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.

Explanation

Let potential of sphere 1 and 2 are `V_(1)` and `V_(2)` respectively. Its radii a and b respectively the charge on them be `Q_(1)` and `Q_(2)` respectively and their capacitance are `C_(1)` and `C_(2)` respectively. Since two spheres are connected with a wire their potential will become equal.
`:.` Let `V_(1) = V_(2)=V` and In Q = CV , V is same
`:. Q prop C `
`:. (Q_(1))/(Q_(2))=(C_(1))/(C_(2))`
But `(C_(1))/(C_(2)) = (a)/(b)`
`:. (Q_(1))/(Q_(2))=(a)/(b)`
The ratio of electric field on the surface of both spheres,
`(E_(1))/(E_(2))=(kQ_(1)//a^(2))/(kQ_(2)//b^(2))=(Q_(1))/(Q_(2))xx(b^(2))/(a^(2))`
`:. (E_(1))/(E_(2))= (a)/(b)xx(b^(2))/(a^(2)) [ because` From equation (1)]
`:. (E_(1))/(E_(2))=(b)/(a)`
But `E= (sigma)/(in_(0))`
`:. (E_(1))/(E_(2))=(sigma_(1))/(sigma_(2)) [ :. ` From equation (2) ]
Hence, the charge density on the surface of sphere is inversely proportional to its radius.
Hence, larger sphere having more plane surface and smaller sphere having sharp surface . A sharp and pointed 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 a sphere is much higher than on its flatter portions.