$B_l = \frac{q}{4\pi\epsilon_0} \frac{a^{2l+1}}{d^{l+1}} \frac{l(1-\kappa)}{l(\kappa+1)+1}$$
For $l=0$: $B_0 = 0$.
For $l \ge 1$: $\lim_{\kappa \to \infty} \frac{l(1-\kappa)}{l(\kappa+1)+1} = \lim_{\kappa \to \infty} \frac{l(1/\kappa - 1)}{l(1+1/\kappa) + 1/\kappa} = \frac{-l}{l} = -1$.
So, in this limit, $B_l = -\frac{q}{4\pi\epsilon_0} \frac{a^{2l+1}}{d^{l+1}}$ for $l \ge 1$.
The induced potential becomes:
$$\lim_{\kappa \to \infty} \Phi_{ind}(r, \theta) = \sum_{l=1}^{\infty} \frac{B_l}{r^{l+1}} P_l(\cos\theta) = -\frac{q}{4\pi\epsilon_0} \sum_{l=1}^{\infty} \frac{a^{2l+1}}{d^{l+1}r^{l+1}} P_l(\cos\theta)$$
This is precisely the potential generated by the image charges for a neutral conducting sphere: an image charge $q' = -qa/d$ at position $b=a^2/d$, and an image charge $-q'$ at the origin to ensure the sphere is neutral. The sum of the potentials from these two image charges gives the expression above.