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:
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:
Show 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 ($8.5 : 12.75 = 2 : 3$).
$8.5 + 12.75 = 21.25\text{ }\mu\text{C}$.
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 ($8.5 : 12.75 = 2 : 3$).
Updated On: Jul 22, 2026
- $12.75\text{ }\mu\text{C}, 8.5\text{ }\mu\text{C}$
- $1.5\text{ }\mu\text{C}, 8.5\text{ }\mu\text{C}$
- $8.5\text{ }\mu\text{C}, 12.75\text{ }\mu\text{C}$
- $8.5\text{ }\mu\text{C}, 1.5\text{ }\mu\text{C}$
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
We are given two isolated conducting spheres, $A$ and $B$, with their radii and a relation between their initial surface charge densities.
We need to find the charges on each sphere after they are brought into electrical contact and then separated.
Step 2: Key Formula and Approach:
1. Find the initial charge of sphere $B$ using the relation $\sigma_A = 2 \sigma_B$.
2. Calculate the total charge of the system, which must be conserved.
3. When connected, the spheres reach a common electric potential ($V$).
The final charges will distribute such that the potential of both spheres is equal:
\[ \frac{Q_A'}{Q_B'} = \frac{R_A}{R_B} \]
Step 3: Detailed Explanation:
• Find initial charge on B ($Q_B$):
Surface charge density of a sphere: $\sigma = \frac{Q}{4\pi R^2}$
Given $\sigma_A = 2 \sigma_B$:
\[ \frac{Q_A}{4\pi R_A^2} = 2 \times \frac{Q_B}{4\pi R_B^2} \implies \frac{Q_A}{R_A^2} = \frac{2Q_B}{R_B^2} \] Given $R_A = 2\text{ cm}$, $R_B = 3\text{ cm}$, and $Q_A = +10\text{ }\mu\text{C}$:
\[ \frac{10}{2^2} = \frac{2Q_B}{3^2} \implies \frac{10}{4} = \frac{2Q_B}{9} \] \[ 2.5 = \frac{2Q_B}{9} \implies 2Q_B = 22.5 \implies Q_B = 11.25\text{ }\mu\text{C} \]
• Calculate total charge ($Q_{\text{total}}$):
\[ Q_{\text{total}} = Q_A + Q_B = 10\text{ }\mu\text{C} + 11.25\text{ }\mu\text{C} = 21.25\text{ }\mu\text{C} \]
• Determine final charges after separation:
Since they are connected, their final charges are proportional to their radii:
\[ Q_A' = \left(\frac{R_A}{R_A + R_B}\right) Q_{\text{total}} = \left(\frac{2}{2 + 3}\right) \times 21.25 = \frac{2}{5} \times 21.25 = 8.5\text{ }\mu\text{C} \] \[ Q_B' = \left(\frac{R_B}{R_A + R_B}\right) Q_{\text{total}} = \left(\frac{3}{2 + 3}\right) \times 21.25 = \frac{3}{5} \times 21.25 = 12.75\text{ }\mu\text{C} \]
Step 4: Final Answer:
The charges on spheres A and B after separation are $8.5\text{ }\mu\text{C}$ and $12.75\text{ }\mu\text{C}$ respectively, which corresponds to Option (C).
We are given two isolated conducting spheres, $A$ and $B$, with their radii and a relation between their initial surface charge densities.
We need to find the charges on each sphere after they are brought into electrical contact and then separated.
Step 2: Key Formula and Approach:
1. Find the initial charge of sphere $B$ using the relation $\sigma_A = 2 \sigma_B$.
2. Calculate the total charge of the system, which must be conserved.
3. When connected, the spheres reach a common electric potential ($V$).
The final charges will distribute such that the potential of both spheres is equal:
\[ \frac{Q_A'}{Q_B'} = \frac{R_A}{R_B} \]
Step 3: Detailed Explanation:
• Find initial charge on B ($Q_B$):
Surface charge density of a sphere: $\sigma = \frac{Q}{4\pi R^2}$
Given $\sigma_A = 2 \sigma_B$:
\[ \frac{Q_A}{4\pi R_A^2} = 2 \times \frac{Q_B}{4\pi R_B^2} \implies \frac{Q_A}{R_A^2} = \frac{2Q_B}{R_B^2} \] Given $R_A = 2\text{ cm}$, $R_B = 3\text{ cm}$, and $Q_A = +10\text{ }\mu\text{C}$:
\[ \frac{10}{2^2} = \frac{2Q_B}{3^2} \implies \frac{10}{4} = \frac{2Q_B}{9} \] \[ 2.5 = \frac{2Q_B}{9} \implies 2Q_B = 22.5 \implies Q_B = 11.25\text{ }\mu\text{C} \]
• Calculate total charge ($Q_{\text{total}}$):
\[ Q_{\text{total}} = Q_A + Q_B = 10\text{ }\mu\text{C} + 11.25\text{ }\mu\text{C} = 21.25\text{ }\mu\text{C} \]
• Determine final charges after separation:
Since they are connected, their final charges are proportional to their radii:
\[ Q_A' = \left(\frac{R_A}{R_A + R_B}\right) Q_{\text{total}} = \left(\frac{2}{2 + 3}\right) \times 21.25 = \frac{2}{5} \times 21.25 = 8.5\text{ }\mu\text{C} \] \[ Q_B' = \left(\frac{R_B}{R_A + R_B}\right) Q_{\text{total}} = \left(\frac{3}{2 + 3}\right) \times 21.25 = \frac{3}{5} \times 21.25 = 12.75\text{ }\mu\text{C} \]
Step 4: Final Answer:
The charges on spheres A and B after separation are $8.5\text{ }\mu\text{C}$ and $12.75\text{ }\mu\text{C}$ respectively, which corresponds to Option (C).
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