Problem 53 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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 1: 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: \(E_1 = 0\)

Step 2: 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) \(E_2 4 \pi (3)^2 = \frac{-6 \times 10^{-9}}{8.85 \times 10^{-12}}\) \(E_2 = \frac{-6 \times 10^{-9}}{8.85 \times 10^{-12} \times 36\pi}\) \(E_2 \approx -11.623 \, N/C\)

Step 3: 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: \(\oint \vec{E_3} \cdot d\vec{A} = \frac{Q_{total}}{\epsilon_0}\) \(q_{total}=q+Q\) \(E_3 4 \pi (5)^2 = \frac{(7 - 6) \times 10^{-9}}{8.85 \times 10^{-12}}\) \(E_3 = \frac{1 \times 10^{-9}}{8.85 \times 10^{-12} \times 100\pi}\) \(E_3 \approx 3.59 \times 10^{-3} \, N/C\)

Step 4: 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) \(\sigma \approx 1.39 \times 10^{-10} \, C/m^2\)

Key Concepts

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism that connects the electric field across a closed surface with the charge contained within it. This powerful law is expressed as:\[\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0}\]Here, the left side represents the electric flux through a closed surface, and \(Q_{enclosed}\) is the net charge inside that surface. The constant \(\epsilon_0\) is the permittivity of free space.

When using Gauss's Law, it is important to choose a Gaussian surface that matches the symmetry of the charge distribution. For a sphere, which is the focus in this exercise, a spherical surface simplifies calculations.

• Within the conducting shell, the charges redistribute to ensure the electric field is zero. Thus, for any point inside the shell, Gauss's Law confirms that no net electric field exists if the point charge is outside.
• Outside the spherical shell, treating all enclosed charges as a single point charge simplifies the electric field determination using Gauss's Law.

Surface Charge Distribution

The distribution of charge on a conductor's surface is crucial for understanding electric fields around it. On a conducting spherical shell, free charges redistribute in response to internal or external influences.

In static conditions, the electric field inside a conductor is zero. This means charges only reside on the surface in a conducting spherical shell:

• A neutralizing charge, equal in magnitude but opposite in sign to any internal charge, is induced on the inner surface of the shell.
• The outer surface contains any additional charge distributed throughout its area.

The surface charge density \( \sigma \) describes how charge is distributed over the surface. For a sphere, it is given by:\[\sigma = \frac{Q}{A} = \frac{Q}{4 \pi b^2}\] where \(Q\) is the total charge and \(b\) is the radius of the spherical surface.

This formula helps calculate how densely packed charges are on the shell’s outer surface.

Conducting Spherical Shell

A conducting spherical shell is a common structure in electrostatics, particularly in problems involving electric fields and charge distributions. Surrounding a charged object, it provides unique electrical properties.

Some important characteristics include:

• **Inner Radius** and **Outer Radius**: Define the thickness of the shell and area over which charge resides.
• A **uniformly distributed charge** on the surface, contingent on the influence of other charges present.
• The ability to shield its inner cavity from external electric fields, maintaining a zero field inside.

When a charge lies at its center, the shell reacts by inducing charges on its surfaces:

• The inner surface aligns charge to neutralize internal fields.
• The outer surface expresses the net charge.

Understanding these properties facilitates calculations using Gauss’s Law, illuminating the way fields behave not only inside but also outside the shell.