At a point near an isolated point charge, the electric potential is 600 V and the magnitude of the electric field is \(200\ \mathrm{N\,C^{-1}}\). The magnitude of the charge is (take \(1/4\pi\varepsilon_0=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2\times10^{-7}\ \mathrm{C}\)
For a point charge, \( V=Q/4\pi\varepsilon_0r \) and \( E=Q/4\pi\varepsilon_0r^{2} \). Dividing the first by the second gives the distance \( r=V/E \). The charge then follows from the potential.
\( \begin{aligned} r &= \frac{V}{E} \\ &= \frac{600}{200} \\ &= 3\ \mathrm{m} \end{aligned} \)
\( \begin{aligned} Q &= 4\pi\varepsilon_0\,rV \\ &= \frac{3\times600}{9\times10^{9}} \\ &= 2\times10^{-7}\ \mathrm{C} \end{aligned} \)
Therefore, the charge is \( 2\times10^{-7}\ \mathrm{C} \), and the point is 3 m away from it.
NCERT (2026-27):
Potential Due to a Point Charge – “… This, by definition is the potential at P due to the charge Q”
\( \displaystyle V(r)=\frac{Q}{4\pi\varepsilon_0 r} \)\( \qquad (2.8) \)