Question

In the given figure shows equipotential surfaces, find the magnitude of electric field in the space (in N/C) (Give integer value)

• A. 400.00
• B. 400
• C. 400.0

Answer 1

Answer: (A), (B), (C)

From the figure, we can see that the distance between any two equipotential surfaces is
$d=10\sin {30}^{\circ }=5\text{cm}$

$\therefore E=|\frac{\Delta V}{d}|=\frac{40-20}{5\times {10}^{-2}}=400\text{N/C}$

Accepted answer : $400$