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ExamsJEE MainPhysics

The electric potential varies with position x (in metres) as V = (5x² + 10x - 9) volt. Find the magnitude of the electric field at x = 1 m.

  1. 20 V/m
  2. 6 V/m
  3. 11 V/m
  4. +23 V/m

Correct answer: 20 V/m

Solution

E = -dV/dx = -(10x + 10). At x = 1, E = -(10+10) = -20 V/m, magnitude 20 V/m.

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