# Capacitors

## Capacitor

### Definition

Capacitance is defined by \(Q=CV\).

Separated charges \(\pm Q\) at potentials \(\pm V\). Thus, we have \(V=V_+-V_-\) and then we get \(C=\frac{Q}{V}=\frac{Q}{V_+-V_-}=\frac{Q}{2V}\).

The work is given as \(W=\frac{1}{2}CV^2\).

How this is calculated: figure out how much work is required to add \(+Q\) to the one conductor and \(-Q\) to the other conductor.

We could also get it by figuring out by moving \(dq\) charges from \(-q\) to \(+q\). We have \(q=C\cdot V(q)\) hence \(V(q)=\frac{q}{C}\). Then, \(dW=V(q)dq\). Hence, \(W=\int_0^Q\frac{q}{C}dq=\frac{1}{2}\frac{Q^2}{C} = \frac{1}{2}CV^2\).

### Notes

#### Near Charge Example

Let us have a neutral conductor. If we bring a positive charge near the conductor, the electrons move close to the positive charge and then we get a depleted region of positive charge on the other side. Note, the total field is still zero since the dipole field cancels out with the applied positive charge’s field.

#### Cavity Charge Example

Consider a neutral conductor with a cavity. Let a positive charge be in the cavity. The electrons move into the inner surface and a depleted positive region appears at the surface. No electric field still in the conductor.

#### Parallel Plate Capacitor

Two large plates of length \(\ell\) and separated by \(d\ll \ell\). Opposite charges \(\pm Q\) on the plates. Recall that the electric field from a sheet charge to be \(\frac{\sigma}{\varepsilon_0}\). So, \(V=\frac{\sigma}{\varepsilon_0}d\). Thus, \(C=\frac{Q}{V}=\frac{\sigma\cdot A}{\frac{\sigma}{\varepsilon_0}d}=\frac{A\varepsilon_0}{d}\).