Question

An electric field in free space is \(\mathbf{E}=\left(5 z^{2} / \epsilon_{0}\right) \hat{\mathbf{a}}_{z} \mathrm{~V} / \mathrm{m}\). Find the total charge contained within a cube, centered at the origin, of \(4-\mathrm{m}\) side length, in which all sides are parallel to coordinate axes (and therefore each side intersects an axis at \(\pm 2\) ).

Short answer

Give a short answer based on the step by step solution above. The total charge contained within the cube is \(Q = 1280\epsilon_0\).

Step-by-step solution

Express the electric field in terms of x, y, and z components

The given electric field has only the z-component and can be expressed as: \(\mathbf{E} = (0, 0, 5z^2/\epsilon_0)\)

Calculate electric flux through each surface of the cube

To find the total electric flux, we'll calculate the electric flux through each of the six faces of the cube. The electric field is only along the z-direction, so for the top and bottom faces, the normal components of the electric field are equal to the z-component. Surfaces parallel to the xy-plane (Top and bottom faces): Flux through each face = \(\int \mathbf{E} \cdot \hat{\mathbf{n}}_{top/bottom} dS = \frac{5z^2}{\epsilon_0} \int_{-2}^{2}\int_{-2}^{2} dxdy\) As for the other four faces, the normal components have no contribution from the electric field, so the electric flux through those faces is zero. Now let's compute the electric flux through top face (z=2): Flux through top face (z=2) = \(\frac{5(2)^2}{\epsilon_0} \int_{-2}^{2}\int_{-2}^{2} dxdy= \frac{20}{\epsilon_0} \int_{-2}^{2}\int_{-2}^{2} dxdy\) Similarly for bottom face (z=-2): Flux through bottom face (z=-2) = \(\frac{5(-2)^2}{\epsilon_0} \int_{-2}^{2}\int_{-2}^{2} dxdy= \frac{20}{\epsilon_0} \int_{-2}^{2}\int_{-2}^{2} dxdy\)

Determine total electric flux

Now let's calculate the total electric flux from the 6 faces of the cube. The total electric flux is the sum of electric fluxes of all individual faces: Total Electric flux = Flux through top face + Flux through bottom face + Flux through 4 side faces Since the electric flux through the side faces is zero, we only need to add the contributions from the top and bottom faces: Total Electric Flux = \(2 \cdot \frac{20}{\epsilon_0} \int_{-2}^{2}\int_{-2}^{2} dxdy\)

Use Gauss's Law to find total charge

According to Gauss's law, the total enclosed charge Q is given by: \(\oint \mathbf{E}\cdot \hat{\mathbf{n}} dS = \frac{Q}{\epsilon_0}\) Applying this to our scenario: Total Electric Flux = \(\frac{Q}{\epsilon_0}\) Therefore, \(Q = Total Electric Flux \cdot \epsilon_0 = 2 \cdot 20 \int_{-2}^{2}\int_{-2}^{2} dxdy \cdot \epsilon_0 = 80 \int_{-2}^{2}\int_{-2}^{2} dxdy \cdot \epsilon_0\) Calculating the integral: \(Q = 80 \left[ (2+2)\cdot(2+2)\right] \cdot \epsilon_0 = 80\cdot 16 \cdot \epsilon_0 = 1280\epsilon_0\)

Express the answer

The total charge contained within the cube is \(Q = 1280\epsilon_0\)

Key concepts

Electric Flux

Electric flux is a measure of the electric field lines passing through an area. Imagine electric field lines as roads and the area as a checkpoint. The number of lines passing through the checkpoint is the electric flux. In mathematical terms, electric flux \( \Phi_E \) through a surface is given by \( \Phi_E = \int \mathbf{E} \cdot d\mathbf{A} \), where \( \mathbf{E} \) is the electric field and \( d\mathbf{A} \) is a differential area vector, perpendicular to the surface.
In our exercise, the electric field \( \mathbf{E} = (5z^2/\epsilon_0) \hat{\mathbf{a}}_z \) allows us to calculate electric flux through the faces of a cube. Since this field only has a component in the z-direction, only the top and bottom faces of the cube contribute to the electric flux. The side faces do not as their normal vectors are perpendicular to the z-axis, thus \( \mathbf{E} \cdot \mathbf{n} = 0 \).
Electric flux is the key tool in applying Gauss's Law, which relates flux to the charge within a surface. Understanding flux as 'field line counting' makes determining enclosed charges in symmetric structures like cubes more intuitive.

Charge Distribution

Charge distribution describes how electric charge is arranged in space. In Gauss's Law, this is crucial because it allows us to relate the total electric flux through a closed surface to the charge contained within that surface. The distribution impacts how we choose surfaces for analysis.
In symmetric charge distributions, like the scenario with our cube, Gauss’s Law becomes especially useful. In our exercise, the electric field depends on the square of \( z \), indicating that the charge influences areas parallel to the xy-plane dominantly.

• Gauss's surfaces (like the cube here) should ideally conform to the symmetry of the charge distribution.
• For the cube, the distribution of the field hints at possible internal charge complexities.

Understanding charge distribution's effects is crucial because it dictates where the field is stronger or weaker, aligning perfectly with chosen Gaussian surfaces to simplify computations.

Electric Field in Free Space

An electric field in free space is a spatial vector quantity signifying how a charged object influences the region around it. Free space has a permittivity \( \epsilon_0 \) that measures its resistance to electric field lines, affecting how charges distribute.
Specific to our scenario, we have \( \mathbf{E}=(5z^2/\epsilon_{0})\hat{\mathbf{a}}_{z} \). This tells us the electric field changes with \( z^2 \), and is directed entirely along the z-axis. Notably, this means any force due to the field acts vertically along this axis.
The electric field in free space has characteristics like:

• Homogeneity: It doesn’t depend on distance (other than \( z^2 \)).
• Uni-directionality: No x or y components are present.

Understanding how electric fields behave in free space as opposed to other mediums is essential for applying laws like Gauss's, where medium properties affect outcomes.

Coordinate Systems in Electromagnetics

Coordinate systems are invisible frameworks used to define positions in space for electromagnetics analysis. The most common one is the Cartesian coordinate system, utilized in our exercise. It aligns with the natural orientation of the problem - cubes aligned with axes.
The key here is the simplicity they add:

• Each side of the cube aligns with a coordinate plane, simplifying the calculation of surface normals and flux.
• The use of standard coordinate systems (like Cartesian) helps in visualizing electric field directions as scalars can be easily interpreted in terms of vector components \( (x, y, z) \).

Utilizing appropriate coordinate systems is crucial in electromagnetics to simplify complex integrations and visualize field behaviors naturally along axis alignments, which is exemplified in the cube problem.