Question

A hollow non-conducting spherical shell has inner radius R1 = 5 cm and outer radius R2 = 19 cm. A charge Q = -35 nC lies at the center of the shell. The shell carries a spherically symmetric charge density rho = Ar for R1 < r < R2 that increases linearly with radius, where A = 16 μC/m4.
a. What is the radial electric field at the point r = 0.5(R1+R2)? Give your answer in units of kN/C. sig.gif?tid=9K89-A2-1C-48-A922-20866
b. What is the radial electric field at the point r = 2R2? Give your answer in units of kN/C.

Answer 1

Answer:

a. E = -13.8 kN/C

b. E = +8.51 kN/C

Explanation:

We will apply Gauss' Law to the regions where the electric field is asked.

Gauss' Law states that if you draw an imaginary surface enclosing a charge distribution, then the electric field through the imaginary surface is equal to the total charge enclosed by this surface divided by electric permittivity.

$\int\vec{E}d\vec{a} = \frac{Q_{\rm enc}}{\epsilon_0}$

a. For this case, we will draw the imaginary surface between the inner and outer shell of the sphere. The total charge enclosed by this surface will be equal to the sum of the charge Q at the center and charge of the shell within the volume from R1 and r.

Here, r = 0.5(R1+R2) = 12 cm.

$E4\pi r^2 = \frac{Q_{\rm enc}}{\epsilon_0}\\Q_{\rm enc} = Q + \rho V_{\rm enc} = Q + (Ar) (\frac{4}{3}\pi (r^3 - R_1^3)) = (-35\times 10^{-9}) + (16\times 10^{-6})(12\times 10^{-2})(\frac{4}{3}\pi((12\times 10^{-2})^3 - (5\times 10^{-2})^3)) = -2.21\times 10^{-8}~C$

$E = \frac{-2.21\times 10^{-8}}{4\pi (12\times10^{-2})^2 \epsilon_0} = -1.38\times 10^4~N/C\\E = -13.8~kN/C$

b. For this case, we will draw the imaginary surface on the outside of the shell.

The total charge enclosed by this surface will be equal to the sum of the charge at the center and the total charge of the shell.

$Q_{\rm enc} = Q + \rho V = Q + (Ar)[\frac{4}{3}\pi (R_2^3 - R_1^3)]\\Q_{\rm enc} = (-35\times 10^{-9}) + [(16\times 10^{-6})(38\times 10^{-2})][\frac{4}{3}\pi((19\times 10^{-2})^3 - (5\times 10^{-2})^3)]\\Q_{\rm enc} = 1.36\times 10^{-7}~C$

$E = \frac{1.36\times 10^{-7}}{4\pi (38\times10^{-2})^2 \epsilon_0} = 8.51\times 10^3~N/C\\E = 8.51~kN/C$