Question

The figure shows equipotential surface concentric at ${}^{\prime }{O}^{\prime }$. The magnitude of electric field at a distance ${}^{\prime }{r}^{\prime }$ meters from ${}^{\prime }{O}^{\prime }$ is

• A. $\frac{9}{r^2}V{m}^{-1}$
• B. $\frac{16}{r^2}V{m}^{-1}$
• C. $\frac{2}{r^2}V{m}^{-1}$
• D. $\frac{6}{r^2}V{m}^{-1}$

Answer 1

The correct option is A $\frac{9}{r^2}V{m}^{-1}$

$\frac{1}{60}+\frac{1}{30}+\frac{1}{20}=\frac{1+2+3}{60}=\frac{6}{60}$

$E_1=\frac{V_1}{d_1}=\frac{20\times {10}^{+2}}{30}$

$E_2=\frac{V_2}{d_2}=\frac{30}{20\times {10}^{-2}}=\frac{30\times 100}{20}$

$E_3=\frac{V_3}{d_3}=\frac{60}{30\times {10}^{-2}}$

$E_1+E_2+E_3=\frac{20}{30}\times {10}^2+\frac{30}{20}\times {10}^2+\frac{60}{30}\times {10}^2$

$={10}^2(\frac{2}{3}+\frac{3}{2}+\frac{6}{3})V/m$

$=100\frac{4+9+18}{6}V/m$

$=100\times \frac{31}{6}$

$=\frac{9}{r^2}V/m$