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.

Let a be the radius of a sphere A, Q_A be the charge on the sphere, and C_A be the capacitance of the sphere.

Let b be the radius of a sphere B, Q_B be the charge on the sphere, and C_B be the capacitance of the sphere.

Since the two spheres are connected with a wire, their potential (V) will become equal.

Let E_A be the electric field of sphere A and E_B be the electric field of sphere B. Therefore, their ratio,

E_A/E_B  = (Q_A /4π ε₀ a² ) x (b² 4π ε₀ )/Q_B

E_A/E_B  = (Q_A / Q_B) x (b² / a² )  ———————-Eq-1

However , Q_A / Q_B =C_AV/ C_B V

and  C_A/ C_B =a/b

Therefore  Q_A / Q_B =a/b  ———————-Eq-2

Putting the values of Eq-2 in Eq-1 ,we obtain

E_A/E_B  = (a/b ) x (b² / a² )=b/a

Therefore ,the ratio of electric fields at the surface is b/a.