A conductive sphere of radius Rs is concentrically placed inside a conductive sp

Question

A conductive sphere of radius Rs is concentrically placed inside a conductive spherical shell with inner radius R1 and outside radius R2 as shown in the figure below. Calculate the electric field as a function of r (the distance from the centre of the sphere), when a charge Q is added to inner sphere.

a) When there is no electrical contact between the sphere and the shell

b) When the two sheres are brought into electrical contact by adding a conductive cable between them.

c) With + and - signs show the charge distribution in each of the above cases.

Explanation / Answer

Relevant equations Flux = EA = Qencl/enaught A = 4?r^2 The electric field in the conducting material of the shell itself is zero. That's a property of a conductor. Use a spherical surface (for a sphere centered at O, having a radius between b and c) as your Gaussian surface. The E-field at all points on the Gaussian surface is zero. (The Gaussian surface lies entirely within conducting material.) Therefore, Gauss's Law tells us that the net charge enclosed within this sphere is zero. There is a charge of q1 on the center sphere. The only other place (inside of our Gaussian surface) where charge can reside is on the inner surface of the spherical shell. Therefore, the charge on the inner surface of the shell (r = b) is -q1, making the charge enclosed equal to zero. The net charge on the shell (the sum of the charges on both inner & outer surfaces) is +q2. But, since the net charge enclosed in our Gaussian surface is zero, there is a charge of -q1 on the inner surface -- to cancel the +q1 charge on the central sphere. The net charge on the shell is given by: qShell=qinner+qouter +q2=-q1+qouter Therefore, +q2++q1=qouter

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.