Example 24.7 A Sphere Inside a Spherical Shell SOLVE IT a < r < b , noting that

Question

Example 24.7  A Sphere Inside a Spherical Shell

SOLVE IT

a < r < b,

noting that the charge inside this surface is +Q (the charge on the solid sphere). Because of the spherical symmetry, the electric field lines must be directed radially outward and be constant in magnitude on the gaussian surface. (Use the following as necessary: ke, r, Q, and a.)The charge on the conducting shell creates zero electric field in the region

r < b,

(for a < r < b)

E1 =  

r > c

, construct a spherical gaussian surface; this surface surrounds a total charge of

qin = Q + (2Q) = Q.

Therefore, model the charge distribution as a sphere with charge

Q

(for r > c)

E3 = 0

Construct a gaussian surface of radius r, where b < r < c, and note that qin must be zero because E3 = 0. Find the amount of charge qinner on the inner surface of the shell:

qin = qsphere + qinner

qinner = qin qsphere = 0 Q = Q

Finalize The charge on the inner surface of the spherical shell must be Q to cancel the charge +Q on the solid sphere and give zero electric field in the material of the shell. Because the net charge on the shell is 2Q, its outer surface must carry a charge Q.

An insulating sphere of radius a and carrying a charge Q surrounded by a conducting spherical shell carrying a charge 2Q.

Explanation / Answer

In region 1 or r<a,

By application of Gauss law E = kQenclosed/r^2

= kQ*(r^3/a^3) /r^2 = kQr/a^3

For region 2, E = kQ/r^2 because enclosed charge is Q

For region 3, E = 0 because Electric field inside conductor is zero.

For region 4, E = k*-Q/r^2

E = - kQ/r^2 because Q enclosed = Q-2Q =-Q

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