Question 20: 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.
Answer:
Let a be the radius of a sphere A, QA be 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 EA be 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
Comments
The second part of this question can be Explained by the phenomenon of Corona Discharge.
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