Question

Two charged spherical conductors of radius R₁ and R₂ are connected by a wire. Then the ratio of surface charge densities of the spheres (σ₁ /σ₂ ) is

• A. \(\frac{R_1^2}{R_2^2}\)
• B. \(\frac{R_1}{R_2}\)
• C. \(\frac{R_2}{R_1}\)
• D. \(\sqrt{(\frac{R_1}{R_2})}\)

Answer 1

The Correct Option is C

To solve this problem, we need to understand the behavior of electric charge distribution between two connected spherical conductors.

When two charged spherical conductors are connected by a wire, they come to the same potential because the wire allows charges to move freely between them until equilibrium is reached.

The potential \( V \) of a charged conductor is given by: 

\(V = \frac{kQ}{R}\)

where:

• \(k\) is Coulomb's constant.
• \(Q\) is the total charge on the conductor.
• \(R\) is the radius of the spherical conductor.

For both spheres to have the same potential \(V_1 = V_2\), we have:

\(\frac{kQ_1}{R_1} = \frac{kQ_2}{R_2}\)

Canceling \( k \) from both sides:

\(\frac{Q_1}{R_1} = \frac{Q_2}{R_2}\)

Rearranging gives us the charge relation:

\(Q_1 = \frac{R_1}{R_2}Q_2\)

The surface charge density \( \sigma \) is given by:

\(\sigma = \frac{Q}{4\pi R^2}\)

Therefore, the surface charge densities for the two spheres are:

\(\sigma_1 = \frac{Q_1}{4\pi R_1^2}\)

\(\sigma_2 = \frac{Q_2}{4\pi R_2^2}\)

Substitute the expression for \(Q_1\) from the charge relation:

\(\sigma_1 = \frac{\frac{R_1}{R_2}Q_2}{4\pi R_1^2}\)

\(\sigma_1 = \frac{Q_2}{4\pi R_2} \cdot \frac{1}{R_1}\)

Thus, the ratio of surface charge densities is:

\(\frac{\sigma_1}{\sigma_2} = \frac{1/R_1}{1/R_2} = \frac{R_2}{R_1}\)

Therefore, the correct answer is:

\(\frac{R_2}{R_1}\)

This confirms that the ratio of surface charge densities is determined by the inverse of the radii ratio of the spheres.