Consider a uniformly positively charged thin spherical shell of radius \( R \). Let \( V_{1} \), \( V_{2} \) and \( V_{3} \) be the potentials at distances from the centre \( r = R \), \( r \gt R \) and \( r \lt R \), respectively. Then
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( V_{1} = V_{3} \gt V_{2} \)
Inside a charged spherical shell the electric field is zero, so the potential does not change there: it is constant and equal to the potential on the surface. Outside, the shell behaves like a point charge at its centre.
\( \displaystyle V_{1} = V_{3} = \frac{1}{4\pi\varepsilon_{0}}\frac{Q}{R} \)
\( \displaystyle V_{2} = \frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r},\quad r \gt R \)
For a positive charge, \( V_{2} \lt V_{1} \) because \( r \gt R \). Hence \( V_{1} = V_{3} \gt V_{2} \).
NCERT (2026-27):
Potential Due to a System of Charges – “… We have seen in Chapter 1 that for a uniformly charged spherical shell, the electric field outside the shell is as if the entire charge is concentrated at the centre. Thus, the potential outside the shell is given by
\( \displaystyle V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r} \quad (r \ge R) \qquad [2.19(a)] \)
where \( q \) is the total charge on the shell and \( R \) its radius. The electric field inside the shell is zero. This implies (Section 2.6) 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
\( \displaystyle V = \frac{1}{4\pi\varepsilon_0}\frac{q}{R} \qquad [2.19(b)] \)”