PH507 Homework
PS1
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.
$$\vec{a} \cdot \vec{r} = a_x x + a_y y + a_z z$$
$$\nabla(\vec{a} \cdot \vec{r}) = \frac{\partial}{\partial x}(a_x x + \dots)\hat{i} + \frac{\partial}{\partial y}(a_y y + \dots)\hat{j} + \frac{\partial}{\partial z}(a_z z + \dots)\hat{k}$$ $$\nabla(\vec{a} \cdot \vec{r}) = a_x\hat{i} + a_y\hat{j} + a_z\hat{k} = \vec{a}$$
PS2
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 $\hat{r} \cdot \hat{k} = \cos\theta$.
$$E_z = \left(\frac{Qr}{4\pi\epsilon_0 R^3}\right) \cos\theta$$
$$F_z = \iiint_{V_{upper}} \rho E_z \, d\tau$$
$$F_z = \int_0^{2\pi} d\phi \int_0^{\pi/2} d\theta \int_0^R dr \, \left(\frac{3Q}{4\pi R^3}\right) \left(\frac{Qr\cos\theta}{4\pi\epsilon_0 R^3}\right) (r^2 \sin\theta)$$
$$F_z = \frac{3Q^2}{16\pi^2\epsilon_0 R^6} (2\pi) \left(\frac{R^4}{4}\right) \left(\frac{1}{2}\right) = \frac{3Q^2 (2\pi R^4)}{128\pi^2\epsilon_0 R^6}$$
$$\vec{F} = \frac{3Q^2}{64\pi\epsilon_0 R^2} \hat{k}$$
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 $\theta=\pi/2$, $u = a^2+d^2$.
$$\int_{(d-a)^2}^{a^2+d^2} u^{-3/2} \, \frac{du}{2ad} = -\frac{1}{ad} \left[ \frac{1}{\sqrt{u}} \right]_{(d-a)^2}^{a^2+d^2}$$ $$= -\frac{1}{ad} \left( \frac{1}{\sqrt{a^2+d^2}} - \frac{1}{\sqrt{(d-a)^2}} \right) = \frac{1}{ad} \left( \frac{1}{d-a} - \frac{1}{\sqrt{a^2+d^2}} \right)$$
$$q'_{hemi} = -\frac{q a (d^2-a^2)}{2} \times \left[ \frac{1}{ad} \left( \frac{1}{d-a} - \frac{1}{\sqrt{a^2+d^2}} \right) \right]$$
$$q' = -q \left[ 1 - \frac{d^2 - a^2}{d\sqrt{d^2 + a^2}} \right]$$