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Question

Four large identical metal plates are placed as shown in the figure. Plate 2 is given a charge Q. All other plates are neutral. Now Plate 1 and 4 are earthed. Area of each plates is A. The charge appearing on the right side of plate 3 is


A
Q4
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B
Q2
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C
Q
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D
Q8
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Solution

The correct option is A Q4
Let us assume x be the charge on the left side of plate 2.

Now, due to induction , charge distribution of all plates is as shown in the figure.

Charge on the right face of the plate 2 be (Qx).

Since, the left face of the plate 3 is opposite to the right face of plate 2 the charge on this face will be (Qx).

Similarly charges on the other faces can be determined for the plates 3 and 4 as shown in figure.


Assume, charges on the left face of plate 1 as y.

Since, the left face of the plate 1 and right face of the plate 4 are earthed, so the potential for both the faces will be same.i.e.,
V1=V4V4V1=0

Hence, the charges on the right face of plate 4 will be equal to the charge on the left face of plate 1 which is y

Let,

V2V1=xdAεo, is the potential difference across the plates 1 and 2.

V3V2=(Qx)2dAεo, is the potential difference across the plates 2 and 3.

V4V3=(Qx)dAεo, is the potential difference across plate 3 and 4.

Substituting the values we get,

xdAεo+(Qx)2dAεo+(Qx)dAεo=0

xd+Q(2d)x(2d)+Qdxd=0

3Q=4x

x=3Q4

Since the charge appearing on the right side of plate 3 is (Qx)

(Qx)=Q3Q4=Q4

Hence, option (b) is the correct answer.

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