Question

A charge of 20 µC is placed on the positive plate of an isolated parallel-plate capacitor of capacitance 10 µF. Calculate the potential difference developed between the plates.

Answer 1

Given :
Capacitance of the isolated capacitor = 10 µF

Charge on the positive plate = 20 µC

Effective charge on the capacitor = `(20-0)/2 = 10  "uC"`

The potential difference between the plates of the capacitor is given by `V = Q/C`

`therefore "Potential difference" = (10  "uC")/(10  "uF")`= `1 "V"`