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Question 31453
JEE Physics · Hard

Electrostatics: Electric Potential MCQ

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Electric PotentialHardQuestion 31453

Question

Potential on axis of a uniformly charged disc (surface charge density σ, radius R) at distance x:
  1. A
    σ/(2ε₀)(√(R²+x²)−x)
    Correct
  2. B
    σx/(2ε₀)
  3. C
    kσR/x
  4. D
    σR/(ε₀x)

Correct answer

σ/(2ε₀)(√(R²+x²)−x)

Explanation

V = σ/(2ε₀)[√(R²+x²) − x]. For x >> R: V → kQ/x (point charge limit). At x=0 (centre): V = σR/(2ε₀).

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