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Figure shows equipotential surfaces concentric at O. The magnitude of electric field (in V/m) at distance r (in meter) measured from O is

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9/r²

6/r²

2/r²

16/r²

answer is B.

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Detailed Solution

V at 10cm arc is, $V = \frac{1}{4 {πε}_{0}} \cdot \frac{q_{1}}{r_{1}}$
$∴ 60 = \frac{1}{4 {πε}_{0}} \frac{q_{1}}{0.1} \to (1)$
E at a distance r is, $E = \frac{1}{4 {πε}_{0}} \frac{q_{1}}{r^{2}} \to (2)$
Solving, E = 6/r² V/m

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