# Force on a point charge due to a sphere

1. Mar 8, 2015

### andre220

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
The electric field due a point charge $$E_\mathrm{pt} = \frac{q}{4\pi\epsilon_0 r^2}$$
$$\Phi(r,\theta) = \sum\limits_{l=0}^\infty \left(A_l r^l + \frac{B_l}{r^{l+1}}\right) P_l\left(\cos{\theta}\right)$$
$$\vec{E} = -\vec{\nabla}\Phi$$

3. 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?

2. Mar 8, 2015

### Staff: Mentor

Where does that come from? I think the idea of the conducting sphere is the non-uniform charge distribution. The electric field and therefore the potential should depend on θ.

3. Mar 8, 2015

### andre220

Right, then I don't see how I can get the constants $A_1, B_1$ with only two boundary conditions.
Thats why I only chose to keep $l = 0$ terms.

4. Mar 8, 2015

### Staff: Mentor

There are formulas how a charge q at some distance L will influence charges on a conducting sphere. It's possible to derive them but I guess that is not necessary here.
Q does not matter for this induced charge.

5. Mar 8, 2015

### andre220

Is this method correct that I am using? Or should I use an image method here?

6. Mar 8, 2015

### TSny

Your method is not incorrect, but it involves an infinite series. I would try the image method first.