The electric potential at a point 20 cm from a point charge is \(-450\ \mathrm{V}\). The potential at infinity is zero. The charge is (Take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-1\times10^{-8}\ \mathrm{C}\)
The potential of a point charge falls as \( 1/r \), \( V=\frac{Q}{4\pi\varepsilon_0r} \). The sign of \( V \) is the sign of \( Q \). So a negative potential means a negative charge.
Rearranging the formula for the charge:
\( \begin{aligned} Q &= 4\pi\varepsilon_0\,rV \\ &= \frac{0.2\times(-450)}{9\times10^{9}} \\ &= -1\times10^{-8}\ \mathrm{C} \end{aligned} \)
NCERT (2026-27):
Potential Due to a Point Charge – “… This, by definition is the potential at P due to the charge Q”
\( V(r)=\frac{Q}{4\pi\varepsilon_0r} \)\( \qquad (2.8) \)
Potential Due to a Point Charge – “Equation (2.8) is true for any sign of the charge Q, though we considered Q > 0 in its derivation. For Q < 0, V < 0, i.e., work done (by the external force) per unit positive test charge in bringing it from infinity to the point is negative. …”