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)

$V=\frac{q}{4\pi {\in }_{0}x}\Rightarrow q=4\pi {\epsilon }_{0}Vx=60\times 0.1\times 4\pi {\epsilon }_0=24\pi {\epsilon }_0$
$\therefore E=\frac{q}{4\pi {\in }_0{x}^2}=\frac{24\pi {\in }_0}{4\pi {\in }_0{x}^2}=\frac{6}{x^2}$
Direction of electric field is radially outward.