The electric field intensity at a point $P$ due to a point charge $q$ kept at a point $Q$ is $40 N C^{- 1}$ and the electric potential at a point $P$ due to the same charge is $20 J C^{- 1}$ . The order of magnitude of charge $q$ is

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$10^{- 6} C$

$10^{- 7} C$

$10^{- 12} C$

$10^{- 9} C$

answer is D.

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Detailed Solution

Electric field of a point charge,

$E = \frac{1}{4 {πε}_{0}} \frac{q}{r^{2}} = 40 N C^{- 1}$

Electric potential of a point charge,

$V = \frac{1}{4 {πε}_{0}} \frac{q}{r} = 20 J C^{- 1}$

The distance $P Q$ is $r = \frac{V}{E} = \frac{20}{40} = 0.5 m$

$∴$ Magnitude of charge $= 4 {πε}_{0} Vr$

$= \frac{1}{9 \times 10^{9}} \times 20 \times 0.5 = 1.11 \times 10^{- 9} C \approx 10^{- 9} C$

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