28) A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, as shown in the figure. The sphere carries an excess charge of -500 nC. A point charge of +300 nC is present at the center. The surface charge density on the inner spherical surface is closest to A) zero. B) +4.0 × 10^-8 C/m². C) +6.0 × 10^-8 C/ m². D) -4.0 × 10^-8 C/ m². E) -6.0 × 10^-8 C/ m².

Question

Quizplus · Accepted Answer

Since the excess charge on the spherical shell is negative, the potential at any point inside the shell is constant and equal to the potential on the surface of the shell. Thus, the potential at the center of the sphere is the same as the potential on the outer surface of the shell, which we can find using the equation for the electric potential of a charged sphere:
V = kQ/R
where k is the Coulomb constant, Q is the charge on the spherical shell, and R is the radius of the shell. Plugging in the values given in the problem, we get:

V = (9×10^9 N·m^2/C^2)(-500×10^-9 C)/(1.20 m)
V = -3.75×10^6 V

Since the point charge at the center of the sphere is positive, it will experience an attractive force toward the negative charge on the spherical shell. This force will cause the positive charge to move toward the inner surface of the shell. Any excess charge on the inner surface of the shell will distribute itself evenly over the surface, since like charges repel each other. Thus, the charge on the inner surface of the shell will be given by:

Q' = -Q(0.80/1.20) = -333.3×10^-9 C

The surface charge density on the inner surface of the shell is then:

σ = Q'/(4πR^2) = -(333.3×10^-9 C)/(4π(0.80 m)^2) = -1.64×10^-7 C/m^2

This is equivalent to -1.64×10^-1 C/m^2, or -1.64×10^5 nC/m^2. The answer closest to this value is D, -4.0 × 10^-8 C/m^2.

28) a Hollow Conducting Spherical Shell Has Radii of 0