Question

Equipotential surfaces are shown in figure. Then the electric field strength will be ______.

Options

• 100 vm^-1 along X-axis 

• 100 Vm^-1 along Y-axis 

• 200 vm^-1 at an angle 120° with X-axis 

• 50 vm^-1 at an angle 120° with X-axis

Answer 1

200 vm^-1 at an angle 120° with X-axis 

Explanation:

Using dV = ${\displaystyle -\overrightarrow{\text{E}}.\text{d}\overrightarrow{\text{r}}}$

⇒ ΔV = -E Δr cos θ

⇒ E = ${\displaystyle \frac{-\Delta \text{V}}{\Delta \text{r}\cos \theta }}$

⇒ E = ${\displaystyle \frac{-(20-10)}{10\times {10}^{-2}\cos 120°}}$

= ${\displaystyle \frac{-10}{10\times {10}^{-2}(-\sin 30°)}}$

= 200 V/m

Direction of E be perpendicular to the equipotential surface i.e. at 120° with X-axis.