A charge of +2.0 × 10^–8 C is placed on the positive plate and a charge of –1.0 × 10^–8 C on the negative plate of a parallel-plate capacitor of capacitance 1.2 × 10^–3 μF. Calculate the potential difference developed between the plates.

Given,

Charge on plate 1, Q₁ = +2.0 × 10^–8C

Charge on plate 2, Q₂ = –1.0 × 10^–8C

Capacitance of the capacitor, C= 1.2 × 10^–3 μF= 1.2 × 10^–9 F

Formula used

We know that for a capacitor with net charge, Q and capacitance, C, the Potential difference deceloped in between the plates, V is,

The charges on the inner plates of the capacitor with plates having charges Q₁ and Q₂ is,

Note: Charges on the outer plates of the capacitor with plates having charges Q1 and Q2 is,

In the given example, the plates has individual charges Q₁ and Q₂. But when they placed as a capacitor, their charges re-arrange and equal and opposite charges will be distributed in each plates. The total net charge, Q_net on the inner sides of each plates will be

Substituting the values,

Hence the inner side of each plates will have a charge of ±1.5 × 10^–8C.

Hence from eqn.1, Potential difference, V is

Or,

Hence the potential difference developed between the plates is 12.5V