Class 12 Physics - Chapter Electrostatic Potential and Capacitance NCERT Solutions | Two charged conducting spheres of radii

Welcome to the NCERT Solutions for Class 12th Physics - Chapter Electrostatic Potential and Capacitance. This page offers a step-by-step solution to the specific question from Exercise 1, Question 20: two charged conducting spheres of radii a and b ar....
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   

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