Question

\(\mathrm{~A}-6.00-\mathrm{nC}\) 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

Question: Calculate the electric field and surface charge distribution due to a point charge of -6.00 nC in the center of a spherical conducting shell. a) Determine the electric field at r = 1.00 m (inside the point charge) b) Determine the electric field at r = 3.00 m (inside the shell) c) Determine the electric field at r = 5.00 m (outside the shell) d) Determine the surface charge distribution, σ, on the outside surface of the shell Answer: a) The electric field at r = 1.00 m (inside the point charge) is 0 N/C. b) The electric field at r = 3.00 m (inside the shell) is -22.28 N/C. c) The electric field at r = 5.00 m (outside the shell) is 1.80 N/C. d) The surface charge distribution on the outside surface of the shell is approximately 3.97 × 10^(-11) C/m².

Step-by-step solution

a) Determine the electric field at r = 1.00 m

Since the electric field inside the point charge is 0 and the sphere has a non-zero charge, by symmetry, there can be no electric field at any point inside the point charge. Therefore, the electric field at r = 1.00 m is 0 N/C.

b) Determine the electric field at r = 3.00 m

For this, we will use Gauss's Law and consider a Gaussian surface of radius r=3.00 m, centered on the point charge. Since the Gaussian surface is inside the conducting shell, the enclosed charge is only the point charge of -6.00 nC: \(Q_\text{enc} = -6.00 \times 10^{-9} \text{C}\) Gauss's Law states, ∮E⋅dA = \(Q_\text{enc}\)/ε₀, but since the electric field is radial, this simplifies to: \(E(4\pi r^2) = \frac{Q_\text{enc}}{\varepsilon_{0}}\) Now we can solve for E: \(E = \frac{Q_\text{enc}}{4\pi \varepsilon_{0}r^2} = \frac{-6.00 \times 10^{-9}}{4\pi \times 8.85\times10^{-12}\times (3.00)^{2}} \approx -22.28 \frac{\mathrm{N}}{\mathrm{C}}\)

c) Determine the electric field at r = 5.00 m

Since the Gaussian surface is now outside the shell, the enclosed charge includes both the point charge and the charge on the shell: \(Q_\text{enc} = -6.00 \times 10^{-9} + 7.00 \times 10^{-9} = 10^{-9} \text{C}\) Using Gauss's Law again: \(E = \frac{Q_\text{enc}}{4\pi \varepsilon_{0}r^2} = \frac{1.00 \times 10^{-9}}{4\pi \times 8.85\times10^{-12}\times (5.00)^{2}} \approx 1.80 \frac{\mathrm{N}}{\mathrm{C}}\)

d) Determine the surface charge distribution, σ, on the outside surface of the shell

First, we'll find the electric field just outside the shell, at \(r = 4.00+\epsilon \mathrm{~m}\). We can use the same enclosed charge calculation from part c: \(E = \frac{Q_\text{enc}}{4\pi \varepsilon_{0}r^2} = \frac{1.00 \times 10^{-9}}{4\pi \times 8.85\times10^{-12}\times (4.00+\epsilon)^{2}}\) Since ε is very small when compared to 4.00 m, we'll approximate the electric field as: \(E \approx \frac{1.00 \times 10^{-9}}{4\pi \times 8.85\times10^{-12}\times (4.00)^{2}} = 4.49 \frac{\mathrm{N}}{\mathrm{C}}\) Now, we can determine the surface charge distribution σ using the relationship between electric field and surface charge density: \(\sigma = \varepsilon_{0} E = (8.85\times10^{-12})(4.49) \approx 3.97 \times 10^{-11} \frac{\mathrm{C}}{\mathrm{m}^2}\)

Key concepts

Gauss's Law

Understanding Gauss's Law is essential when dealing with electric fields and charge distributions. This pivotal law in electromagnetism, formulated by Carl Friedrich Gauss, relates the electric field in a closed surface to the electric charge enclosed within that surface.

Gauss's Law is mathematically expressed as \( \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_{0}} \), where \( \vec{E} \) is the electric field, \( d\vec{A} \) is the differential area vector on the closed surface, \( Q_{\text{enc}} \) is the total charge enclosed, and \( \varepsilon_{0} \) represents the permittivity of free space. This integral can be easier to compute by taking advantage of symmetrical situations where the electric field is uniform in magnitude and direction over the surface.

Applying Gauss's Law helps to simplify complex problems into manageable calculations. For example, considering a spherical charge distribution, we can choose a Gaussian surface that's also a sphere concentric with the charge distribution, assuring us that the electric field is constant over our surface and allowing us to calculate the electric field easily in terms of the total enclosed charge. This approach was applied in step 2 of the provided solution, where a spherical Gaussian surface was selected to calculate the electric field inside a conducting shell.

Electric Charge

The concept of electric charge is fundamental to the understanding of electromagnetism. Electric charge is a property of matter that causes it to experience a force when near other electrically charged matter. There are two types of charges, positive and negative, and charges with the same sign repel each other while those with opposite signs attract.

Likewise, the unit of charge is the coulomb (C), and in the context of Gauss's Law, the total charge enclosed by a Gaussian surface is a central variable. As illustrated in the step-by-step solution, calculating the electric field at various points requires knowledge of the enclosed charge. For instance, inside a conducting shell (step 2), only the charge of the point charge is considered, whereas outside the shell (step 3), the total charge is the sum of the point charge and the charge on the shell itself.

This distinction is crucial as the enclosed charge changes depending on the position of the Gaussian surface relative to the charge distribution. It's important for students to not only grasp this concept but also to have the ability to compute the total charge in different scenarios to accurately apply Gauss's Law.

Surface Charge Distribution

Surface charge distribution, often denoted as \( \sigma \), is a measure of how much electric charge is distributed over a given surface area. Understanding this can provide insight into how charges are distributed on conductors and the resulting electric fields that are created. It's especially relevant in problems dealing with regions of transition, such as the surface of a conducting shell.

In electrostatic equilibrium, conductors carry their entire excessive charge on their surface, causing a distribution of surface charge that affects the electric field in and around the conductor. In step 4 of the solution, we calculated the surface charge distribution on the outer surface of a conducting shell by utilizing the electric field just outside the shell and the relationship \( \sigma = \varepsilon_{0} E \). Because the electric field was derived using Gauss's Law, we could easily relate it to the charge density.

When considering improvement advice for students, it's important to emphasize not only the concept of surface charge distribution itself but also how it relates to the broader principles of electric fields and electrostatics. By exploring how \( \sigma \) varies on different materials and shapes, students can gain a deeper understanding of electric fields and their calculation.