Question

Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain, why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions?

Answer 1

Let, the charge possessed by two conducting spheres of radii a and b be q₁ and q₂ respectively.

Potential on the surface of first sphere, V₁ = \(\frac{1}{4 \pi \varepsilon_0}.\frac{q_1}{a}\)

Potential on the surface of second sphere, V₂ = \(\frac{1}{4 \pi \varepsilon_0}.\frac{q_2}{a}\)

Till the potentials of two conductors become equal the flow of charges continue.

Therefore, the ratio of the electric field of first sphere to that of the second sphere is b:a.

The surface charge densities of the two spheres are given as

This implies,
The surface charge densities are inversely proportional to the radii of the sphere.