Sphere 1 with radius has positive charge . Sphere 2 with radius is far from sphere 1 and initially uncharged. After the separated spheres are connected with a wire thin enough to retain only negligible charge, (a) is potential of sphere 1 greater than, less than, or equal to potential of sphere 2? What fraction of ends up on (b) sphere 1 and (c) sphere 2? (d) What is the ratio of the surface charge densities of the spheres?

The radius of sphere 1 s .

The radius of sphere 2 is .

The electric potential due to the point charge is,

Here,is the coulomb’s constant, is the charge, and is the distance from the charge.

The potential of sphere 1 is,

….. (1)

Here, is the charge on sphere 1.

The potential of sphere 2 is,

….. (2)

Here, is the charge on sphere 2.

The total electric charge is,

….. (3)

The charges must be the same for both spheres when both spheres are connected by a thin wire. The electric potential is directly proportional to the charge. Hence, the electric potential on both spheres must be the same.

V₁=V₂

Here, V₁ is the electric potential on sphere 1 and V₂ is the electric potential on sphere 2.
Hence, the electric potential of sphere 1 is equal to the electric potential on sphere 2.

The relation between the radius of the sphere 1 and sphere 2 is,

The relation between the potential of the sphere 1 and sphere 2 is,

….. (4)

Substitute for in the equation (3) and solve for .

Therefore, the charge on the sphere 1 in the fraction of is .

Substitute for in the equation (4) and solve for .

Hence, the charge on the sphere 2 in the fraction of is .

The surface charge density of sphere 1 is,

The surface charge density of sphere 2 is,

Take the ratio of the surface charge densities of sphere 1 and 2.

Substitute for and for in the above equation.

Hence, the ratio of the surface charge densities of the spheres is .