# Charge on grounded conducting sphere in uniform E-field

• How does one deal with a grounded conducting sphere in uniform electric field ie-what is the charge distribution on the sphere?

• If two charged conducting spheres are in each other's electric field then will the field of one affect the electric potential inside the second sphere?

In order to find the potential function, one usually employs the methods used here. Then, once you have this function for the potential:

$V(r)=\frac{2q}{4\pi\epsilon_0 a^2}(r-\frac{R^3}{r^2})cos(\theta)$

Find the electric field at the surface of the grounded sphere of radius R

$E(R)=\frac{\partial{V}}{\partial r}\bigg|_{r=R}=\frac{2q*cos(\theta)}{4\pi\epsilon_0 a^2}*3=\frac{6q*cos(\theta)}{4\pi\epsilon_0 a^2}$

$\frac{\sigma}{\epsilon_0}=E(R)$

$\Rightarrow \sigma=\frac{6q*cos(\theta)}{4\pi a^2}$

Let's see what happens when a charged sphere is brought near another charged conducting sphere, or really any charge configuration changes the electric field in the environment of the charged conductiong sphere in question. The elctric field everywhere will change (except, of course inside the sphere) and there are lots of nifty tricks to figure out the new electric field, the method of image charges being the most prominent. Now, consider a regular conducting sphere with a charge q and a radius R sitting somewhere with no external fields. The potential at its surface and inside it $=\frac{q}{4\pi\epsilon_0R}$, a constant. That means that you can add this constant to the (new) potential function for the region outside without changing the electric field outside, because the gradient of a constant is $0$. Now, inside the sphere, $E=0$, which means the potential inside is a constant function (gradient of a constant is zero). You can set that constant to be whatever you want and $=\frac{q}{4\pi\epsilon_0R}$ looks pretty good, unless you want a discontinuity in the potential function.
The most analytical way to face the problem of the conducting sphere in a uniform field is solving the Laplace's equation for the potential - $\nabla^2V=0$, which comes immediately from Gauss' law in free space, $\vec\nabla.\vec{E}=\rho/\epsilon_0$, along with the definition of the potential, $\vec{E}=-\vec\nabla{V}$. Solving Laplace's equation is a well known important problem and, after that, all that's left is applying the boundaries conditions for the specific problem, refining the general solution. See, for example, this article.
The potential inside the second sphere won't change ultimately. The physical reason is that, since it's a conducting material, the electrostatic condition forbids the existence of a non-zero field inside the sphere because, if there was one, the free charges (of a conducting material) would move and this wouldn't be electrostatics. That's why the whole sphere must be an equipotential, since if there was a variation in the potential within the sphere there would be a resulting field in some point. In fact, would be enough knowing that the surface of the sphere (a shell) is an equipotential because there's a general property of the Laplace's equation solutions: they don't admit local minima or maxima. And, more than that, since the sphere is grounded its potential is, by definition, $0$.