The electric potential at a point 12 cm from a point charge is 450 V. The magnitude of the electric field at a point 6 cm from the same charge is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(1.5\times10^{4}\ \mathrm{V\,m^{-1}}\)
The potential of a point charge falls as \( 1/r \), but its field falls as \( 1/r^{2} \). First find \( Q/4\pi\varepsilon_0 \) from the potential at 12 cm. Then use it to get the field at 6 cm.
\( \begin{aligned} \frac{Q}{4\pi\varepsilon_0} &= Vr \\ &= 450\times0.12 \\ &= 54\ \mathrm{V\,m} \end{aligned} \)
\( \begin{aligned} E &= \frac{Q}{4\pi\varepsilon_0r^{2}} \\ &= \frac{54}{(0.06)^{2}} \\ &= 1.5\times10^{4}\ \mathrm{V\,m^{-1}} \end{aligned} \)
Halving the distance doubles the potential, but it makes the field four times as large.
NCERT (2026-27):
Potential Due to a Point Charge – “… the electrostatic potential ( ∝ 1/r) and the electrostatic field ( ∝ 1/r2 ) varies with r.”