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: Plan:
Use the fact that equal potential means $\sigma R$ is equal on both spheres.

Step 2: Steps:
For a sphere, $V = \frac{\sigma R}{\varepsilon_0}$. After connection, $\sigma_a R = \sigma_b(2R)$, so the small sphere has density $2\sigma_1$ when the large one has $\sigma_1$.
Charge conservation: $4\pi R^2\sigma + 16\pi R^2\sigma = 4\pi R^2(2\sigma_1) + 16\pi R^2\sigma_1$. That is $20\sigma = 8\sigma_1 + 16\sigma_1 = 24\sigma_1$, so $\frac{\sigma_1}{\sigma} = \frac{20}{24} = \frac56$.

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