A point charge of \(4\ \mathrm{nC}\) is placed at the origin. Point A is 3 cm and point B is 6 cm from the charge. The potential difference \(V_A-V_B\) is (Take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(600\ \mathrm{V}\)
The potential at a distance \( r \) from a point charge \( Q \) is \( V=Q/4\pi\varepsilon_0r \). For a positive charge the potential falls as we move away, so the nearer point A is at the higher potential.
\( \begin{aligned} V_A &= \frac{9\times10^{9}\times4\times10^{-9}}{0.03} \\ &= 1200\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} V_B &= \frac{9\times10^{9}\times4\times10^{-9}}{0.06} \\ &= 600\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} V_A-V_B &= 1200-600 \\ &= 600\ \mathrm{V} \end{aligned} \)
NCERT (2026-27):
Potential Due to a Point Charge – “… This, by definition is the potential at P due to the charge Q”
\( V(r)=\frac{Q}{4\pi\varepsilon_0r} \)\( \qquad (2.8) \)