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// mum ` and in the -ve X-direction
B. `5/3 V // mum`and in the +ve X-direction
C. `10/9 V//mum` and in the –ve X-direction
D. `10/9 V// mum` and in the +ve X-direction

Answer

D

Explanation

`E= (delta V)/(delta x) hat i-(delta V)/(delta y) hat j-(delta V)/(delta z) hat k implies E_x= - (delta v)(delta x)=-d/(dx) [20/(x^2-4)]=(40x)/(x^2-4)^2`
`implies E_x at x=4 mu m=10/9 V mu m`