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 reuslt obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

Explanation

The charge flows from the sphere at higher potential to the other at lower potential till their potentials become equal. After sharing the charges on two spheres would be.
`(Q_(1))/(Q_(2)) = (C_(1) V)/(C_(2)V)` Where `C_(1). C_(2)` are the capacities of two spheres.
But `(C_(1))/(C_(2)) = (a)/(b) :. (Q_(1))/(Q_(2)) = (a)/(b)`
Ratio of surface density of charge on the two spheres
`(sigma_(1))/(sigma_(2)) = (Q_(1))/(4 pi a^(2)) xx (4 pi b^(2))/(Q_(2)) = (Q_(1))/(Q_(2)). (b^(2))/(a^(2)) = ((a)/(b)) ((b^(2))/(a^(2))) = (b)/(a)`
Hence ratio of electric fields at the surface of two spheres
`E_(1))/(E_(2)) = (sigma_(1))/(sigma_(2)) = (b)/(a)`
A sharp and potential end can be treated as a sphere of very small radius and a flat portion behaves as a sphere of much larger radius. Therefore, charge density on sharp and pointed ends of conductor is much higher than on its flatter portions.