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The potential at a point x ( measured in...

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) vol t`

A

` (10//9)` volt //mumand in the + ve `x` direction

B

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

C

`( 5 //3 ) volt //` mum and in the `+ ve 9 x` direction

D

`(10 //9 ) volt //`volt //mum in the - ve `x` direction

Text Solution

Verified by Experts

The correct Answer is:
A

` v (x) = (20)/(x^2 -4), E = - (dv)/(dx) =- d/(dx) ((20)/(x^2 -4)) = (20)/((x^2 - 4)^@) (2 x)`
`E at x = 4 mu m, ((20)(2 xx 4))/(144) = (10)/9` volt// mu m
Also as `x` increases , `V` decreases So , `E` is along `+ ve x -` axis .
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