Question

A few spherical equipotential surfaces are shown in the figure. The electric field at any point, at a distance $x$ from the centre, is

• A. $\frac{6}{x^{2}}$ perpendicular to the plane of paper
• B. $\frac{600}{x^{2}}$ perpendicular to the plane of paper
• C. $\frac{6}{x^{2}}$ radially
• D. $\frac{600}{x^{2}}$ radially

Answer 1

Answer: (C)

$V=\frac{q}{4 \pi \in _{0} x} \, \Rightarrow \, q=4\pi ε_{0}Vx=60\times \, 0.1\times 4\pi ε_{0}=24\pi ε_{0}$
$\therefore \, E=\frac{q}{4 \pi \in _{0} x^{2}}=\frac{24 \pi \in _{0}}{4 \pi \in _{0} x^{2}}=\frac{6}{x^{2}}$
Direction of electric field is radially outward.