A thin spherical shell of radius 10 cm carries a charge of \(4\ \mathrm{nC}\) spread uniformly over it. A point charge of \(2\ \mathrm{nC}\) is placed at its centre. The electric potential at a point P inside the shell, 5 cm from the centre, is (take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(720\ \mathrm{V}\)
The potential at P is the sum of two parts, one from the point charge and one from the shell. Inside a charged thin shell, the potential of the shell is the same everywhere and equals its value at the surface, \( Q/4\pi\varepsilon_0R \).
Here \( q=2\times10^{-9}\ \mathrm{C} \) at \( r=0.05\ \mathrm{m} \), and \( Q=4\times10^{-9}\ \mathrm{C} \) with \( R=0.10\ \mathrm{m} \).
\( \begin{aligned} V &= \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{r}+\frac{Q}{R}\right) \\ &= 9\times10^{9}\left(\frac{2\times10^{-9}}{0.05}+\frac{4\times10^{-9}}{0.10}\right) \\ &= 9\times10^{9}\times(4+4)\times10^{-8} \\ &= 720\ \mathrm{V} \end{aligned} \)
NCERT (2026-27):
Potential Due to a System of Charges – “… By the superposition principle, the potential V at P due to the total charge configuration is the algebraic sum of the potentials due to the individual charges”
\( V=V_1+V_2+\dots+V_n \qquad (2.17) \)
Potential Due to a System of Charges – “… The electric field inside the shell is zero. This implies … that potential is constant inside the shell (as no work is done in moving a charge inside the shell), and, therefore, equals its value at the surface, which is”
\( V=\frac{1}{4\pi\varepsilon_0}\frac{q}{R} \qquad [2.19(b)] \)