# A Parallel-plate Capacitor Has Plate Area 25⋅0 Cm2 and a Separation of 2⋅00 Mm Between the Plates. the Capacitor is Connected to a Battery of 12⋅0 V. (A) Find the Charge on the Capacitor. - Physics

Sum

A parallel-plate capacitor has plate area 25⋅0 cm2 and a separation of 2⋅00 mm between the plates. The capacitor is connected to a battery of 12⋅0 V. (a) Find the charge on the capacitor. (b) The plate separation is decreased to 1⋅00 mm. Find the extra charge given by the battery to the positive plate.

#### Solution

Given :

Area of the plate , A = 25  "cm"^2 = 25 xx 10^-4  "m"^2

Separation between the plates, d = 2  "mm" = 2 xx 10^-3  "m"

Potential difference between the plates, V = 12 V

The capacitance of the given capacitor is given by

C = (∈_0A)/d

= ((8.85 xx 10^-12) xx (25 xx 10^-4))/((2 xx 10^-3))

= 11.06 xx 10^-12  "F"

(a) Charge on the capacitor is given by
Q = CV

= 11.06 xx 10^-12 xx 12

= 1.33 xx 10^-10 "C"

(b) When the separation between the plates is decreased to 1 mm , the capacitance C^' can be calculated as :

C^' = (∈_0A)/d

= ((8.85 xx 10^-12) xx (25 xx 10^-4))/(1 xx 10^-3)

= 22.12 xx 10^-12  "F"

Charge on the capacitor is given by

Q^' = C^'V

= 22.12 xx 10^-12 xx 12

= 2.65 xx 10^-10 "C"

Extra Charge

= (2.65 xx 10^-10 - 1.32 xx 10^-10) C

= 1.33 xx 10^-10 "C"

Concept: Capacitors and Capacitance
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 9 Capacitors
Q 5 | Page 165