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))` volts. The electric field E at `x=4 mu m` is given by

Options

A. `5/3 V/mu m` and in the -ve x deirection
B. `5/3 V/mu m` and in the +ve x deirection
C. `10/9 V/mu m` and in the -ve x deirection
D. `10/9 V/mu m` and in the +ve x deirection

Answer

D

Explanation

`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`