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Electrostatic Potential and Capacitance
two charged conducting spheres of radii a and b ar...

Question:
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, Q_A be the charge on the sphere, and C_A be the capacitance of the sphere. Let b be the radius of a sphere B, Q_Bbe the charge on the sphere, and C_B be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (V) will become equal.

Let E_Abe the electric field of sphere A and E_B 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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