A problem of an equipotential hollow sphere

Question

I was given the following exercise:

There are two concentric hollow spheres of radii $a$ and $b$ as shown. The inner sphere is connected to a constant potential $V_0$ while the outer sphere has uniform surface charge density $\rho$. Find the potential in all the space.

I have the correct answer from my professor which is:

• if $r > b$ :

$$V = \frac{1}{r} \left[ \left(V_0 - \frac{\rho b}{\epsilon_0}\right) a + \frac{\rho b^2}{\epsilon_0}\right]$$

• if $a < r < b$: $$V = \frac{a}{r}\left(V_0 -\frac{\rho b}{\epsilon_0}\right)+\frac{\rho b}{\epsilon_0}$$

• else $V = V_0$

My attempt

However, I can't find the correct procedure. I started making $V(\infty) = 0$. Then, I tried with the image method, so I replaced the inner surface with a point charge located at the center (where r = 0) to find an equivalent problem in the region $r > a$. Since I know that $V(a) = V_0$, I have that:

$$ V_0 = - \int_\infty^a E dr = \int_a^b \frac{k q}{r^2} dr + \int_b^\infty \frac{\rho}{\epsilon_0} \frac{b^2}{r^2} dr = k q \left(\frac{1}{a} - \frac{1}{b} \right) + \frac{\rho b}{\epsilon_0}$$

$$k q = \left(V_0 - \frac{\rho b}{\epsilon_0} \right) \left(\frac{1}{a} - \frac{1}{b} \right)^{-1}$$

Now, I find again the potential, starting with the region $r > b$

$$ V(r) = - \int_\infty^r \left(\frac{kq}{r^2} + \frac{\rho}{\epsilon_0} \frac{b^2}{r^2} \right) dr = \frac{1}{r} \left( kq + \frac{\rho b^2}{\epsilon_0}\right) = \frac{1}{r} \left[ \left(V_0 - \frac{\rho b}{\epsilon_0} \right) \left(\frac{1}{a} - \frac{1}{b} \right)^{-1} + \frac{\rho b^2}{\epsilon_0}\right]$$

Which is different from the answer. The term $\left(\frac{1}{a} - \frac{1}{b} \right)^{-1}$ should be just simply $a$ instead. And for $a < r < b$, the contribution of the outer sphere in the electric field is zero, so I only count the field of the charge:

$$ V(r) = \int_a^b \frac{k q}{r^2} dr + V(b) = k q \left(\frac{1}{a} - \frac{1}{b} \right) + \frac{1}{b} \left[ \left(V_0 - \frac{\rho b}{\epsilon_0} \right) \left(\frac{1}{a} - \frac{1}{b} \right)^{-1} + \frac{\rho b^2}{\epsilon_0}\right]$$

The image method is not valid inside both surfaces because the electric field is not the same in that region. But since the electric field is zero, it is trivial to note that $V(r)$ has to be equal to $V_0$.

I can't figure out if I'm wrong or not. Does anyone know if my method is not correct or if I just made a mistake in some computation?

Answer 1

In your first equation you calculate the potential of the inner sphere by integrating the electrical field from infinity to the inner sphere radius. You correctly split the integral into two parts, from $a$ to $b$ and from $b$ to infinity. The field between $a$ and $b$ is indeed just $kq/r^2$. However, the field outside the outer sphere is a superposition of the fields from both spheres. Thus, the field there is not just $\rho b^2/\epsilon_0r^2$, but $$ E_{outside} = \frac{\rho}{\epsilon_0} \frac{b^2}{r^2} + \frac{k q}{r^2} $$

Thus, your first equation should be $$ V_0 = - \int_\infty^a E dr = \int_a^b \frac{k q}{r^2} dr + \int_b^\infty \left(\frac{\rho}{\epsilon_0} \frac{b^2}{r^2} + \frac{k q}{r^2}\right) dr = \int_a^\infty \frac{k q}{r^2} dr + \int_b^\infty \frac{\rho}{\epsilon_0} \frac{b^2}{r^2} dr $$ Physically it means that the field from the inner sphere contributes to the potential change in the whole space $r > a$, not just in $b > r > a$.

Thus, you'll get $$ V_0 = \frac{k q}a + \frac{\rho b}{\epsilon_0} $$ with the corresponding changes in your following calculations.

Answer 2

Your method is basically correct.

I think that the error is that in the top equation you have the limits of integration for the first integral as $\int_a^b$ but those should be $\int_a^\infty$. Notice the following contradiction. If you look at the second part you are using that component of the field as if it extends outside the sphere of radius $b$, but you have assumed that it disappears outside the sphere of radius $b$ in your first integral.

In the last part you say that "for $a<r<b$, the contribution of the outer sphere to the electric field is zero" but of course it contributes to the potential. The first term and the last term in both the equations of your teacher are just the normal charged sphere terms.

Another way to solve this problem, as suggested in the comments by Andrew, would be to work out what charge on the inner sphere of radius $a$ would be necessary to cancel the charge of the outer sphere and make the potential inside the inner sphere just $V_0$, and use that to calculate the middle term. I actually did this at first and was going to answer your question with that, but I thought I had better check what you did before posting that and you did it a different way.