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Asked by Varada Last Modified
Answer 
Apoorva Purohit
Let a be the radius of a sphere A, QAbe the charge on the sphere, and CA be the capacitance of the sphere. Let b be the radius of a sphere B, QB be the charge on the sphere, and CB be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (V) will become equal.
Let EAbe the electric field of sphere A and EB be the electric field of sphere B. Therefore, their ratio,
Putting the value of (2) in (1), we obtain
Therefore, the ratio of electric fields at the surface is
.
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 the conductor is much higher than on its flatter portions.
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