Boundary Value Problems

Step 1: Set up the boundary conditions

We need $\Phi = 0$ on the plane $z = 0$, $\Phi \to 0$ as $r \to \infty$, and the field must match a point charge $q$ at $(0, 0, d)$ for $z > 0$.

Step 2: Place an image charge

An image charge $-q$ at $(0, 0, -d)$ makes $\Phi = 0$ at $z = 0$ by symmetry. For $z > 0$:

$$\Phi(\mathbf{r}) = \frac{q}{4\pi\epsilon_0}\left[\frac{1}{\sqrt{\rho^2 + (z-d)^2}} - \frac{1}{\sqrt{\rho^2 + (z+d)^2}}\right]$$

Step 3: Compute induced surface charge

$$\sigma = -\epsilon_0\frac{\partial\Phi}{\partial z}\bigg|_{z=0} = -\frac{qd}{2\pi(\rho^2 + d^2)^{3/2}}$$

Step 4: Verify total induced charge

$$Q_{\text{ind}} = \int_0^\infty \sigma \cdot 2\pi\rho\,d\rho = -qd\int_0^\infty\frac{\rho\,d\rho}{(\rho^2 + d^2)^{3/2}} = -q$$

Step 5: Force on the charge

The force on $q$ is that due to the image charge at distance $2d$:

$$F = -\frac{q^2}{4\pi\epsilon_0(2d)^2} = -\frac{q^2}{16\pi\epsilon_0 d^2}$$

This attractive force causes the charge to be pulled toward the conductor.

Step 1: Set up the geometry

Consider a box $0 \leq x \leq a$, $0 \leq y \leq b$, $0 \leq z \leq c$. Five faces are grounded ($\Phi = 0$) and the top face ($z = c$) has $\Phi = V_0(x,y)$.

Step 2: Separation of variables

The boundary conditions at $x = 0, a$ and $y = 0, b$ select $X_n = \sin(n\pi x/a)$ and $Y_m = \sin(m\pi y/b)$. The $z$-equation gives $Z'' = \gamma_{nm}^2 Z$ where $\gamma_{nm} = \pi\sqrt{n^2/a^2 + m^2/b^2}$.

Step 3: Apply $\Phi = 0$ at $z = 0$

This selects $Z_{nm}(z) = \sinh(\gamma_{nm}z)$ (vanishes at $z = 0$).

Step 4: Match the top boundary condition

$$V_0(x,y) = \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}A_{nm}\sin\frac{n\pi x}{a}\sin\frac{m\pi y}{b}\sinh(\gamma_{nm}c)$$

Step 5: Determine coefficients by orthogonality

$$A_{nm} = \frac{4}{ab\sinh(\gamma_{nm}c)}\int_0^a\int_0^b V_0(x,y)\sin\frac{n\pi x}{a}\sin\frac{m\pi y}{b}\,dx\,dy$$

The complete solution is $\Phi = \sum_{nm}A_{nm}\sin(n\pi x/a)\sin(m\pi y/b)\sinh(\gamma_{nm}z)$.

Step 1: Write Laplace's equation in cylindrical coordinates

$$\frac{1}{\rho}\frac{\partial}{\partial\rho}\left(\rho\frac{\partial\Phi}{\partial\rho}\right) + \frac{1}{\rho^2}\frac{\partial^2\Phi}{\partial\phi^2} + \frac{\partial^2\Phi}{\partial z^2} = 0$$

Step 2: Assume $\Phi = R(\rho)\Psi(\phi)Z(z)$

The $z$-equation gives $Z'' = k^2 Z$ (exponential or trigonometric). The $\phi$-equation gives $\Psi'' = -m^2\Psi$ with $m$ integer (periodicity).

Step 3: Bessel's equation for the radial part

$$\rho^2 R'' + \rho R' + (k^2\rho^2 - m^2)R = 0$$

This is Bessel's equation of order $m$ with solutions $J_m(k\rho)$ (regular at origin) and $N_m(k\rho)$ (singular at origin).

Step 4: Modified Bessel functions for imaginary argument

When $k^2 < 0$ (i.e., $Z$ is trigonometric in $z$), the radial equation has modified Bessel function solutions $I_m(\kappa\rho)$ and $K_m(\kappa\rho)$.

Step 5: Orthogonality of Bessel functions

For a cylinder of radius $a$ with $\Phi = 0$ at $\rho = a$, the allowed $k$ values are $k_{mn} = x_{mn}/a$ where $J_m(x_{mn}) = 0$. The Bessel functions satisfy the orthogonality relation:

$$\int_0^a J_m(k_{mn}\rho)J_m(k_{mn'}\rho)\rho\,d\rho = \frac{a^2}{2}[J_{m+1}(x_{mn})]^2\delta_{nn'}$$