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What is the magnitude of a point charge which produces an electric field of 2 N/C at a distance of 60 cm $(1 / 4 {πε}_{0} = 9 \times 10^{9} N - m^{2} / C^{2})$

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$8 \times 10^{- 11} C$

$6 \times 10^{- 10} C$

$3 \times 10^{- 11} C$

$2 \times 10^{- 12} C$

answer is A.

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Detailed Solution

$E = \frac{1}{4 {πε}_{0}} \cdot \frac{Q}{r^{2}} \Rightarrow 2 = 9 \times 10^{9} \times \frac{Q}{(0 .6)^{2}} \Rightarrow Q = 8 \times 10^{- 11} C$

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