# 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, $$C=\frac{Q}{V}=\frac{\sigma\cdot A}{\frac{\sigma}{\varepsilon_0}d}=\frac{A\varepsilon_0}{d}$$.

## Method of Images (Uniqueness Theorem)

Created: 2023-06-25 Sun 02:32

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