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 :

CONCEPT:

The electric potential is defined as the amount of energy that is needed to move the electric charge is known as electric potential. It is denoted as "V".

\(V = \frac{1}{{4\pi {\varepsilon _o}}}\frac{q}{r}\)

Where q is the charge and r is the radius.

CALCULATION:

 Here, we are given two charged spherical conductors of radius R₁ and R₂ which are connected by the wire.

The figure is shown below,

Now, the potential of the first spherical conductor is,

\(V_1 = \frac{1}{{4\pi {\varepsilon _o}}}\frac{q_1}{R_1}\)

The potential for the second spherical conductor is,

\(V_2 = \frac{1}{{4\pi {\varepsilon _o}}}\frac{q_2}{R_2}\)

when two spherical conductors are connected with a conducting wire, their potential will be the same.

\(\therefore \frac{1}{{4\pi { \in _0}}}\frac{{{Q_1}}}{{{R_1}}} = \frac{1}{{4\pi { \in _0}}}\,\frac{{{Q_2}}}{{{R_2}}}----(1)\)

As we know, the surface charge density (\(\sigma \)) is,

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

Let us make equation (1) in the form of  \(\sigma\) we get,

\( \frac{1}{{4\pi { \in _0}}}\,\frac{{{Q_1}}}{{{R_1}}}\, \times \,\frac{{{R_1}}}{{{R_1}}} = \frac{1}{{4\pi { \in _0}}}\frac{{{Q_2}}}{{{R_2}}} \times \frac{{{R_2}}}{{{R_2}}}\)

\( \Rightarrow \frac{{{Q_1}{R_1}}}{{4\pi R_1^2{ \in _0}}} = \frac{{{Q_2}{R_2}}}{{4\pi R_2^2{ \in _0}}}\)

\(\Rightarrow \frac{{{\sigma _1}{R_1}}}{{{ \in _0}}} = \frac{{{\sigma _2}{R_2}}}{{{ \in _0}}}\)

\(\Rightarrow \frac{{{\sigma _1}}}{{{\sigma _2}}} = \frac{{{R_2}}}{{{R_1}}}\)

Hence, option 3) is the correct answer.