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 the problem of finding the ratio of surface charge densities of two charged spherical conductors interconnected by a wire, we need to apply the concept of charge distribution in electrostatics.

When two conductors are connected by a wire, they reach an equipotential state. This means that the electric potential \( V \) on both conductors is the same:

\(V_1 = V_2\)

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

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

where \( k \) is the Coulomb's constant, \( Q \) is the charge on the sphere, and \( R \) is its radius.

For Sphere 1, the potential is:

\(V_1 = \frac{kQ_1}{R_1}\)

For Sphere 2, the potential is:

\(V_2 = \frac{kQ_2}{R_2}\)

Equating the potentials because they are connected by a wire:

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

Cancelling out \( k \) yields:

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

This equation implies that:

\(Q_1 R_2 = Q_2 R_1\)

The surface charge density \( \sigma \) is defined as charge per unit area:

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

Then the surface charge densities are,

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

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

The ratio of surface charge densities is:

\(\frac{\sigma_1}{\sigma_2} = \frac{Q_1}{4\pi R_1^2} \times \frac{4\pi R_2^2}{Q_2}\)

Simplifying further, using the equation \( Q_1 R_2 = Q_2 R_1 \):

\(\frac{\sigma_1}{\sigma_2} = \frac{R_2^2}{R_1^2} \times \frac{Q_1}{Q_2} = \frac{R_2^2}{R_1^2} \times \frac{R_1}{R_2} = \frac{R_2}{R_1}\)

Therefore, the correct ratio of the surface charge densities \(\sigma_1/\sigma_2\) is:

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