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ExamsJEE AdvancedPhysics

A plastic disk is uniformly charged on one side with surface charge density sigma. Three quadrants of the disk are removed, leaving only one quadrant of inner radius r and outer radius R. With V = 0 at infinity, the electric potential at point P located at the center of the original disk (on the axis, at the center) due to the remaining quadrant is:

  1. sigma/(2*epsilon₀) * [(r² + R)^(1/2) - r]
  2. sigma/(2*epsilon₀) * [R - r]
  3. sigma/(8*epsilon₀) * [(r² + R²)^(1/2) - r]
  4. None of these

Correct answer: None of these

Solution

The potential at the center of a uniformly charged annular disk (inner radius r, outer radius R) is sigma/(2*epsilon₀)*(sqrt(r²+R²)-r). Since only one quadrant remains, the potential is 1/4 of this value: sigma/(8*epsilon₀)*(sqrt(r²+R²)-r). Option C has a typo (r²+R² under root is correct) but the prefactor matches — re-examining: option C is sigma/(8*epsilon₀)*[(r²+R²)^(1/2)-r] which is correct.

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