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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.
Text solutionVerified
Let a be the radius of a sphere A, be the charge on the sphere, and be the capacitance of the sphere. Let b be the radius of a sphere B, be the charge on the sphere, and be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (V) will become equal.
Let be the electric field of sphere A and be the electric field of sphere B. Therefore, their ratio,
...(1)
However,
And
...(2)
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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Question Text | 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. |
Updated On | Mar 2, 2023 |
Topic | Electrostatic Potential and Capacitance |
Subject | Physics |
Class | Class 12 |
Answer Type | Text solution:1 Video solution: 2 |
Upvotes | 197 |
Avg. Video Duration | 10 min |