(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)