A capacitor having a capacitance of 100μF is charged to a potential d

1.	A capacitor having a capacitance of 100μF is charged to a potential difference of 50V. (a) What is the magnitude of the charge on each plate? (b) The charging battery is disconnected and a dielectric of dielectric constant 2.5 is inserted. Calculate the new potential difference between the plates, (c) What charge would have produced this potential .difference in absence of the dielectric slab, (d) Find the charge induced at a surface of the dielectric slab.
Answer» Capacitance = 100μF = 10^–4F P.d = 30V (a) q = CV = 10^–4 × 50 = 5 × 10^–3 c = 5mc Dielectric constant = 2.5 (b) New C = C' = 2.5 × C = 2.5 × 10^–4F New p.d = q/c₁ [∴’q’ remains same after disconnection of battery] = (5 x 10^-3)/(2.5 x 10^-4) = 20V. (c) In the absence of the dielectric slab, the charge that must have produced C × V = 10^–4 × 20 = 2 × 10^–3 c = 2mc (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 - 1/2.5) = 5 x 10^-3 x 3/5 = 3 x 10^-3 = 3mc.

A capacitor having a capacitance of 100μF is charged to a potential difference of 50V. (a) What is the magnitude of the charge on each plate? (b) The charging battery is disconnected and a dielectric of dielectric constant 2.5 is inserted. Calculate the new potential difference between the plates, (c) What charge would have produced this potential .difference in absence of the dielectric slab, (d) Find the charge induced at a surface of the dielectric slab.

Answer»

Capacitance = 100μF = 10^–4F

P.d = 30V

(a) q = CV = 10^–4 × 50 = 5 × 10^–3 c = 5mc

Dielectric constant = 2.5

(b) New C = C' = 2.5 × C = 2.5 × 10^–4F

New p.d = q/c₁ [∴’q’ remains same after disconnection of battery]

= (5 x 10^-3)/(2.5 x 10^-4) = 20V.

C × V = 10^–4 × 20 = 2 × 10^–3 c = 2mc

(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 - 1/2.5) = 5 x 10^-3 x 3/5 = 3 x 10^-3 = 3mc.