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JEE PHYSICS MCQ: ELECTRIC POTENTIAL AND CAPACITANCE

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The potential at a point distance xx (measured in μm\mu \mathrm{m}) due to some charges situated on the xx axis is given by V(x)=20x2−5VV(x)=\frac{20}{x^2-5} \mathrm{V}. The electric field at x=5μmx=5 \mu \mathrm{m} is given by:
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Correct option: A
510Vμm−1\frac{5}{10} \mathrm{V \mu} \mathrm{m}^{-1} and in positive x{x}-direction
Option A (510Vμm−1\frac{5}{10} \mathrm{V \mu} \mathrm{m}^{-1} and in positive x{x}-direction) is correct because it accurately calculates the electric field at x=5μmx=5 \mu \mathrm{m} by differentiating the given potential function.

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