Kevin Nash
Last Activity: 10 Years ago
Sol. area = a = 20 cm^2 = 2 × 10^–2 m^2
d = separation = 1 mm = 10^–3 m
Ci = ε base 0 *2 * 10^-3/10^-3 = 2ε base 0 Cf = ε base 0 * 2 * 10^-3/2 * 10^-3 = ε base 0
q base I = 24 ε base 0 qf = 12 ε base 0 So, q flown out 12 ε base 0. Ie, q base I – q base f.
(a) So, q = 12 × 8.85 × 10^–12 = 106.2 × 10^–12 C = 1.06 × 10^–10 C
(b) Energy absorbed by battery during the process
= q × v = 1.06 × 10^–10 C × 12 = 12.7 × 10^–10 J
(c) Before the process
E base i = (1/2) × Ci × v^2 = (1/2) × 2 × 8.85 × 10^–12 × 144 = 12.7 × 10^–10 J
After the force
E base i = (1/2) × Cf × v^2 = (1/2) × 8.85 × 10^–12 × 144 = 6.35 × 10^–10 J
(d) Workdone = Force × Distance
= 1/2 q^2/ε base 0A = 1 * 10^3 = 1/2*12 * 12 * ε base 0 * ε base 0 * 10^-3/ε base 0 * 2 * 10^-3
(e) From (c) and (d) we have calculated, the energy loss by the separation of plates is equal to the work done by the man on plate. Hence no heat is produced in transformer.