Question -
Answer -
Let a be the radius of asphere A, QA be the charge on the sphere, and CA bethe capacitance of the sphere. Let b be the radius of a sphereB, QB be the charge on the sphere, and CB bethe capacitance of the sphere. Since the two spheres are connected with a wire,their potential (V) will become equal.
Let EAbe the electric field ofsphere 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 ofelectric fields at the surface is
A sharp and pointed end can be treated as asphere of very small radius and a flat portion behaves as a sphere of muchlarger radius.Therefore, charge density on sharp and pointed ends of theconductor is much higher than on its flatter portions.