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

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5/3 Volt/ μm and in the –ve x direction

5/3 Volt/ μm and in the +ve x direction

10/9 Volt/ μm and in the –ve x direction

10/9 Volt/ μm and in the +ve x direction.

answer is D.

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Detailed Solution

Vx=20x2−4 E=-dVdx=20(x2−4)2(2x−0)=160144=109

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