# Energy of a Charged Sphere

In this problem, you’ll be calculating the energy of a charged sphere. This can be pretty hard if you’ve never done anything like it before. But it’s super easy once you know the trick. Give it a shot!

You have a solid metal ball with a total charge $Q$ and a radius $r$. What is it’s energy?

## one way to find the energy of a charged sphere

The first thing to notice is that all the charge goes to the surface. This is because the electric field in a conductor in equilibrium is zero, so if you pick any small region inside the sphere, there is no net enclosed charge. Since this region is arbitrary, there can be no charge inside the sphere at all.

The next thing to figure out is what you mean by potential energy? This is the hardest part of the problem so spend a bit of time thinking about it…

OK. By definition, the potential energy is the work you have to do to assemble this system. But assemble it from what? Since we’re only considering the charge, our initial state is a neutral sphere with the rest of the charge infinitely far away from each other so that they do not interact. Now imagine slowly adding charge bit by bit. It will distribute itself uniformly on the surface of the sphere and send out an electric field. But we know that the electric field outside a uniformly charged sphere is the same if all the charge were located at the centre!

Aren’t familiar with this? Here’s a quick proof. Let’s say you’re at a radius $R$ from the centre of the sphere, and assume that $R$ is greater than the radius of the sphere $r$. Let’s say the total charge on the sphere is $Q$. Then by Gauss’s Law, the electric field at $R$ satisfies

$\displaystyle E(R) \cdot 4 \pi R^2 = \frac{Q}{\epsilon_0} \Rightarrow E(R) = \frac{Q}{4 \pi \epsilon_0 R^2}$

This is the same electric field if all the charge were localized at the centre of the sphere, so we’re done!

Back to the problem. Let’s say that you bring in a small point charge $dq$ when the ball already has a charge $q$ (notice $q = 0$ when we’re just starting out). Then there’s a force on $dq$ due to the electric field from $q$! In other words, we’re doing work! Since $dq$ is always outside the sphere (except for an instant in the end when it is exactly on the sphere), we can treat $q$ as a charge at the centre the entire time. In the end, $dq$ is a distance $r$ away from $q$. So the work we have to do is just the potential energy between two point charges! Mathematically,

$\displaystyle dW = \frac{q \, dq}{4 \pi \epsilon_0 r}$

Since $q$ goes from $0$ to $Q$, we can integrate this to get the total work done

$\displaystyle \int dW = W = \int_{0}^{Q} \frac{q \, dq}{4 \pi \epsilon_0 r} = \frac{Q^2}{8 \pi \epsilon_0}$

Notice that this energy is proportional to $Q^2$, not $Q$. So if you put twice the charge, you actually have to do four times the work. That’s pretty cool!

Keep on reading if you’re familiar with the energy density in an electric field.

# another way to find the energy of a charged sphere

Like before, we know the electric field inside the sphere is zero and outside the sphere is

$E(R) = \frac{Q}{4 \pi \epsilon_0 R^2}$

Since energy density is $1/2 \epsilon_0 E^2$, we can find the total energy of a charged sphere via an integration over all space.

$\displaystyle U = \int_{all \, space} 1/2 \epsilon_0 E^2 dV = \int_{inside \, sphere}1/2 \epsilon_0 E^2 dV + \int_{outside \, sphere}1/2 \epsilon_0 E^2 dV$

But since the electric field is zero inside the sphere, this simplifies to

$\displaystyle U = \int_{outside \, sphere}1/2 \epsilon_0 E^2 dV$

The most natural way to integrate is by breaking space up into little shells of radius $R$ and thickness $dR$. Then the volume element is

$\displaystyle dV = 4 \pi R^2 dR$

If this formula is new to you, just imagine that you cut up the sphere and place it flat on a plane. The area is just the surface area of a sphere: $4 \pi R^2$. The thickness is $dR$. Therefore the volume is the product of the two.

Let’s carry out the integral.

$\displaystyle U = \int_{r}^{\infty} 1/2 \epsilon_0 E(R)^2 4 \pi R^2 dR$

Substituting our expression for $E(R)$, we get

$\displaystyle U = \int_{r}^{\infty} 1/2 \epsilon_0 (\frac{Q}{4 \pi \epsilon_0 R^2})^2 4 \pi R^2 dR$

Simplifying,

$\displaystyle U = \frac{Q^2}{8 \pi \epsilon_0} \int_{0}^{\infty} \frac{dR}{R^2}$

$\displaystyle \Rightarrow U = \frac{Q^2}{8 \pi \epsilon_0 R}$

just like before.

# is a charged sphere a capacitor?

You may notice that the energy of a charged sphere is just the energy of a capacitor with capacitance $C = 4 \pi \epsilon_0 r$. Here’s the formula for the energy of a capacitor in case you forget

$U = \frac{Q^2}{2C}$

This leads to the interesting interpretation of an isolated charged sphere as simply a capacitor with the other plate infinitely far away (imagine another sphere with an infinite radius around our sphere). It’s also an easy way to remember the formula for the energy of a charged sphere!

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## 2 thoughts on “Energy of a Charged Sphere”

1. Anonymous says: