Q. 203.6( 9 Votes )

# 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 the 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 the sphere B, Q_{B} be the charge on the sphere, and C_{B} be the capacitance of the sphere. Since, the two spheres are connected with wire, their potential V is equal.

Electric field generated by a charge, E =

Where,

q = charge

d = distance from origin

ϵ_{0} = permittivity of space

Let E_{A} be the electric field for sphere A and E_{B} be the electric field for sphere B. Therefore their ratio,

=

However, = and =

⇒ =

∴ = =

Hence, the ration of electric fields at surface is

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