A \(-6.00-n C\) point charge is located at the center of a conducting spherical shell. The shell has an inner radius of \(2.00 \mathrm{~m},\) an outer radius of \(4.00 \mathrm{~m},\) and a charge of \(+7.00 \mathrm{nC}\) a) What is the electric field at \(r=1.00 \mathrm{~m} ?\) b) What is the electric field at \(r=3.00 \mathrm{~m} ?\) c) What is the electric field at \(r=5.00 \mathrm{~m} ?\) d) What is the surface charge distribution, \(\sigma,\) on the outside surface of the shell?

Short Answer

Expert verified
Answer: The electric field values and surface charge distribution are as follows: a) At r = 1m (inside the point charge): \(E_1 \approx 0 \, N/C\) b) At r = 3m (inside the spherical shell): \(E_2 \approx -11.623 \, N/C\) c) At r = 5m (outside the spherical shell): \(E_3 \approx 3.59 \times 10^{-3} \, N/C\) d) Surface charge distribution on the outside surface of the shell: \(\sigma \approx 1.39 \times 10^{-10} \, C/m^2\)

Step by step solution

01

Find the electric field at r = 1m (inside the point charge)

The electric field inside a point charge is uniformly zero. Therefore, the electric field at r = 1m is: \(E_1 = 0\)
02

Find the electric field at r = 3m (inside the spherical shell)

Applying Gauss's law for a Gaussian surface within the spherical shell (sphere of radius r = 3m), the electric field at r = 3m is: \(\oint \vec{E_2} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0}\) \(q_{enclosed}=q\) (The point charge q is enclosed by the Gaussian surface) \(E_2 4 \pi (3)^2 = \frac{-6 \times 10^{-9}}{8.85 \times 10^{-12}}\) \(E_2 = \frac{-6 \times 10^{-9}}{8.85 \times 10^{-12} \times 36\pi}\) \(E_2 \approx -11.623 \, N/C\)
03

Find the electric field at r = 5m (outside the spherical shell)

Applying Gauss's law for a Gaussian surface outside the spherical shell (sphere of radius r = 5m), the electric field at r = 5m is: \(\oint \vec{E_3} \cdot d\vec{A} = \frac{Q_{total}}{\epsilon_0}\) \(q_{total}=q+Q\) \(E_3 4 \pi (5)^2 = \frac{(7 - 6) \times 10^{-9}}{8.85 \times 10^{-12}}\) \(E_3 = \frac{1 \times 10^{-9}}{8.85 \times 10^{-12} \times 100\pi}\) \(E_3 \approx 3.59 \times 10^{-3} \, N/C\)
04

Calculate the surface charge distribution on the outside surface of the shell

The shells charge Q is uniformly distributed on its outer surface. The surface charge density \(\sigma\) can be calculated as follows: \(\sigma = \frac{Q}{A}\) Here, \(A=4\pi b^2\) is the surface area of the outer surface of the spherical shell. \(\sigma = \frac{7 \times 10^{-9}}{4 \pi (4)^2}\) \(\sigma \approx 1.39 \times 10^{-10} \, C/m^2\) The electric field and surface charge distribution are as follows: a) \(E_1 \approx 0 \, N/C\) b) \(E_2 \approx -11.623 \, N/C\) c) \(E_3 \approx 3.59 \times 10^{-3} \, N/C\) d) \(\sigma \approx 1.39 \times 10^{-10} \, C/m^2\)

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