Two metallic spheres of radii 1 cm and 3 cm are given charges of 4 × 10⁻² C and respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is:

• A: 2 × 10⁻² C
• B: 3 × 10⁻² C
• C: 4 × 10⁻² C
• D: 1 × 10⁻² C

To find the final charge on the larger sphere after connecting the two metallic spheres with a conducting wire, we need to consider the principle of charge distribution. When two conductors are connected, they will share their total charge in proportion to their capacitance, which depends on their sizes (radii).

Given Data

• Radius of smaller sphere (r1) = 1 cm
• Charge on smaller sphere (Q1) = 4 × 10⁻² C
• Radius of larger sphere (r2) = 3 cm
• Charge on larger sphere (Q2) = 0 C (initially)

Calculating Capacitance

The capacitance of a sphere is directly proportional to its radius. Therefore, we can express the capacitance ratio as:

Capacitance ratio = r1 : r2 = 1 : 3

Total Charge

The total charge when the spheres are connected is:

Total charge (Q_total) = Q1 + Q2 = 4 × 10⁻² C + 0 = 4 × 10⁻² C

Charge Distribution

Let the final charge on the larger sphere be Q_final. The charges will distribute according to their capacitance:

Q_final = (Total charge) × (Capacitance of larger sphere) / (Total capacitance)

Since the capacitance of the smaller sphere is 1 and that of the larger is 3, the total capacitance is:

Total capacitance = 1 + 3 = 4

Final Charge Calculation

Now, we can calculate the final charge on the larger sphere:

Q_final = (4 × 10⁻² C) × (3 / 4) = 3 × 10⁻² C

Final Answer

The final charge on the larger sphere is 3 × 10⁻² C, which corresponds to option B.