Two concentric hollow conducting spheres of radii 10 cm and 20 cm carry charges \( 2\,\mu\mathrm{C} \) and \( 4\,\mu\mathrm{C} \) respectively. The electric potential at their common centre 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) → \( 3.6\times10^{5}\,\mathrm{V} \)
The field inside a charged spherical shell is zero. So the potential is the same everywhere inside it and equals its value at the surface, \( q/4\pi\varepsilon_0R \). The common centre lies inside both shells, and potentials are scalars, so the two contributions add by the superposition principle.
In SI units the data are:
\( q_1=2\times10^{-6}\,\mathrm{C} \)
\( R_1=10\,\mathrm{cm}=0.1\,\mathrm{m} \)
\( q_2=4\times10^{-6}\,\mathrm{C} \)
\( R_2=20\,\mathrm{cm}=0.2\,\mathrm{m} \)
\( \begin{aligned} V &= \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{R_1}+\frac{q_2}{R_2}\right) \\ &= 9\times10^{9}\times\left(\frac{2\times10^{-6}}{0.1}+\frac{4\times10^{-6}}{0.2}\right)\,\mathrm{V} \\ &= 9\times10^{9}\times4\times10^{-5}\,\mathrm{V} \\ &= 3.6\times10^{5}\,\mathrm{V} \end{aligned} \)
NCERT (2026-27):
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”
\( \displaystyle V=\frac{1}{4\pi\varepsilon_0}\,\frac{q}{R} \qquad [2.19(b)] \)