I am studying electromagnetism and I am a bit confused about the following problem:

We have a charged ($+Q$) sphere (radius $R_0$) surrounded by a conductive shell (radius $R_1$ and $R_2$) which we keep at a steady potential $V_0$. I need to find the electric field everywhere at space.

My problem is for $r>R_2$, I don't understand how to account for the spherical shell when using Gauss's law.

Any help would be appreciated.


You can solve this without Gauss' Law, just using precalculus physics.

Your situation is spherically-symmetric. That means for any sphere of radius $r>R_2$ the electric field points either inwards or outwards. All we have to do is find the magnitude and direction. Without loss of generality I'll assume that $V_0$ is positive. (If $V_0$ is negative then just reverse everything.) Since the conductive shell is kept at steady potential $V_0$ that means the electric field points outward.

Now we have to figure out the magnitude of the electric field. There are several ways of approaching this problem. I'm going to use the one I happen to prefer.

The $+Q$ charge is a red herring. It doesn't affect our answer. That's because keeping the conductive shell at steady potential $V_0$ shields the rest of the world from the effects of $+Q$. So all we have to consider is a conductive shell of radius $R_2$ kept at steady potential $V_0$.

According to the shell theorem, a spherical charge viewed from outside the sphere can be treated as a point charge. Thus, from the perspective of a point $r>R_2$, the sphere kept at potential $V_0$ can be treated as a point charge $q$ located at the real sphere's origin. All we have to do is is figure out what the charge of this imaginary point charge is.

Fortunately, we have an equation for the electric potential produced by a point charge.

$$ V(\vec r) = \frac 1{4\pi\epsilon_0}\frac qr $$

We know the potential is $V_0$ at radius $R_2$.

$$ V_0 = \frac 1{4\pi\epsilon_0}\frac q{R_2} $$

This lets us solve for our imaginary charge $q$.

$$ q=4\pi\epsilon_0V_0R_2 $$

Lastly all we have to do is plug this charge $q$ into Coulomb's Law.

$$E=\frac 1{4\pi\epsilon_0}\frac{|q|}{r^2}$$ $$E=\frac 1{4\pi\epsilon_0}\frac{|4\pi\epsilon_0V_0R_2|}{r^2}$$ $$E=\frac{V_0R_2}{r^2}$$

  • $\begingroup$ Thank you very much ! $\endgroup$ – Lagrange Jun 10 at 21:43
  • $\begingroup$ Welcome to Stack Exchange! $\endgroup$ – lsusr Jun 10 at 21:44

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