Figure shows equipotential surfaces concentric at ‘O’, the magnitude of electric field at distance r (in meter) measured from O

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$\frac{9}{r^{2}} {Vm}^{- 1}$

$\frac{2}{r^{2}} {Vm}^{- 1}$

$\frac{6}{r^{2}} {Vm}^{- 1}$

$\frac{16}{r^{2}} {Vm}^{- 1}$

answer is B.

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Detailed Solution

$\begin{array}{rcl} V at 10 cm & = & \frac{1}{4 {πε}_{0}} \frac{q}{r}; \\ 60 & = & \frac{1}{4 {πε}_{0}} \frac{q}{0.1} \\ \frac{q}{4 {πε}_{0}} & = & 60 \times 0.1 - - (1) \\ E at r & = & \frac{1}{4 {πε}_{0}} \frac{q}{r^{2}}; \\ E (r) & = & \frac{60 x 0.1}{r^{2}} = \frac{6}{r^{2}} V / m \end{array}$

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