(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.

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(a) Define the capacitance of a capacitor. (b) State three factors on which the ...

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(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

μ F

and 6

μ 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,

C a p a c i t a n c e = \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} Q V

But

V = \frac{Q}{C}

∴ W = \frac{Q}{2} \times \frac{Q}{C} = \frac{Q^{2}}{2 C}

∴ Energy stored W = \frac{Q^{2}}{2 C}

(d)

\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}

\frac{1}{6} + \frac{1}{4} = \frac{10}{24}

C = 2.4 μ F

Q = \frac{24}{10} \times 100 \times 10^{- 6} = 240 μ C

240 \times 10^{- 6} C

(ii) p.d across 4

μ F, V_{4} = \frac{Q}{C}

V_{4} = \frac{240 \times 10^{- 6}}{4 \times 10^{- 6}} = 60 V

p.d across 6

μ F, V_{6} = \frac{Q}{C}

\frac{240 \times 10^{- 6}}{6 \times 10^{- 6}} = 40 V

(iii) Energy stored =

\frac{Q^{2}}{2 C}

\frac{(240 \times 10^{- 6})^{2}}{2 \times 2.4 \times 10^{- 6}}

1.2 \times 10^{- 2} J

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