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.

Question

Philoid · Accepted Answer

Explanation:

Given,

Charge on plate 1, Q₁ = 20 μC

Charge on plate 2, Q₂ = 0C Since no charge is given to the other plate and the setup is isolated)

Capacitance of the capacitor, C= 10 μF

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 question, the charges on the inner plates , according to above formulas,

Hence from eqn.1, the potential difference

Hence the potential difference developed in between the plates is 1V.