Question

Some equipotential surface is shown in the figure. What can you say about the magnitude and the direction of the electric field?

Answer 1

(a) The electric field is always perpendicular to the equipotential surface. (As shown in the figure)

So, the angle between $\overrightarrow{E}and\overrightarrow{dx}$ = $90°+30°$
Change in potential in the first and second equipotential surfaces, dV = 10 V
So,$\overrightarrow{E}.\overrightarrow{dx}=-dV$
$$\begin{gathered} \Rightarrow Edx\cos (90°+30°)=-dV \\ \Rightarrow E(10\times {10}^{-2})\cos 120°=-10 \\ \Rightarrow E=200\mathrm{V}/\mathrm{m} \end{gathered}$$
The electric field is making an angle of 120$°$ with the x axis.

(b) The electric field is always perpendicular to the equipotential surface.

So, the angle between $\overrightarrow{E}$and $\overrightarrow{dr}$ = 0$°$
Potential at point A, $V_A=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q}{r}=60\mathrm{V}$
$$\begin{gathered} \Rightarrow \frac{q}{4\mathrm{\pi }{\epsilon }_0}=60\times r \\ \Rightarrow \frac{q}{4\mathrm{\pi }{\epsilon }_0}=0.6 \end{gathered}$$
So, electric field, $E=\frac{q}{4\mathrm{\pi }{\epsilon }_0}\times \frac{1}{r^2}=\frac{0.6}{r^2}$
The electric field is radially outward, decreasing with increasing distance.