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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. `5/3 V/mu m` and in the -ve x deirectionB. `5/3 V/mu m` and in the +ve x deirectionC. `10/9 V/mu m` and in the -ve x deirectionD. `10/9 V/mu m` and in the +ve x deirection |
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Answer» Correct Answer - D `V(x)=20/(x^(2)-4)` The electric field `vec(E)` along all such lines where potential is a function of position So `(delV)/(delx)=((x^(2)-4)(0)-20(2x))/((x^(2)-4)^(2))=(-40x)/((x^(2)-4)^(2))` `vec(E)=(40x)/((x^(2)-4)^(2))(hat(i)) At x=4 mu m` `vec(E)=(40(4))/((4^(2)-4)^(2))=160/144=(10/9 hat(i))V//m` |
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