The electrostatic potential on the surface of uniformly charged spherical shell of radius R=10 ~cm is 120 V . The potential at the centre of shell, at a distance r=5 ~cm from centre, and at a distance r=15 cm from the centre of the shell respectively, are:

Electrostatics PYQ — JEE Main 2025 · 3 Apr · Shift 1 · Q68 with Solution

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Correct answer: B

For a spherical shell of radius $R$ :

Outside the shell $(r \ge R)$ , the potential is like that of a point charge at the centre: $V(r)=\frac{kQ}{r}$
On the surface $(r=R)$ : $V(R)=\frac{kQ}{R}$
Inside the shell $(r<R)$ , the potential is constant and equal to the surface potential: $V(r)=\frac{kQ}{R}$

Given: $R=10\text{ cm}, \quad V(R)=120\text{ V}$

So, $\frac{kQ}{R}=120$

Thus, everywhere inside the shell, potential is: $V=120\text{ V}$

The centre lies inside the shell, so: $V_{\text{centre}}=120\text{ V}$

Since $5\text{ cm} < 10\text{ cm}$ , this point is also inside the shell.

Therefore, $V(5\text{ cm})=120\text{ V}$

Since $15\text{ cm} > 10\text{ cm}$ , this point is outside the shell.

Using $V(r)=\frac{kQ}{r}$

and since $\frac{kQ}{R}=120,$ we get $kQ = 120R$

Hence, $V(15)=\frac{120R}{15}$

Using $R=10$ cm: $V(15)=120\times \frac{10}{15}=120\times \frac{2}{3}=80\text{ V}$

The potentials respectively are: $120\text{ V},\ 120\text{ V},\ 80\text{ V}$

So the correct option is: B