Question

The surface charge density of an isolated sphere A of radius $2\text{ cm}$ having a charge of $+10\text{ }\mu\text{C}$ is twice the surface charge density of another sphere B of radius $3\text{ cm}$. If the two spheres are joined and then separated, the charges on the sphere A and B after separation are respectively:

• A. $12.75\text{ }\mu\text{C}, 8.5\text{ }\mu\text{C}$
• B. $1.5\text{ }\mu\text{C}, 8.5\text{ }\mu\text{C}$
• C. $8.5\text{ }\mu\text{C}, 12.75\text{ }\mu\text{C}$
• D. $8.5\text{ }\mu\text{C}, 1.5\text{ }\mu\text{C}$

Hint:

Total charge is conserved, so check the sum of the options:
$8.5 + 12.75 = 21.25\text{ }\mu\text{C}$.
This sum is consistent with the initial total charge.
Also, the final charges must be in the ratio of their radii ($2:3$). Only Option (C) meets this criterion ($8.5 : 12.75 = 2 : 3$).

Answer 1

The Correct Option is C

Step 1: Find the charge on sphere B before contact.
Surface charge density is $\sigma = \frac{Q}{4\pi R^2}$, and we are told $\sigma_A = 2\sigma_B$. Writing this out: \[ \frac{Q_A}{R_A^2} = 2\frac{Q_B}{R_B^2} \quad\Rightarrow\quad Q_B = \frac{Q_A R_B^2}{2R_A^2} = \frac{10 \times 3^2}{2 \times 2^2} = 11.25\ \mu\text{C} \]
Step 2: Find the total charge, which stays conserved. \[ Q_{\text{total}} = Q_A + Q_B = 10 + 11.25 = 21.25\ \mu\text{C} \]
Step 3: Use the direct sharing formula for two joined spheres.
Once connected, both spheres sit at one common potential, so the final charge on each simply splits the total in the ratio of the radii: \[ Q_A' = Q_{\text{total}}\cdot\frac{R_A}{R_A+R_B}, \qquad Q_B' = Q_{\text{total}}\cdot\frac{R_B}{R_A+R_B} \]
Step 4: Substitute the numbers. \[ Q_A' = 21.25\times\frac{2}{5} = 8.5\ \mu\text{C}, \qquad Q_B' = 21.25\times\frac{3}{5} = 12.75\ \mu\text{C} \] \[ \boxed{Q_A' = 8.5\ \mu\text{C}, \ Q_B' = 12.75\ \mu\text{C}} \]