Deepak Patra
Last Activity: 10 Years ago
Sol. Capacitance = 100 μF = 10^–4 F
P.d = 30 V
(a) q = CV = 10^–4 × 50 = 5 × 10^–3 c = 5 mc
Dielectric constant = 2.5
(b) New C = C’ = 2.5 × C = 2.5 × 10^–4 F
New p.d = q/c^1 [∴ ‘q’ re,aoms sa,e after disconnection of battery]
= 5 * 10^-3/2.5 * 10^-4 = 20 V
(c) In the absence of the dielectric slab, the charge that must have produced
C × V = 10^–4 × 20 = 2 × 10^–3 c = 2 mc
(d) Charge induced at a surface of the dielectric slab
= q (1 –1/k) (where k = dielectric constant, q = charge of plate)
= 5 × 10^-3 (1- ½.5) 5 * 10^-3 * 3/5 = 3 * 10^-3 = 3 mc.