If the electric potential at a certain distance from a point charge is 900V and electric intensity is 225 Vm^–1. The charge is

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4× 10^–7 C

2× 10^–7 C

4× 10^–3 C

2× 10^–5 C

answer is A.

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Detailed Solution

$\begin{array}{l} E = \frac{V}{d} \Rightarrow d = \frac{V}{E}; d = \frac{900}{225} = 4 m; V = \frac{1}{4 {πε}_{0}} \frac{q}{r} \\ \Rightarrow 900 = 9 \times 10^{9} \times \frac{q}{4} \Rightarrow q = 4 \times 10^{- 7} C \end{array}$

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