Q133 / 150 · MHT CET 2026 (16 April Shift 2) · Physics
Two spherical hollow spheres of radii '$R_1$' and '$R_2$' are charged with same charge. Let '$\sigma_1$' and '$\sigma_2$' are their respective charge densities, then the ratio $\sigma_1$ to $\sigma_2$ is
- A$R_1 : R_2$
- B$R_2 : R_1$
- Correct answerC$R_2^2 : R_1^2$
- D$R_1^2 : R_2^2$
Solution
Let the charge on both the hollow spheres be $Q$.
The surface charge density $\sigma$ of a sphere of radius $R$ is given by $\sigma = \dfrac{Q}{4\pi R^2}$.
For the first sphere of radius $R_1$, the charge density is $\sigma_1 = \dfrac{Q}{4\pi R_1^2}$.
For the second sphere of radius $R_2$, the charge density is $\sigma_2 = \dfrac{Q}{4\pi R_2^2}$.
Taking the ratio of their charge densities:
$\dfrac{\sigma_1}{\sigma_2} = \dfrac{\dfrac{Q}{4\pi R_1^2}}{\dfrac{Q}{4\pi R_2^2}} = \dfrac{R_2^2}{R_1^2}$
Thus, the ratio $\sigma_1 : \sigma_2$ is $R_2^2 : R_1^2$.
Answer: $R_2^2 : R_1^2$
The surface charge density $\sigma$ of a sphere of radius $R$ is given by $\sigma = \dfrac{Q}{4\pi R^2}$.
For the first sphere of radius $R_1$, the charge density is $\sigma_1 = \dfrac{Q}{4\pi R_1^2}$.
For the second sphere of radius $R_2$, the charge density is $\sigma_2 = \dfrac{Q}{4\pi R_2^2}$.
Taking the ratio of their charge densities:
$\dfrac{\sigma_1}{\sigma_2} = \dfrac{\dfrac{Q}{4\pi R_1^2}}{\dfrac{Q}{4\pi R_2^2}} = \dfrac{R_2^2}{R_1^2}$
Thus, the ratio $\sigma_1 : \sigma_2$ is $R_2^2 : R_1^2$.
Answer: $R_2^2 : R_1^2$
Hence, the correct answer is option: 'C'.