# Electric potential

1. Jul 22, 2009

### bodensee9

A sphere of radius R has charge Q distributed uniformly over its surface. How large a sphere contains 90% of the energy stored in the electrostatic field of this charge distribution?

I assume that this is a spherical shell? So, to find energy, would we have
$$U = \frac{1}{8\pi}\int E(r)^{2}dV$$
Then, to find E(r), inside the sphere we have no charge since it's a shell?
But outside the sphere we have
$$\frac{Q}{r^{2}}\vec{r}$$
Then, if we integrate, would we find that the energy is
$$\frac{1}{2}\int\frac{Q^{2}\ast4r^{2}\pi}{r^{4}}dr$$
which is
$$\frac{Q^{2}}{2R}$$
So then if we have 90% of the potential energy then we have
$$\frac{9Q^{2}}{20R}$$
So would the large sphere have radius 20/9R? Or how do we find such a radius?
Actually, if this is not a spherical shell, but a solid sphere with charge Q distributed uniformly on its surface, would the answer still be the same? Like can we treat the inside as having on charge? Many thanks.

2. Jul 22, 2009

### Sam_Goldberg

Hi bodensee.

This is my first time helping someone out, so I'll try my best. First, you have a 1/2 in front of the integral, but I believe it should be 1/(8pi) from the original equation. Remember that a differential element of volume is not dr; rather, it is 4pi r2 dr. Second, you should check your limits of integration. Since, as you correctly pointed out, there is no E-field inside of the sphere (due to Gauss's law), you need to start you integration at a certain lower limit, perhaps the radius of the charged sphere itself? Next, you need an upper limit. Since you need to get 90% of the total energy, why don't you first find the total energy? What would be the upper limit for that? Then figure out how far you need to go to enclose 90% of the energy of the field.

As for your final question, isn't a spherical shell and a charge uniformly distributed on the surface the same thing? (Again, think of Gauss's law.)

3. Jul 22, 2009

### bodensee9

Right, um, I'm not very good at typing with latex, but i think if you do the integral, you will still have Q^2/2R as the total field. So I think 90% would be 9Q^2/20R.

4. Jul 22, 2009

### Sam_Goldberg

Well, there are two radii you need to consider. One is the radius of the charged sphere, R. For the total energy I assume you integrated from R to infinity? Then for 90% you don't know the radius of the enclosing imaginary surface, so you may want to just call your upper limit D or something and compare that with the total energy. 90% of the total energy should be the energy when you integrate from R to D.

5. Jul 22, 2009

### diazona

Well, let's see: you're using Gaussian units, right?
$$U = \frac{1}{8\pi}\int E^{2}\mathrm{d}V$$
So we agree on that.
$$\vec{E} = \frac{Q}{r^2}\hat{r}$$
Also on that. The integral would be
$$U = \frac{1}{8\pi}\int_R^{\infty} \frac{Q^2}{r^4}4\pi r^2\mathrm{d}r$$
so you need that 8 pi factor - and the limits of integration - as Sam pointed out but otherwise, OK.
$$U = \frac{Q^2}{2}\int_R^{\infty} \frac{1}{r^2}\mathrm{d}r = \frac{Q^2}{2R}$$
So that looks good. And you're right that 90% of the potential energy would be 9Q2/20R.

Now, since you integrated all the way out to infinity, you're enclosing 100% of the potential energy. But look at the question: it asks how far out would you have to go to get 90% of the potential energy? In this case you wouldn't integrate all the way out to infinity; you stop the integral (set its upper limit) at the point where you reach 90% of the potential energy. Then you can do the math to figure out what that point is.

6. Jul 22, 2009

### bodensee9

Oh .. okay, thanks!