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

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

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

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

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

answer is B.

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Detailed Solution

V at $10 cm = \frac{1}{4 {πε}_{0}} \frac{q}{r}; 60 = \frac{1}{4 {πε}_{0}} \frac{q}{0 .1}$
E at $r = \frac{1}{4 {πε}_{0}} \frac{q}{r^{2}}; E (r) = \frac{60 \times 0 .1}{r^{2}} = \frac{6}{r^{2}} v / m$

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