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.