Question

A solid sphere of radius \(R_1\) and volume charge density \(\rho = \frac{\rho_0}{r}\) is enclosed by a hollow sphere of radius \(R_2\) with negative surface charge density \(\sigma\), such that the total charge in the system is zero. \(\rho_0\) is a positive constant and \(r\) is the distance from the centre of the sphere. The ratio \(\frac{R_2}{R_1}\) is

• A. \(\frac{\sigma}{\rho_0}\)
• B. \(\sqrt{\frac{2\sigma}{\rho_0}}\)
• C. \(\sqrt{\frac{\rho_0}{2\sigma}}\)
• D. \(\frac{\rho_0}{\sigma}\)

Hint:

For total charge = 0, set volume charge of inner sphere equal to surface charge of outer sphere (with sign).

Answer 1

The Correct Option is C

Step 1: Find total charge inside solid sphere.
Given volume charge density:
\[ \rho(r) = \frac{\rho_0}{r} \]
Total charge:
\[ Q_1 = \int \rho \, dV \]
In spherical coordinates:
\[ dV = 4\pi r^2 dr \]
So:
\[ Q_1 = \int_0^{R_1} \frac{\rho_0}{r} \cdot 4\pi r^2 dr = 4\pi \rho_0 \int_0^{R_1} r \, dr \]
\[ = 4\pi \rho_0 \left[\frac{r^2}{2}\right]_0^{R_1} = 4\pi \rho_0 \cdot \frac{R_1^2}{2} = 2\pi \rho_0 R_1^2 \]
Step 2: Find charge on hollow sphere surface.
Surface charge density is negative: \(-\sigma\).
Charge on hollow sphere:
\[ Q_2 = -\sigma \cdot 4\pi R_2^2 \]
Step 3: Total charge is zero.
\[ Q_1 + Q_2 = 0 \]
\[ 2\pi \rho_0 R_1^2 - 4\pi \sigma R_2^2 = 0 \]
Step 4: Solve for ratio.
\[ 2\pi \rho_0 R_1^2 = 4\pi \sigma R_2^2 \]
\[ \rho_0 R_1^2 = 2\sigma R_2^2 \]
\[ \frac{R_2^2}{R_1^2} = \frac{\rho_0}{2\sigma} \Rightarrow \frac{R_2}{R_1} = \sqrt{\frac{\rho_0}{2\sigma}} \]
Final Answer:
\[ \boxed{\frac{R_2}{R_1} = \sqrt{\frac{\rho_0}{2\sigma}}} \]