Problem 56 - Fundamentals of Physics

Question

The electric field in a particular space is \(E=(x+2) \mathrm{i} \mathrm{N} / \mathrm{C}\), with \(x\) in meters. Consider a cylindrical Gaussian surface of radius \(20 \mathrm{~cm}\) that is coaxial with the \(x\) axis. One end of the cylinder is at \(x=0 .\) (a) What is the magnitude of the electric flux through the other end of the cylinder at \(x=2.0 \mathrm{~m} ?\) (b) What net charge is enclosed within the cylinder?

Step-by-Step Solution

Answer

(a) Electric flux is \(0.16\pi\, N\cdot m^2/C\); (b) Net charge is \(4.44\times 10^{-12}\pi\, C\).

Step 1: Formula for Electric Flux

The electric flux through a surface is given by the formula \(\Phi_E = \int \mathbf{E} \cdot d\mathbf{A} \). In this case, the surface is a flat disk at \(x=2.0\, m\), perpendicular to the x-axis, so \(d\mathbf{A} = \mathbf{i} dA\) and the electric field also points in the \(\mathbf{i}\) direction.

Step 2: Calculate Area of End Cap

The area of the circular end cap at \(x=2.0\, m\) is \(A = \pi r^2\) where \(r = 0.20\, m\). Calculate this to find \(A.\) \[A = \pi (0.20)^2 = 0.04\pi\, m^2\].

Step 3: Evaluate Electric Field at x=2.0 m

Substitute \(x=2.0\, m\) into the electric field function \(E(x) = (x+2) \mathbf{i}\, N/C\). This gives:\[E(2) = (2 + 2) = 4\, N/C\].

Step 4: Calculate Electric Flux at x = 2.0 m

Substitute the electric field and area into the flux formula:\[\Phi_E = E \cdot A = 4\, N/C \times 0.04\pi\, m^2 = 0.16\pi\, N\cdot m^2/C\].

Step 5: Net Charge Enclosed Using Gauss's Law

From Gauss's Law, the net charge \(Q_{enc}\) enclosed by the cylinder is given by \(\Phi_E = \frac{Q_{enc}}{\varepsilon_0}\). Solving for \(Q_{enc}\):\[Q_{enc} = \Phi_E \times \varepsilon_0 = 0.16\pi\, N\cdot m^2/C \times 8.85 \times 10^{-12}\, C^2/N\cdot m^2 = 4.44\times 10^{-12}\pi\, C\].

Key Concepts

Gaussian surface

A Gaussian surface is an imaginary surface used in calculating flux, particularly in statics and electrodynamics. It helps simplify the application of Gauss's law by providing a path through which the electric flux can be calculated.
The surface is especially useful because it allows us to consider a symmetrical area based on the given distribution of charges. In this problem, the Gaussian surface is a cylindrical shape aligning with the x-axis.

• The radius of the cylinder is 0.20 m, which matches the size of typical Gaussian surfaces adapted to the problem's symmetry.
• The cylinder perfectly aligns with variations in the electric field, simplifying the integration process when calculating the electric flux.
• Since one end of our Gaussian surface is fixed at x=0, any change in electric characteristics can be neatly calculated at the other end, x=2.0 m.

This setup makes it possible to analyze how the electric field interacts with this closed surface, helping us calculate the net charge enclosed.

electric field

An electric field describes the force a charged particle would experience at a given point in space. In this exercise, the electric field is expressed as \(E=(x+2) \, \mathbf{i}\, \mathrm{N}/\mathrm{C}\), an expression that depends on the variable \(x\) to show its change along the x-axis.

• The formula indicates that the electric field has only an i-component, meaning it is aligned with the x-axis and perpendicular to the respective surface of interest on our Gaussian surface.
• As \(x\) changes position from 0 to 2.0 m, the field is recalculated to reveal the point-specific electric flux passing through the surface.
• This vector representation helps visualize how strong and in what direction the field acts on a charged object placed within the field.

Plugging \(x = 2.0\, \mathrm{m}\) into the formula yields \(E(2) = 4 \, \mathrm{N}/\mathrm{C}\), a crucial value for calculating electric flux.

Gauss's law

Gauss's law is a powerful principle that relates the electric flux through a closed surface to the enclosed charge. It states that the net electric flux through a closed surface is proportional to the enclosed electric charge. In mathematical terms, \(\Phi_E = \frac{Q_{enc}}{\varepsilon_0}\).

• The formula is rearranged in this exercise to solve for \(Q_{enc}\), the net charge enclosed by the cylinder.
• Knowing the electric flux from previous calculations, we can use Gauss's law to find this total charge enclosed by the surface.
• \(\varepsilon_0\) is the permittivity of free space, a constant value used to quantify how much electric field is permitted through space, which equals approximately \(8.85 \times 10^{-12} \, \mathrm{C}^2/\mathrm{N} \cdot \mathrm{m}^2\).

This application allows us to see how charges within the cylinder impact the external electric field in electromagnetic scenarios.

net charge

Net charge refers to the total charge within a closed system, accounting for all positive and negative charges present. It is either calculated directly or derived from the application of Gauss's law, which is what we do here.

• In this problem, using the derived electric flux \(\Phi_E = 0.16\pi \, \mathrm{N} \cdot \mathrm{m}^2/\mathrm{C}\), we find the net charge using Gauss's law.
• The enclosed charge \(Q_{enc}\) is then calculated by multiplying the electric flux by the permittivity of free space, \(\varepsilon_0\).
• This results in a total net charge of \(Q_{enc} = 4.44\times 10^{-12}\pi \, \mathrm{C}\).

The net charge gives valuable insight into how many excess positive or negative charges reside within this Gaussian surface, aiding in understanding the broader electrical dynamics at play.