Eluipotential surfcaces are shown in figure. Then the electric field strength will be

Options

A. `100 Vm^(-1)` along X-axis
B. `100 Vm^(-1)` along Y-axis
C. `200 Vm^(-1)` at an angle `120^(@)` with X-axis
D. `50 Vm^(-1)` at an angle `120^(@)` with X-axis

Answer

C

Explanation

Using `dv = - vecE. vecdr`
`rArr DeltaV = - E.Deltar cos theta`
`rArr = (-DeltaV)/(Deltar cos theta)`
`rArr E = (-(20 - 10))/(10 xx 10^(-2) cos 120^(@))`
`= (-10)/(10 xx 10^(-2)(-sin 30^(@)) ) = (-10)/(-1//2) = 200 V//m`
Direction of `E` be perpendicular to the equipotential surface i.e., at `120^(@)` with x-axis.