The potential at a point  x ( measured in \mu m) due to some charges situated on the x-axis is given by

V(x)= 20/(x^{2}-4)volt

The electric field E at x = 4 \mu m is given by 

  • Option 1)

    (10/9)volt /\mu m \: and\: in \: the\: +ve \: x \: direction

  • Option 2)

    (5/3)volt /\mu m \: and\: in \: the\: -ve \: x \: direction

  • Option 3)

    (5/3)volt /\mu m \: and\: in \: the\: +ve \: x \: direction

  • Option 4)

    (10/9)volt /\mu m \: in \: the\: -ve \: x \: direction

 

Answers (2)

As we learnt in

Relation between field and potential -

E=\frac{-dv}{dr}

- wherein

\frac{dv}{dr} -   Potential gradient.

 

 Given Potential

V(x)= 20/(x^{2}-4)volt

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

Positive sign indicate E is +ve x direction


Option 1)

(10/9)volt /\mu m \: and\: in \: the\: +ve \: x \: direction

Correct

Option 2)

(5/3)volt /\mu m \: and\: in \: the\: -ve \: x \: direction

Incorrect

Option 3)

(5/3)volt /\mu m \: and\: in \: the\: +ve \: x \: direction

Incorrect

Option 4)

(10/9)volt /\mu m \: in \: the\: -ve \: x \: direction

Incorrect

N neha

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