Question:

The electric field intensity at a point $P$ due to point charge $q$ kept at point $Q$ is $24 \,N \,C^{-1}$ and the electric potential at point $P$ due to same charge is $12\, J \,C^{-1}$. The order of magnitude of charge $q$ is

Updated On: Jul 7, 2022
  • $10^{-6} \, C$
  • $10^{-7} \, C$
  • $10^{-10} \, C$
  • $10^{-9} \, C$
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The Correct Option is D

Solution and Explanation

Electric field of a point charge, $ E=\frac{1}{4\pi\varepsilon_{0}} \frac{q}{r^{2}}=24\, N \, C^{-1}$ Electric potential of a point charge, $V=\frac{1}{4\pi\varepsilon_{0}}\frac{q}{r}=12 \, J \,C^{-1}$ The distance $PQ$ is $r=\frac{V}{E}=\frac{12}{24}=0.5\, m $ $\therefore$ Magnitude of charge $q'=4\pi\varepsilon_{0} \, Vr $ $=\frac{1}{9\times10^{9}}\times12 \times0.5$ $=0.667 \times 10^{-9} \, C$ $\approx 10^{-9} \, C$
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Concepts Used:

Electric Field

Electric Field is the electric force experienced by a unit charge. 

The electric force is calculated using the coulomb's law, whose formula is:

\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)

While substituting q2 as 1, electric field becomes:

 \(E=k\dfrac{|q_{1}|}{r^{2}}\)

SI unit of Electric Field is V/m (Volt per meter).