(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 the energy W stored in a charged capacitor of capacitance C carrying a charge Q on either plate.
(d) Two capacitors of capacitance 4\(\mu F\) and 6\(\mu F\) are connected in series to a 100V d.c supply. Draw the circuit diagram and calculate the (i) charge on either plate of each capacitor (ii) p.d. across each capacitor; (iii) energy of the combined capacitors.
Explanation
(a) Capacitance of a capacitor is the ratio of the charge Q on either plate of the capacitor to the p.d in volts between the plates. Mathematically,
\(Capacitance = \frac{Q}{V}\)
Q - Charge ; V - voltage.
(b) Three factors on which the capacitance of a parallel - plate capacitor depends on are: (1) Common area of the plates (2) Dielectric between the plates (3) Separation between the plates.
(c) Work done W = QV
Average work done = \(\frac{1}{2} QV\)
But \(V = \frac{Q}{C}\)
\(\therefore W = \frac{Q}{2} \times \frac{Q}{C} = \frac{Q^{2}}{2C}\)
\(\therefore \text{Energy stored W = } \frac{Q^{2}}{2C}\)
(d) 
\(\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}\)
= \(\frac{1}{6} + \frac{1}{4} = \frac{10}{24}\)
\(C = 2.4 \mu F\)
\(Q = \frac{24}{10} \times 100 \times 10^{-6} = 240 \mu C\)
= \(240 \times 10^{-6} C\)
(ii) p.d across 4\(\mu F, V_{4} = \frac{Q}{C}\)
\(V_{4} = \frac{240 \times 10^{-6}}{4 \times 10^{-6}} = 60V\)
p.d across 6\(\mu F, V_{6} = \frac{Q}{C}\)
= \(\frac{240 \times 10^{-6}}{6 \times 10^{-6}} = 40V\)
(iii) Energy stored = \(\frac{Q^{2}}{2C}\)
= \(\frac{(240 \times 10^{-6})^{2}}{2 \times 2.4 \times 10^{-6}}\)
= \(1.2 \times 10^{-2} J\)