# Negative energy capacitor?

1. Oct 4, 2013

### johne1618

The electrostatic energy in a capacitor is negative is it not?

If one had enough positive and negative charge on the plates so that the electrostatic energy was greater in magnitude than the rest mass energy then the system as a whole would have a negative total energy.

Would this be possible and how would such a system behave?

In general relativity negative pressure behaves like negative gravitational mass.

But if the capacitor is static then the negative electrostatic pressure must be balanced by the positive pressure in the structure of the capacitor so that the overall gravitational mass doesn't change.

Maybe that's why the total mass/energy must remain constant as the capacitor is charged up?

Last edited: Oct 4, 2013
2. Oct 4, 2013

### phyzguy

No it's not. The energy of a capacitor can be written as:
$$Energy = \frac{C V^2}{2}$$
or
$$Energy = \frac{\epsilon_0}{2}\int E^2 dV$$

Both are positive quantities.

3. Oct 4, 2013

### johne1618

But if you constructed a capacitor by bringing positive and negative plates from infinity to a separation $d$ you would gain energy in the process. I must admit that's not the normal way one charges a capacitor though.

4. Oct 4, 2013

### phyzguy

Yes, you would gain energy. So the electric field energy of a capacitor constructed like this would be less than the electric field energy was when the two plates were at infinite separation. However, it would still be positive. Try calculating the electric field energy of two charges at inifinite separation and compare it to the electric field energy of a dipole created by having those two charges at a separation d. You will see that the energy of the second configuration is less, but they are both positive.

5. Oct 5, 2013

### johne1618

I think I see what you mean.

The total electrostatic energy of a system of two spheres with charges $q$ and $-q$ is:

$$U = \frac{kq^2}{2r} + \frac{kq^2}{2r} - \frac{kq^2}{d}$$

where the radius of the spheres is $r$ and the separation $d$.

$$U = k q^2 \left( \frac{1}{r}-\frac{1}{d} \right)$$

As $d>r$ then the total electrostatic energy $U$ must be greater than zero.

Last edited: Oct 5, 2013
6. Oct 5, 2013

### dauto

You would have to first create a negative plate at infinity on one side and a positive plate on infinity on the other side. They don't come for free and they require energy