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Question

Physics Question on Electric Field

The electric field intensity at a point PP due to point charge qq kept at point QQ is 24NC124 \,N \,C^{-1} and the electric potential at point PP due to same charge is 12JC112\, J \,C^{-1}. The order of magnitude of charge qq is

A

106C10^{-6} \, C

B

107C10^{-7} \, C

C

1010C10^{-10} \, C

D

109C10^{-9} \, C

Answer

109C10^{-9} \, C

Explanation

Solution

Electric field of a point charge, E=14πε0qr2=24NC1 E=\frac{1}{4\pi\varepsilon_{0}} \frac{q}{r^{2}}=24\, N \, C^{-1} Electric potential of a point charge, V=14πε0qr=12JC1V=\frac{1}{4\pi\varepsilon_{0}}\frac{q}{r}=12 \, J \,C^{-1} The distance PQPQ is r=VE=1224=0.5mr=\frac{V}{E}=\frac{12}{24}=0.5\, m \therefore Magnitude of charge q=4πε0Vrq'=4\pi\varepsilon_{0} \, Vr =19×109×12×0.5=\frac{1}{9\times10^{9}}\times12 \times0.5 =0.667×109C=0.667 \times 10^{-9} \, C 109C\approx 10^{-9} \, C