Consider the parallel plate capacitor connected to the battery as shown in the figure
The capacitance “C” of the parallel plate capacitor is given as
C = εοA/d
WhereA = Area = 10 square centimeter = 10 × 10−4 meterεο = Permittivity of free space = 8.85 × 10-12d = distance between the parallel plate capacitor = 5mm = 5 × 10−3 meter
q = Charge stored in the capacitor = 20 pC = 20 × 10−12
C = (8.85 × 10-12 × 10 × 10−4) ⁄ 5 × 10−3
C = 1.77 × 10−12 F
The self-capacitance of a conductor is defined by
C = q ⁄ V
Hence the voltage across the capacitor
V = q ⁄ V = (20 × 10−12) ⁄ (1.77 × 10−12)
V = 11. 3 Volts
EXAMIANS
