Q. 325.0( 2 Votes )

# 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.

Answer :

__Explanation:__

Given,

Charge on plate 1, Q_{1} = 1 μ*C*

Charge on plate 2, Q_{2} = 2 μ*C*

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 Q_{1} and Q_{2} 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.

Rate this question :

A parallel-plate capacitor has plate area 20 cm^{2}, plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ resistor. Find the energy for the capacitor 8.9 μs after the connections are made.

How many time constants will elapse before the energy stored in the capacitor reaches half of its equilibrium value in a charging RC circuit?

HC Verma - Concepts of Physics Part 2