The correct option is
A Q4Let 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
(Q−x).
Since, the left face of the plate
3 is opposite to the right face of plate
2 the charge on this face will be
−(Q−x).
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=V4⇒V4−V1=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,
V2−V1=−xdAεo, is the potential difference across the plates
1 and
2.
V3−V2=(Q−x)2dAεo, is the potential difference across the plates
2 and
3.
V4−V3=(Q−x)dAεo, is the potential difference across plate
3 and
4.
Substituting the values we get,
−xdAεo+(Q−x)2dAεo+(Q−x)dAεo=0
⇒−xd+Q(2d)−x(2d)+Qd−xd=0
⇒3Q=4x
⇒x=3Q4
Since the charge appearing on the right side of plate
3 is
(Q−x)
∴(Q−x)=Q−3Q4=Q4
Hence, option (b) is the correct answer.