Answer :

Given,

The charge given to the middle plate Q) is 1.0 μC

The capacitance between the plates, C is 50 nF=50× 10^{–3} μF

__Formula used__

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,

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

a)

The charge given to the plate Q will be distributed equally on the either sides of plates as shown in figure. Hence the upper and lower sides of plate Q will be charged to +0.5 μC.

Here, we get two capacitors namingly as P-Q and Q-R.

In capacitor P-Q, the upper plate is neither connected to any battery nor given any charges. So the total charge on the plate is 0*C*. But when it is made into a capacitor plate, a charge is induced in it from the plate Q. The amount of the charge can be calculated from the eqn.2,

Which is,

Where Q_{1} is the charge on one plate Q= 1.0 μ*C*

And Q_{2} is the charge on plate P = 0*C*

Hence by substituting in the above equation, we get,

Hence the inner surfaces get a charge of ±0.5μC on each plates. Since the plate Q is positively charged, Plate P will get -0.5μC charge.

But we know that the net charge on plate P is zero. Hence to nutralise the inner surface charge, the outer surface will get a charge of +0.5μC

b) From the above calculation, we found that the inner surfaces of the capacitor P-Q has a charge of ±0.5μC.

The potential difference between the plates can be found by the eqn.1, as

Hence the potential difference between the upper and middle plates of the arrangements is 10V.

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