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`

Options

A. `(5)/(3) vol t//mu m` and in `+ x` direction.
B. `(10)/(9) V//mu m` and in neg `x` direction.
C. `(10)/(9) V//mu m` and in `+ x` direction
D. `(5)/(3) V//mu m` and inneg. `x` direction.

Answer

C

Explanation

Here, `V(x) = (20)/(x^(2) - 4) = 20 (x^(2) -4)^(-1)`.
`E = -(dV)/(dx) = -20 (-1) (x^(2) -4)^(-2) (2x)`
`= (40x)/((x^(2) - 4)^(2))`
At `x = 4 muC, E = (40xx4)/((4^(2) - 4)^(2)) = (160)/(144) = (10)/(9) V//mu m`
`vec(E)` is along the positive x-direction.