(a) Define the capacitance of a capacitor. (b) State three factors on which the capacitance of a parallel plate capacitor depends. (c) Derive a formula for ...

(a) Capacitance of a capacitor

The capacitance of a capacitor is the ratio of the magnitude of the charge \(Q\) on either plate to the potential difference \(V\) between the plates:

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

where \(Q\) is the charge (coulomb) and \(V\) is the p.d. (volt). Its SI unit is the farad (F).

(b) Factors affecting the capacitance of a parallel-plate capacitor

1. The common (overlapping) area \(A\) of the plates, with \(C \propto A\).
2. The distance of separation \(d\) between the plates, with \(C \propto \dfrac{1}{d}\).
3. The permittivity \(\varepsilon\) (nature) of the dielectric medium between the plates, with \(C \propto \varepsilon\).

These combine as \(C = \dfrac{\varepsilon A}{d}\).

(c) Energy \(W\) stored in a charged capacitor

When the capacitor already holds a charge \(q\), the p.d. across it is \(v = \dfrac{q}{C}\). To move a further small charge \(dq\) onto the plates the work done is

\[ dW = v\,dq = \frac{q}{C}\,dq \]

The total work to charge the capacitor from \(0\) to the final charge \(Q\) is therefore

\[ W = \int_{0}^{Q} \frac{q}{C}\,dq = \frac{1}{C}\cdot\frac{Q^{2}}{2} = \frac{Q^{2}}{2C} \]

Using \(Q = CV\), the same result may be written as

\[ W = \frac{Q^{2}}{2C} = \frac{1}{2}QV = \frac{1}{2}CV^{2} \]

(d) Two capacitors 4\(\mu\text{F}\) and 6\(\mu\text{F}\) in series across 100 V

Circuit diagram: the two capacitors are joined end to end (in series) in a single loop with the 100 V d.c. supply.

Effective (combined) capacitance of capacitors in series:

\[ \frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{1}{4} + \frac{1}{6} = \frac{3+2}{12} = \frac{5}{12} \] \[ \Rightarrow C = \frac{12}{5} = 2.4\,\mu\text{F} \]

(i) Charge on either plate of each capacitor. In a series circuit the charge is the same on every capacitor and equals the charge supplied to the combination:

\[ Q = CV = 2.4\times10^{-6} \times 100 = 2.4\times10^{-4}\,\text{C} = 240\,\mu\text{C} \]

Hence each capacitor carries \(Q = 240\,\mu\text{C}\) on either plate.

(ii) P.d. across each capacitor.

\[ V_{4} = \frac{Q}{C_{4}} = \frac{2.4\times10^{-4}}{4\times10^{-6}} = 60\,\text{V} \] \[ V_{6} = \frac{Q}{C_{6}} = \frac{2.4\times10^{-4}}{6\times10^{-6}} = 40\,\text{V} \]

Check: \(V_{4} + V_{6} = 60 + 40 = 100\,\text{V}\), which equals the supply voltage.

(iii) Energy of the combined capacitors.

\[ W = \frac{Q^{2}}{2C} = \frac{(2.4\times10^{-4})^{2}}{2 \times 2.4\times10^{-6}} = \frac{5.76\times10^{-8}}{4.8\times10^{-6}} = 1.2\times10^{-2}\,\text{J} \]

Equivalently \(W = \tfrac{1}{2}QV = \tfrac{1}{2}(2.4\times10^{-4})(100) = 0.012\,\text{J}\).

Therefore the energy stored in the combination is \(W = 1.2\times10^{-2}\,\text{J}\).