Two large, parallel conducting plates X and Y, kept close to each other, are given charges Q1β and Q2β(Q1β>Q2β). The four surfaces of the plates are A, B, C and D, as shown in figure. Then :
This question has multiple correct options
A
The charge on A is 21β(Q1β+Q2β).
B
The charge on B is 21β(Q1ββQ2β).
C
The charge on C is -21β(Q1ββQ2β).
D
The charge on D is 21β(Q1β+Q2β)
Hard
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Updated on : 2022-09-05
Solution
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Correct options are A) , B) , C) and D)
Let charge for surface A =Q1ββq, for surface B =q, for surface C =Q2ββqβ² and for surface D =qβ². As the field inside X plate is zero, 2AΟ΅0β1β(Q1ββqβqβQ2β+qβ²βqβ²)=0 β2q=Q2ββQ1ββq=21β(Q1ββQ2β) similarly for Y plate, 2AΟ΅0β1β(qβ²βQ2β+qβ²βqβQ1β+q)=0 or 2qβ²=Q1β+Q2ββqβ²=21β(Q1β+Q2β) Thus, charge on A is Q1ββq=Q1ββ21β(Q1ββQ2β)=21β(Q1β+Q2β). charge on B is q=21β(Q1ββQ2β) charge on C is Q2ββqβ²=Q2ββ21β(Q1β+Q2β)=β21β(Q1ββQ2β) Charge on D is qβ²=21β(Q1β+Q2β)