- #1

andre220

- 75

- 1

## Homework Statement

An insulated conducting sphere of radius ##R##, carrying a total charge of ##Q##, is in the field of a point charge ##q## of the same sign. Assume ##q\ll Q##. Calculate and plot the force exerted by the sphere on ##q## as a function of distance from the center. In particular, for ##q/Q = 0.1##, calculate the point at which the direction of force reverses.

## Homework Equations

The electric field due a point charge $$E_\mathrm{pt} = \frac{q}{4\pi\epsilon_0 r^2}$$

$$\Phi(r,\theta) = \sum\limits_{l=0}^\infty \left(A_l r^l + \frac{B_l}{r^{l+1}}\right) P_l\left(\cos{\theta}\right)$$

$$\vec{E} = -\vec{\nabla}\Phi$$

## The Attempt at a Solution

So since the solution is wanted as a function of distance from the center of the sphere, I am using the legendre solutions to laplace's equation. The boundary conditions are ##\Phi(R,\theta) = 0## and ##E_\mathrm{in} -E_\mathrm{out} = -\sigma/\epsilon_0 \implies E_\mathrm{point} - \nabla_r \Phi =-\frac{4\pi R^2}{\epsilon_0}##

Then once the boundary conditions are satisfied then the force is just ##F = qE_\mathrm{sphere}##.

So for the potential of the sphere:

$$\Phi(r) = A_0 + \frac{B_0}{r}$$

$$\Phi(R) = 0 = A_0 + \frac{B_0}{R}\implies B_0 = -A_0 R\implies \Phi(r) = A_0\left(1-\frac{R}{r}\right)$$

$$E_\mathrm{sphere} = -\frac{2 A_0 R}{r^2}$$

Then, applying the boundary conditions at the surface of the sphere:

$$\frac{q}{4\pi\epsilon_0 R^2} + \frac{2 A_0}{R} = -\frac{4\pi R^2 Q}{\epsilon_0}\implies A_0 = \frac{-q + 16\pi^2 R^4 Q}{8\pi\epsilon_0 R}$$

Then,

$$E_\mathrm{sphere} = \frac{1}{r^2}\left(\frac{q + 16\pi^2 Q R^4}{4\pi\epsilon_0}\right)$$

Then from here I would just do force by multiplying by ##q## of the point charge and then find the point at which the force reverses.

Alot to take in, but does this seem correct thus far?