Question

Some equipotential surfaces are shown in figure given below. The magnitude of electric field will be

• A. $1000\text{V/m}$
• B. $500\text{V/m}$
• C. $300\text{V/m}$
• D. $1200\text{V/m}$

Answer 1

The correct option is A $1000\text{V/m}$

Since the equipotential planes are equally spaced, the electric field will be uniform.
The direction of $\overrightarrow{E}$ will be perpendicular to the equipotential surfaces, and it will flow from high potential to low potential as shown in the figure.


From relation,

$\Delta V=Ed$

Where,
$d$ = Perpendicular distance between two equipotential surfaces

$\Rightarrow 60=E(10\sin {37}^{\circ }\times {10}^{-2})$

$\Rightarrow E=\frac{60}{10\sin {37}^{\circ }\times {10}^{-2}}=\frac{60\times 100}{10\times \left(\frac{3}{5}\right)}$

$\therefore E=1000\text{V/m}$

Therefore, option $\left(c\right)$ is the correct answer.