InterviewSolution
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InterviewSolution
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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` directionB. `( 5 //3)` volt //mu m and in the - ve `x` directionC. `( 5 //3 ) volt //` mum and in the `+ ve 9 x` directionD. `(10 //9 ) volt //`volt //mum in the - ve `x` direction |
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Answer» Correct Answer - 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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