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.
plate are A=25 cm2
=25×10−4 m2,
separation d=2 mm=2×10−3 m
potential v=12 V
(a) Capacitance C=∈0 Ad
=8.88×10−12×25×10−42×10−3
=11.06×10−12 F
charge q1=CV
=11.06×10−12×12
=1.32×10−10 C
(b) d=1 m m=1×10−8 m
So, C=∈0 Ad
=8.88×2×10−31×10−12
=22.06×10−12 F
q2=CV
=22.06′×10−12×12
=2.65×10−10 C
Extra charge
=(2.65×10−10−1.32×10−10) C
=1.33×10−10 C