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 the sphere A, QA be the charge on the sphere, and CA be the capacitance of the sphere. Let b be the radius of the sphere B, QB be the charge on the sphere, and CB be the capacitance of the sphere. Since, the two spheres are connected with wire, their potential V is equal.


Electric field generated by a charge, E =


Where,


q = charge


d = distance from origin


ϵ0 = permittivity of space


Let EA be the electric field for sphere A and EB be the electric field for sphere B. Therefore their ratio,


=


However, = and =


⇒ =


∴ = =


Hence, the ration of electric fields at surface is


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