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The potential at a point ( measured in m ) due to some charges situated on the x - axis is given by v $(x) = 20 / (x^{2} - 4)$ volts .The electric field E at $x = 4 μ$ m is given by:

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5/3 volt/ $μ$ m and in the + ve x- direction

10/9 volt/ $μ$ m and in the +ve x- direction

5/3 volt/ $μ$ m and in the – ve x- direction

10/9 volt/ $μ$ m and in the -ve x- direction

answer is A.

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Detailed Solution

$\begin{array}{l} E = \frac{- dV}{dx} = \frac{- d}{dx} [\frac{20}{x^{2} - 4}] = - 20 (- 1) {(x^{2} - 4)}^{- 2} (2 x) \\ = \frac{40 \times 4}{{(x^{2} - 4)}^{2}} = \frac{40}{12 \times 12} = \frac{10}{9} V / μm along +ve x direction \end{array}$

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