Capacitance and energy stored in a capacitor can be calculated or determined from a graph of charge against potential. Charge and discharge voltage and current graphs for capacitors.

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A capacitor is a gap in a circuit with space for charge on the 'plates' shown as the horizontal lines.

When a capacitor is charged, electrons on the lower plate repel electrons from the upper plate, which then move to the positive terminal of the supply.

The voltage applied \(V_{c}\) to charge the capacitor (circuit 1 below) is measured with a voltmeterand charge accumulated \(Q\) is measured by removing the charged capacitor from circuit 1 and connecting it to a coulomb meter (circuit 2).

By varying the charging voltage and measuring the associated charge\(Q\) a graph can be drawn.

The gradient of this graph is equal to the capacitance of the capacitor.

\[C= \frac{Q}{V}\]

And the area under the graph is the energy stored by the capacitor.

\[E= \frac{1}{2}QV\]

This can be combined with the equation for capacitance above to give two alternative equations for energy stored.

\(E= \frac{1}{2}\frac{Q^2}{C}\) and

\[E= \frac{1}{2}C{V^2}\]

- Question
A \(2200 \mu F\) capacitor is charged up with a \(1.5 V\) cell.

(a) What charge is stored?

(b) What energy is stored?

\[C= 2200\mu F\]

\[=2200 \times {10^{ - 6}}F\]

\[V = 1.5 V\]

\[Q=?\]

a) \(Q=CV\)

\[=2200 \times {10^{ - 6}} \times 1.5\]

\[=0.0033C\]

b) \(E= \frac{1}{2}C{V^2}\)

\[= \frac{1}{2} \times 2200 \times {10^{-6}} \times {1.5^2}\]

\[= 0.00248J\]

- The S.I. unit of charge is the Coulomb, \(C\)
- The S.I. unit of capacitance is the Farad, \(F\)
- A capacitor of capacitance 1 Farad will store 1 Coulomb of charge with the potenial difference across it is 1 volt