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The potential at a point x (measured in μm) due to two charges situated on the x-axis is given by V(x) = 20/(x² − 4) volt. The electric field E at x = 4 μm is given by

1. (10/9) volt/μ m and in the +ve x direction
2. (5/3) volt/μ m and in the −ve x direction
3. (5/3) volt/μ m and in the +ve x direction
4. (10/9) volt/μ m and in the −ve x direction

Correct answer: (10/9) volt/μ m and in the +ve x direction

Solution

V = 20/(x^2-4) -> dV/dx = -40x/(x^2-4)^2. At x=4: dV/dx = -160/144 = -10/9. So E = -dV/dx = +10/9 V/um, directed along +x.

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