Question

Find the charge on a capacitor of plate area 25 cm and distance 2.0mm connected to 12V battery?

Answer 1

Answer:

1.33*10^-10 C

EXPLATION

Given area A = 25 cm = 25 * 10^-4

voltage v = 12 v

separation between the plates d=2mm

Q= cv

• the capacitance of the parallel palte capacitor is given by c= apcilon A / d
• where,
• c is the capacitance of the parallel plate capacitor
• apcillon is the permittivity of the free space,

• c = 8.85 * 10^-12 * 25 *10^-4 / 2*10^-3
• = 11.06*10^-12 F

1. Q1 = CV

• 11.06*10^-12 * 12
• =1.33*10^-10

b) d is decreased to 1.0 mm we have to calculate the extra charge given by battery to positive plate

• c= apcillonA / d
• = 8.85*10^-12 * 25*10^-4/ 1×10^ -3
• = 22.12 × 10^-12 F

Q2 = 22.12×10^-12× 12

= 2.652 × 10^ -10 c

extra charge is = Q2 - Q1

(2.652 -1.33) ×10^-10

=1.33×10^-10

charge on the capacitor when d = 2mm is 1.33×10^10 C

when d is decreased at 1.0 mm the extra charge given by battery is = 1.33×10^-10