Q56 / 75 · JEE Main 2025 (3 Apr Shift 1) · Physics
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 $\mathrm{r}=15 \mathrm{~cm}$ from the centre of the shell respectively, are :
- Correct answerA$120 \mathrm{~V}, 120 \mathrm{~V}, 80 \mathrm{~V}$
- B$40 \mathrm{~V}, 40 \mathrm{~V}, 80 \mathrm{~V}$
- C$0 \mathrm{~V}, 0 \mathrm{~V}, 80 \mathrm{~V}$
- D$0 \mathrm{~V}, 120 \mathrm{~V}, 40 \mathrm{~V}$
Solution
Potential inside shell is equal to potential on surface
$\mathrm{V}_{\mathrm{in}}=\mathrm{V}_{\text {surface }}=\frac{\mathrm{kQ}}{\mathrm{R}}=120 \mathrm{~V}$
at $\mathrm{r}=15 \mathrm{~cm}$
$\mathrm{V}=\frac{\mathrm{kQ}}{\mathrm{r}}=\frac{120 \times 10}{15}=80 \mathrm{~V}$
$\mathrm{V}_{\mathrm{in}}=\mathrm{V}_{\text {surface }}=\frac{\mathrm{kQ}}{\mathrm{R}}=120 \mathrm{~V}$
at $\mathrm{r}=15 \mathrm{~cm}$
$\mathrm{V}=\frac{\mathrm{kQ}}{\mathrm{r}}=\frac{120 \times 10}{15}=80 \mathrm{~V}$
Hence, the correct answer is option: 'A'.