JEE MAIN - Physics (2025 - 3rd April Morning Shift - No. 18)
The electrostatic potential on the surface of uniformly charged spherical shell of radius $\mathrm{R}=10 \mathrm{~cm}$ is 120 V . The potential at the centre of shell, at a distance $\mathrm{r}=5 \mathrm{~cm}$ from centre, and at a distance $\mathrm{r}=15$ cm from the centre of the shell respectively, are:
$0 \mathrm{~V}, 120 \mathrm{~V}, 40 \mathrm{~V}$
$120 \mathrm{~V}, 120 \mathrm{~V}, 80 \mathrm{~V}$
$40 \mathrm{~V}, 40 \mathrm{~V}, 80 \mathrm{~V}$
$0 \mathrm{~V}, 0 \mathrm{~V}, 80 \mathrm{~V}$
Explanation
The potential inside a uniformly charged spherical shell is equal to the potential on its surface. This means:
$ V_{\text{in}} = V_{\text{surface}} = \frac{kQ}{R} = 120 \, \text{V} $
This maintains the potential at both the center of the shell and any point inside it up to the surface. Therefore, the potential at the center of the shell and at any point inside the shell (like at a distance $ r = 5 \, \text{cm} $) is also $ 120 \, \text{V} $.
For a point outside the shell, at a distance $ r = 15 \, \text{cm} $, the potential is given by the formula:
$ V = \frac{kQ}{r} = \frac{120 \times 10}{15} = 80 \, \text{V} $
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