# Charge not in center of spherical cavity of a conductor

I have seen the well known example of a charge $Q$ placed in the center of a spherical cavity of radius $R$of a conductor. We can then say that the inner wall has a charge of $-Q$ and we can find the electric field $E$ by applying Gauss' law for a sphere of radius $r\lt R$.

So what if the charge Q is not in the center of the sphere but at some point of radius $a\lt R$ ? The charge of the inner wall will again be $-Q$ (we find this by applying Gauss' law for a sphere of radius $R' \gt R$ and noting that inside conductors $E=0$).

What is the best way to compute the electric field and the electric potential function though? Is it with respect to a position vector (x,y,z) or can we use other coordinates to simplify? I guess using the radius doesn't make sense as the problem is not symmetric anymore. Assume that everything outside the cavity is from the same conducting material, with potential $V_0=0$

• What do you mean by "what is the best way to compute the electric field and the electric potential function" ? Are you asking how do we compute the electric field and potential at a point on the spherical conductor or everywhere ? – HsMjstyMstdn May 5 '17 at 17:41

For finding the potential, we can do it by using an image charge $q'$ placed appropriately, such that the potential from $Q$ and $q'$ is zero everywhere on the surface of the conducting spherical cavity.
From a well known example, if we have a conducting sphere of radius R and a point charge with distance d from the center, then we can achieve zero potential on the surface of the sphere by placing an image charge $Q'=-QR/d$ at a distance $d'= R^2/d$. So if we invert the charges and their positions we have what we're looking for:
We place an image charge $q'=-Qa/R$ at distance $d=R^2/a$ from the center of the caavity. Now the potential on the surface is zero, as is the real potential everywhere outside the sphere. Now we can use the formula $$V(x,y,z)={1\over 4 \pi \epsilon_0 }({q\over r_q}+{q'\over r_q'})$$ to find the potential function inside the cavity, taking (0,0,0) to be the canter: $$V(x,y,z)={1\over 4 \pi \epsilon_0 }({q\over \sqrt{(x-a)^2+y^2+z^2}}-{qR\over a \sqrt{(x-d)^2+y^2+z^2}})$$ for $\sqrt{x^2+y^2+z^2} \le R$
After that we easily find the field from $E=-\nabla V$