Class 12 Physics, CBSE and UP Board, Electrostatic Potential and Capacitance, Additional Exercise,Chapter-2 Exercise 2.20, NCERT Solutions for Class 12 Physics
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.
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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,
EA/EB = (QA /4π ε0 a2 ) x (b2 4π ε0 )/QB
EA/EB = (QA / QB) x (b2 / a2 ) ———————-Eq-1
However , QA / QB =CAV/ CB V
and CA/ CB =a/b
Therefore QA / QB =a/b ———————-Eq-2
Putting the values of Eq-2 in Eq-1 ,we obtain
EA/EB = (a/b ) x (b2 / a2 )=b/a
Therefore ,the ratio of electric fields at the surface is b/a.