Problem 76 - University Physics with Modern Physics

Question

Two infinite, uniformly charged, flat nonconducting surfaces are mutually perpendicular. One of the surfaces has a charge distribution of \(+30.0 \mathrm{pC} / \mathrm{m}^{2}\), and the other has a charge distribution of \(-40.0 \mathrm{pC} / \mathrm{m}^{2}\). What is the magnitude of the electric field at any point not on either surface?

Step-by-Step Solution

Answer

Question: Compute the magnitude of the electric field at any point not on either surface created by two mutually perpendicular infinitely large flat nonconducting surfaces. The first surface has a positive uniform charge distribution of 30.0 nC/m², and the second surface has a negative uniform charge distribution of 40.0 nC/m². Answer: The magnitude of the electric field at any point not on either surface is approximately 2.83 N/C.

Step 1: Calculate electric fields due to each surface individually

Using Gauss' Law, the electric field (E) for an infinite charged plane with a surface charge density (σ) can be given by: E = \(\frac{σ}{2ε_0}\) Here, ε₀ = 8.85 x 10⁻¹² C²/N.m² is the vacuum permittivity. Let's denote the electric field due to the first surface with the positive charge distribution as \(E_1\), and the electric field due to the second surface with negative charge distribution as \(E_2\). Now, we can calculate these fields as: \(E_1 = \frac{+30.0 \times 10^{-12}}{2(8.85 \times 10^{-12})}\), \(E_2 = \frac{-40.0 \times 10^{-12}}{2(8.85 \times 10^{-12})}\)

Step 2: Combine the electric fields vectorially

Since the surfaces are mutually perpendicular, their electric fields will also be perpendicular to each other. We can use the Pythagorean theorem to find the magnitude of the resultant electric field (\(E_{total}\)): \(E_{total} = \sqrt{E_1^2 + E_2^2}\) Calculate the values of \(E_1\) and \(E_2\): \(E_1 \approx \frac{30 \times 10^{-12}}{2(8.85 \times 10^{-12})} \approx 1.69 \,\text{N/C}\), \(E_2 \approx \frac{-40 \times 10^{-12}}{2(8.85 \times 10^{-12})} \approx -2.26 \,\text{N/C}\) Now, substitute these values and compute the magnitude of the resultant electric field: \(E_{total} \approx \sqrt{(1.69 \,\text{N/C})^2 + (-2.26 \,\text{N/C})^2} \approx 2.83 \,\text{N/C}\) Thus, the magnitude of the electric field at any point not on either surface is approximately 2.83 N/C.

Key Concepts

Gauss' Law

Gauss' Law is a powerful tool in electromagnetism that helps calculate electric fields for different charge distributions. It states that the net electric flux passing through a closed surface is directly proportional to the total enclosed charge. This can be expressed mathematically as:

• \( \Phi_E = \frac{Q_{enc}}{\epsilon_0} \)

Here, \( \Phi_E \) is the electric flux, \( Q_{enc} \) is the enclosed charge, and \( \epsilon_0 \) is the vacuum permittivity.
Gauss' Law is useful in situations involving high symmetry, such as with spherical, cylindrical, or planar charge distributions. In the exercise, Gauss’ Law helps us find the electric field due to an infinite plane charge distribution. By using the formula:

• \( E = \frac{\sigma}{2\epsilon_0} \)

we efficiently determine the field created by each surface. The field from an infinite plane is constant and perpendicular to the plane, which simplifies calculations when analyzing combinations of charge distributions.

Surface Charge Density

Surface charge density, \( \sigma \), is a measure of how much charge exists on a surface per unit area. It is expressed in units of coulombs per square meter (C/m\(^2\)). Surface charge density is crucial when dealing with planar charge distributions, like in the case of electrically charged surfaces in the exercise.

• \( \sigma = \frac{Q}{A} \)

In the exercise, two surfaces are described with charge densities of \(+30.0\, \mathrm{pC/m^2}\) and \(-40.0\, \mathrm{pC/m^2}\). Surface charge density affects the electric field produced by using it in formulas derived from Gauss' Law.
Because the surfaces are infinite and uniformly charged, the electric field produced is uniform and doesn’t depend on the distance from the surface, making calculations straightforward. Understanding \( \sigma \) helps in grasping how distributions of charges affect surrounding electric fields.

Vacuum Permittivity

Vacuum permittivity, symbolized as \( \epsilon_0 \), is a fundamental constant in electromagnetism, representing the ability of a vacuum to allow electric field lines to pass through it. It is also known as the electric constant and has a fixed value of approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N.m}^2 \). Vacuum permittivity appears in several key electromagnetic equations, including Gauss' Law.

• Influences how electric fields interact in free space.
• Part of calculations determining the strength of electric fields due to charges.

In the given exercise, \( \epsilon_0 \) factors into the formula \( E = \frac{\sigma}{2\epsilon_0} \), which allows calculating the electric field near an infinite plane. This concept ensures that calculated fields represent physical phenomena accurately. Understanding vacuum permittivity is essential for interpreting the behavior of electric fields in various contexts, especially in theoretical physics and engineering.