# A charge of 1 μC is given to one plate of a parallel-plate capacitor of capacitance 0.1 μF and a charge of 2 μC is given to the other plate. Find the potential difference developed between the plates.

Explanation:

Given,

Charge on plate 1, Q1 = 1 μC

Charge on plate 2, Q2 = 2 μC

Capacitance of the capacitor, C= 0.1 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,

So, the charge, Q by substituting the given values, is

Hence from eqn.1, the potential difference

Hence the potential difference developed in between the plates is 5V.

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