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Question: The electric field intensity at point P due to point charge q kept at point Q is \(24NC^{- 1}\)and t...

The electric field intensity at point P due to point charge q kept at point Q is 24NC−124NC^{- 1}and the electric potential at point P due to same charge is 12JC−112JC^{- 1}. The order of magnitude of charge q is

A

10−6C10^{- 6}C

B

10−7C10^{- 7}C

C

10−10C10^{- 10}C

D

10−9C10^{- 9}C

Answer

10−9C10^{- 9}C

Explanation

Solution

: Electric field of point charge,

E=14πε0qr2=24NC−1E = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{r^{2}} = 24NC^{- 1}

Electric potential of a point charge,

V=14πε0qr=12JC−1V = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{r} = 12JC^{- 1}

The distance PQ is

r=VE=1224=0.5mr = \frac{V}{E} = \frac{12}{24} = 0.5m

∴\therefore Magnitude of charge

q′=4πε0Vrq' = 4\pi\varepsilon_{0}Vr

=19×109×12×0.5=0.667×10−9C= \frac{1}{9 \times 10^{9}} \times 12 \times 0.5 = 0.667 \times 10^{- 9}C

≈10−9C\approx 10^{- 9}C