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Question: Answered & Verified by Expert
The sketch below shows cross-sections of the equipotential surface between two charged conductors that are shown in solid black. Some points on the equipotential surfaces near the conductors are marked as A, B, C.......... The arrangement lies in the air. Take ϵ0=8.85×10-12 C2 N-1 m-2
The surface charge density of the plate is equal to

PhysicsElectrostaticsNEET
Options:
  • A 8.85×10-10 C m-2
  • B -8.85×10-10 C m-2
  • C 17.7×10-10 C m-2
  • D -17.7×10-10 C m-2
Solution:
1314 Upvotes Verified Answer
The correct answer is: 8.85×10-10 C m-2

E=40-100.3=100 V m-1

(near the plate the electric field has to be uniform

  it is almost due to the plate).

For conducting plate

E=σε0     σ=ε0E

Therefore     σ=8.85×10-12×100

= 8.85×10-10 C m-2

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