A parallel-plate capacitor has plate area 25.0 cm2 and separation of 2.00 mm between the plates. The capacitor is connected to a battery of 12.0V.
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.
a) Given Area
Voltage V=12V
Separation between the plates d=2mm
Formula used :
Charge of the capacitor can be calculated as
The capacitance of the parallel plate capacitor is given by
Where,
C is the capacitance of the parallel plate capacitor
D is the separation between the capacitor plates
A is the area of a circular plate capacitor
ε₀ is the permittivity of the free space, ε₀=8.85×10-12 F/m
b) d is decreased to 1.0mm. We have to calculate the extra charge given by the battery to the positive plate.
Extra charge is =
Charge on the capacitor when d = 2mm is = 
When d is decreased to 1.00 mm the extra charge given by the battery is =