#### The potential at a point  x ( measured in ) due to some charges situated on the x-axis is given byThe electric field E at x = 4 is given by  Option 1) Option 2) Option 3) Option 4)

As we learnt in

Relation between field and potential -

- wherein

Given Potential

$Electric\: \: field\: \: E= \frac{-dV}{dx}= \frac{-d}{dx}\left ( \frac{20}{x^{2}-4} \right )= \frac{40x}{\left ( x^{2} -4\right )^{2}}$

$At \: \: x= 4\mu m$

$\therefore \: \: \: \: E= \frac{40\times 4}{\left [ 16-4 \right ]^{2}}= \frac{160}{144}= \frac{10}{9}V/\mu m$

Option 1)

Correct

Option 2)

Incorrect

Option 3)

Incorrect

Option 4)

Incorrect