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# 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 :The charge on B is 12(Q1−Q2).The charge on A is 12(Q1+Q2).The charge on C is -12(Q1−Q2).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
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#### Let charge for surface A =Q1−q, for surface B =q, for surface C =Q2−q′ and for surface D =q′.As the field inside X plate is zero, 12Aϵ0(Q1−q−q−Q2+q′−q′)=0−2q=Q2−Q1⇒q=12(Q1−Q2)similarly for Y plate, 12Aϵ0(q′−Q2+q′−q−Q1+q)=0or 2q′=Q1+Q2⇒q′=12(Q1+Q2)Thus, charge on A is Q1−q=Q1−12(Q1−Q2)=12(Q1+Q2).charge on B is q=12(Q1−Q2)charge on C is Q2−q′=Q2−12(Q1+Q2)=−12(Q1−Q2)Charge on D is q′=12(Q1+Q2)

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