Question

A parallel plate capacitor with air as dielectric between the plates has a separation of 5mm and plate area 50cm2 is connected to a 200v source. calculate the charge on plates....

Answer 1

Explanation:

C=

d

ϵ

o

A

According to the values given C=1.8×10

−11

F

Now Q=CV

Hence charge on the positive plate will be 1.8×10

−9

C

(a) Dielectric constant, k = 6

C=1.8×10

−11

F

new capacitance, C

′

=kC=108 pF

V=100volts

q

′

=V×C

′

=1.08×10

−8

C

V remains the same.

(b) C

′

=kC=6×1.8×10

−11

=108pF

q remains the same.

q=1.8×10

−9

C

V

′

=q/C

′

=16.67 V