A metal ball of radius R1 is placed in the centre of two concentric metal sphere

Question

A metal ball of radius R1 is placed in the centre of two concentric metal spheres of radii R2 and R3 respectively, Fig. 6. The ball is charged to a charge Q1, and the outer sphere is charged to a charge Q2. The inner sphere is earthed, with the contact wire not touching the outer sphere.

a) Find the charge q induced on the inner sphere.

b) Find how the charge q is distributed between the inner and the outer surfaces of the sphere.

c) Sketch the graph of the electric field potential as a function of the distance r from the ball centre, assuming Q1 > 0 and Q2 > 0.

Explanation / Answer

A is between R1 and R2 at a distance a frm centre.. B is between R2 and R3 at a distance b frm centre.. then Pick a Gaussian surface (sphere since we are in spherical symmetry) center at the point charge and of radius a. This surface contains only the point charge, so qencl = q1. The formula for E gives EA =k q1/a2 For an electrostatic situation, inside of aconductor, there is no charge; i.e., qinside = 0.Also, Einside = 0 and there is no flux inside,finside = 0.ThusEB = 0 :Notice also that since the electric field at Bis zero, the total enclosed charge is zero, or q1 + q'2 = 0. Thereforeq'2 = -q1 : This verifies that the charge on the innersurface of a conducting shell is -q1, whereq1 is the charge is the charge enclosed by theshell. Find E at C, where OC = c. Here the Gaussian surface is a sphere cen-tered at the point charge q1 and of radius c.The enclosed charge in this sphere is all thecharge, or q1 + q2. The electric field at C is EC = k(q1 + q2)/c2..

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