Question

Two metallic solid spheres A and B,have radius R and 3R,respectively. The solid spheres are charged and kept isolated. Then,the two spheres are connected to each other through a thin conducting wire. The ratio of the final charge on the spheres A to B is:

• A. 1:1
• B. 1:3
• C. 3:1
• D. 1:9
• E. 9:1

Answer 1

The Correct Option is B

Two isolated charged spheres with radii R and 3R are connected by a conducting wire.

When connected, the potentials equalize: $$V_A = V_B$$ $$\frac{kQ_A}{R} = \frac{kQ_B}{3R}$$

Charge Ratio is $$\frac{Q_A}{Q_B} = \frac{R}{3R} = \frac{1}{3}$$

Alternatively, since capacitance C ∝ radius: $$\frac{C_A}{C_B} = \frac{R}{3R} = \frac{1}{3}$$ And Q = CV, so when V equalizes: $$\frac{Q_A}{Q_B} = \frac{C_A}{C_B} = \frac{1}{3}$$

Answer 2

1. Understand the concept of charge distribution:

When two charged conductors are connected by a conducting wire, charge flows between them until they reach the same potential.

2. Recall the formula for the potential of a sphere:

The electric potential (V) of a conducting sphere with charge Q and radius R is given by:

$$V = \frac{kQ}{R}$$

where k is Coulomb's constant.

3. Set the final potentials equal:

Let $Q_A$ and $Q_B$ be the initial charges on spheres A and B, and $Q_A'$ and $Q_B'$ be their final charges after they are connected. Since the spheres reach the same potential after connection:

$$V_A' = V_B'$$

$$\frac{kQ_A'}{R} = \frac{kQ_B'}{3R}$$

4. Solve for the charge ratio:

Simplify the equation:

$$Q_A' = \frac{Q_B'}{3}$$

$$\frac{Q_A'}{Q_B'} = \frac{1}{3}$$