C = \frac{Q}{V}

For a solid conductive sphere of radius $R$ and charge $Q$ :

C = \frac{R}{k}

Capacitors

Total charge of a capacitor is ALWAYS zero.

In an ideal capacitor, outside, we have $E = 0$ , and between the plates:

⟹ E = \frac{σ}{ε} = \frac{Q}{ε A} Δ V = \frac{Q d}{ε A} C = \frac{ε A}{d} U = \frac{1}{2} \frac{Q ^{2}}{C} = \frac{1}{2} C Δ V^{2} = \frac{1}{2} Q Δ V

U = \int_{V} \frac{1}{2} ε E^{2} d V

If we charge a capacitor and then vary $C$ , we have two cases:

Constant $Δ V$ :

Δ U = \frac{1}{2} Δ C Δ V^{2}

Constant $q$ :

Δ U = \frac{1}{2} Δ (\frac{1}{C}) Q^{2}

If we place capacitors in parallel, we get:

C_{e q} = Σ C_{i}

And in series:

\frac{1}{C _{e q}} = Σ \frac{1}{C _{i}}

Engineering Notes