Question

A charge of + 2.0 x 10^-8C is placed on the positive plate and a charge of - 1.0 x 10^-4C

Answer 1

The voltage across the capacitor is V = 12.5 V

Explanation:

Given data

• The charge on positive plate is q1 = + 2.0* 10 to the power -8 C
• The charge on the negative plate is q2 = -1.0*10 to the power - 8C
• The capacitance of capacitor is C = 1.2*10 to the power -9F

The expression for finding the net charge across the capacitor is ,

q= (q1-q2)/2

Substitute all the values in above equation.

q=(2.0*10 to the power -8 - (-1.0 *10 to the power -8))/2

q= 1.5* 10 to the power -8C

The expression for finding the voltage across the capacitor is

V = q/C

Substitute all the values in above equation.

V = 1.5 * 10 to the power -8/ 1.2 * 10 to the power -9

V = 12.5V

Thus the voltage across the capacitor is V = 12.5 V