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Question

Two large, parallel conducting plates X and Y, kept close to each other, are given charges Q1 and Q2(Q1>Q2). The four surfaces of the plates are A, B, C and D, as shown in figure. Then :
142406_9d8619e90c864a1cb94d9e9ca9d4c171.png
  1. The charge on B is 12(Q1Q2).
  2. The charge on A is 12(Q1+Q2).
  3. The charge on C is -12(Q1Q2).
  4. The charge on D is 12(Q1+Q2)

A
The charge on C is -12(Q1Q2).
B
The charge on D is 12(Q1+Q2)
C
The charge on A is 12(Q1+Q2).
D
The charge on B is 12(Q1Q2).
Solution
Verified by Toppr

Let charge for surface A =Q1q, for surface B =q, for surface C =Q2q and for surface D =q.
As the field inside X plate is zero, 12Aϵ0(Q1qqQ2+qq)=0
2q=Q2Q1q=12(Q1Q2)
similarly for Y plate, 12Aϵ0(qQ2+qqQ1+q)=0
or 2q=Q1+Q2q=12(Q1+Q2)
Thus, charge on A is Q1q=Q112(Q1Q2)=12(Q1+Q2).
charge on B is q=12(Q1Q2)
charge on C is Q2q=Q212(Q1+Q2)=12(Q1Q2)
Charge on D is q=12(Q1+Q2)

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142406_9d8619e90c864a1cb94d9e9ca9d4c171.png
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