Question

Two isolated metallic solid spheres of radii $R$ and $2 R$ are charged such that both have same charge density $\sigma$. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is $\sigma^{\prime}$ The ratio $\frac{\sigma^{\prime}}{\sigma}$ is :

• A. $\frac{5}{3}$
• B. $\frac{9}{4}$
• C. $\frac{5}{6}$
• D. $\frac{4}{3}$

Answer 1

The Correct Option is C

The relationship between the charges and radii is given by:

\[ \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_1'}{R} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_2'}{2R} \]

Simplifying, we find:

\[ Q_2' = 2Q_1' \]

Using the charge conservation equation:

\[ Q_1' + Q_2' = Q_1 + Q_2 \]

Substitute \(Q_2' = 2Q_1'\):

\[ Q_1' + 2Q_1' = 20 \pi R^2 \sigma \]

\[ 3Q_1' = 20 \pi R^2 \sigma \]

\[ Q_1' = \frac{20 \pi R^2 \sigma}{3} \]

Substitute \(Q_2' = 2Q_1'\):

\[ Q_2' = \frac{40 \pi R^2 \sigma}{3} \]

The surface charge densities are related by:

\[ \sigma' = \frac{Q_2'}{4 \pi (2R)^2} \]

\[ \sigma' = \frac{\frac{40 \pi R^2 \sigma}{3}}{16 \pi R^2} \]

\[ \sigma' = \frac{40}{3} \cdot \frac{1}{16} \cdot \sigma \]

\[ \sigma' = \frac{5}{6} \cdot \sigma \]

Final Answer:

\(\frac{\sigma'}{\sigma} = \frac{5}{6}\)