An insulating sphere of radius 0.250 m has uniform charge density 7.40×10⁻⁹ C/m³. A small object that can be treated as a point charge is released from rest just outside the surface of the sphere. The small object has a positive charge 3.70×10⁻⁶ C.

How much work does the electric field of the sphere do on the object as the object moves to a point very far from the sphere?

 W = 1.611 10⁻⁵ (1 / 0.250 - 1 / R_max)

 If R_max is infinite,  W = 6.444 10⁻⁵-5 J

Explanation:

The work is defined as

          W = F.dr

In this case the outside is repulsive since the two bodies have the same charge signand the angle is zero

The force is given by Coulomb's law

           F = k Q q / r²

The charge can be found from the density

            ρ = Q / V

            ρ = Q / (4/3 π R³)

            Q = ρ 4/3 π R³

The work is

             W = k ρ 4/3 π R³ q ∫ 1 / r² dr

We integrate

             W = 8,988 10⁹ 7.40 10⁻⁹ 4/3 π 0.250³ 3.70 10⁻⁶ (-1 / r)

We evaluate between the lower limit r = R and the upper limit r = R_max

              W = 16.11 10⁻⁶ (-1 / R_max + 1 / R)

              W = 1.611 10⁻⁵ (1 / 0.250 - 1 / R_max)

If R_max is infinite, the work is

            W = 1.611 10⁻⁵ 1 / 0.250

            W = 6.444 10⁻⁵-5 J

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .