Question

Two isolated metallic spheres of radii \(R\) and \(2R\) are charged such that both have the same charge density \(σ\). The spheres are then connected by a thin conducting wire. If the new charge density of the larger sphere is \(σ_1\). The ratio \(σ_1\) to \(σ\) is

• A. \(9:4\)
• B. \(4:3\)
• C. \(5:3\)
• D. \(5:6\)

Hint:

After connection, potentials are equal, so charges share in proportion to radii.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The spheres have radii \(R\) and \(2R\) and the same surface charge density \(\sigma\). Charges: \(q_1 = 4\pi R^2\sigma\) and \(q_2 = 4\pi(2R)^2\sigma = 16\pi R^2\sigma\). Total charge \(= 20\pi R^2\sigma\).

Step 2: After connection:
Connected spheres reach the same potential, \(\frac{kq_1'}{R} = \frac{kq_2'}{2R}\), so \(q_2' = 2q_1'\).
Total is conserved: \(3q_1' = 20\pi R^2\sigma\), so \(q_2' = \frac23\times20\pi R^2\sigma = \frac{40\pi R^2\sigma}{3}\).

Step 3: New density:
\[ \sigma_1 = \frac{q_2'}{4\pi(2R)^2} = \frac{40\pi R^2\sigma/3}{16\pi R^2} = \frac{5\sigma}{6} \]
So \(\sigma_1:\sigma = 5:6\).

Final Answer:
The ratio \(\sigma_1:\sigma\) is \(5:6\), option (D). \[ \boxed{5:6} \]