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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 surfacesof 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 EAbe the electric field of sphere A and EB be the electric field of sphere B. Therefore, their ratio,

image232 NCERT Solutions class-12 physics Electrostatic Potentia and Capacitance Part-2

image233 NCERT Solutions class-12 physics Electrostatic Potentia and Capacitance Part-2

However image234 NCERT Solutions class-12 physics Electrostatic Potentia and Capacitance Part-2

And image235 NCERT Solutions class-12 physics Electrostatic Potentia and Capacitance Part-2

image236 NCERT Solutions class-12 physics Electrostatic Potentia and Capacitance Part-2

Putting the value of (2) in (1), we obtain

image237 NCERT Solutions class-12 physics Electrostatic Potentia and Capacitance Part-2

Therefore, the ratio of electric fields at the surface isimage238 NCERT Solutions class-12 physics Electrostatic Potentia and Capacitance Part-2.

 

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