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, \(W=\int_0^Q\frac{q}{C}dq=\frac{1}{2}\frac{Q^2}{C} = \frac{1}{2}CV^2\).


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)

Author: Christian Cunningham

Created: 2023-06-25 Sun 02:32