I have been working on a problem about finding the electrostatic potential energy stored on a capacitor of concentric spheres with inner radius $a$ and outer radius $b$ and with charge $Q$. I've got to first calculate the energy using the capacitance and then integrating the energy density.

I did the following: first of all I've used Gauss' Law to find the electric field a distance $r$ from the center of both spheres with $a < r < b$. As expected I've got the field of a point charge $Q$ at the center. Then I've integrated the field along a segment joining the two spheres and I've got the following difference of potential


Then I've found the capacitcante using $Q=CV$ and finally I've found the energy using $U=Q^2/2C$. That's fine, I've got the value:


Now the second part, I should find the same value integrating the energy density. So, the energy density is $\mathcal{u}=\epsilon_0E^2/2$ and hence it is:

$$\mathcal{u}=\frac{1}{2}\epsilon_0 \frac{1}{16\pi^2\epsilon_0^2}\frac{Q^2}{r^4}$$

Since I must integrate on the region between the two spheres I've used spherical coordinates and computed the following integral getting the energy:

$$U_2=\int_0^\pi\int_0^{2\pi}\int_a^b \frac{1}{2}\epsilon_0 \frac{1}{16\pi^2\epsilon_0^2}\frac{Q^2}{r^4}r^2\sin\phi dr d\theta d\phi$$

And this integral gave simply:

$$U_2 =\frac{1}{8\pi\epsilon_0}\frac{Q^2(b-a)}{ab}$$

But wait a moment, I've got $U_2 = -U_1$ instead of $U_2 = U_1$ as expected. I've calculated it once again and once again and I've got the same problem. Can someone point out what's happening? Where's my mistake?

Thanks in advance for your help!


1 Answer 1


Your mistake is that in the expression $$ Q=CV $$ the symbol $V$ here represents the magnitude of the potential difference between the spheres. Thus, since $b>a$ here, you need to switch the order of $b$ and $a$ in the first expression you wrote down for $V$ if you want to plug it into the expression defining capacitance (in other words, you need to take its absolute value).

  • $\begingroup$ @user1620696 Sure thing! $\endgroup$ Commented Apr 26, 2013 at 2:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.