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If the potential at certain distance from a point charge is 600V and the intensity of electric field is $300 N C^{- 1} .$ The magnitude of the charge in coulomb is ___

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$1.33 \times 10^{- 7}$

$2 \times 10^{- 6}$

$2 \times 10^{- 7}$

$33 \times 10^{- 6}$

answer is A.

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Detailed Solution

$V = \frac{1}{4 π \in_{0}} . \frac{q}{r}, E = \frac{1}{4 π \in_{0}} . \frac{q}{r^{2}}, \frac{V}{E} = \frac{q}{r} \frac{r^{2}}{q} = r$

$\Rightarrow r = \frac{600}{300} = 2 m, V = 9 \times 10^{9} \frac{q}{r} \Rightarrow q = \frac{V r}{9 \times 10^{9}}$

$q = \frac{600 \times 2}{9 \times 10^{9}} = \frac{4}{3} 10^{- 7} C = 1.33 \times 10^{- 7} C$

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