Question

A cube has an edge length of \(1.00 \mathrm{~m} .\) An electric field acting on the cube from outside has a constant magnitude of \(150 \mathrm{~N} / \mathrm{C}\) and its direction is also constant but unspecified (not necessarily along any edges of the cube). What is the total charge within the cube?

Short answer

Answer: The expression for the total charge within the cube is given by: $$Q_{enc} = K \times 8.85 \times 10^{-12} \mathrm{C}$$ where \(K\) is the sum of the cosine of the angles between the electric field and the normal vector of each face. The exact value of \(K\) cannot be determined due to the unspecified direction of the electric field.

Step-by-step solution

1. Understand Gauss's Law

Gauss's law states that the total electric flux through a closed surface is equal to the total charge enclosed by that surface divided by the vacuum permittivity constant \(\epsilon_0\). Mathematically, it can be written as: $$\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\epsilon_0}$$ where \(\oint \mathbf{E} \cdot d\mathbf{A}\) represents the integral of electric field \(\mathbf{E}\) on a differential surface vector \(d\mathbf{A}\) over the closed surface, \(Q_{enc}\) is the total charge enclosed by the surface, and \(\epsilon_0\) is the vacuum permittivity constant \((\approx 8.85 \times 10^{-12} \mathrm{C^2/Nm^2})\).

2. Calculate the Electric Flux through Each Face of the Cube

To find the total electric flux through the closed surface (the cube), we have to calculate the electric flux through each of the six faces of the cube and sum them. Since the electric field is constant, the total flux through each face can be calculated by multiplying the magnitude of the electric field with the area of the face and the cosine of the angle between the electric field direction and the normal vector of the face. For an arbitrary face i, the flux can be calculated as: $$\Phi_i = EA_i\cos{\theta_i}$$ where \(\Phi_i\) is the electric flux through the i-th face, \(E\) is the magnitude of the electric field (150 N/C), \(A_i\) is the area of the i-th face (since it's a cube, all faces have an area of \((1.00 \mathrm{m})^2 = 1.00 \mathrm{m^2}\)), and \(\theta_i\) is the angle between the electric field and the normal vector of the i-th face.

3. Sum the Electric Fluxes and Apply Gauss's Law

The total electric flux through the closed surface (the cube) can be calculated by summing the flux through each face: $$\oint \mathbf{E} \cdot d\mathbf{A} = \sum_{i=1}^{6} \Phi_i$$ Now, apply Gauss's law to find the total charge enclosed by the cube: $$Q_{enc} = \epsilon_0 \oint \mathbf{E} \cdot d\mathbf{A} = \epsilon_0 \sum_{i=1}^{6} \Phi_i$$

4. Find the Total Charge

Substitute the given values and calculate the total charge enclosed by the cube: $$Q_{enc} = (8.85 \times 10^{-12} \mathrm{C^2/Nm^2}) \sum_{i=1}^{6} (150 \mathrm{N/C})(1.00 \mathrm{m^2})\cos{\theta_i}$$ Since we cannot find the exact value of \(\cos{\theta_i}\) for each face due to the unspecified direction of the electric field, we cannot find the precise total charge within the cube. However, the result will be of the form: $$Q_{enc} = K \times 8.85 \times 10^{-12} \mathrm{C}$$ where \(K\) is the sum of the cosine of the angles between the electric field and the normal vector of each face.

Key concepts

Electric Field

Imagine standing in a field on a windy day, feeling the breeze push against you from a specific direction; the electric field concept is somewhat akin to this, but instead of wind, it's about the force experienced by charges. It is a vector quantity, defining the force per unit charge exerted on an infinitesimally small positive test charge at any point in space. For example, when we say that the electric field has a magnitude of 150 N/C, it means that a test charge placed in this field would experience a force of 150 newtons for every coulomb of charge.
It's important to note that the electric field's strength and direction can change depending on where you measure it. Whether near a charged balloon or between the plates of a capacitor, the electric field guides how charges move. However, in our scenario with the cube, we were given a constant electric field, which simplifies calculations as the force remains the same no matter where you are within the cube.

Electric Flux

You can think of electric flux as the number of electric field lines passing through a surface. It’s a scalar quantity representing the electric field’s flow through a given area. The more lines that pass through, the greater the flux. The formula
\[\begin{equation} \$\Phi_i = EA_i\cos{\theta_i}\$\$ \end{equation}\]
where \(\Phi_i\) represents the electric flux, is a way to quantify this concept based on the magnitude of the electric field (\(E\)), the area it's passing through (\(A_i\)), and the angle (\(\theta_i\)) at which it intersects the surface. In the scenario of the cube from the exercise, because the direction of the electric field isn't specified, it's impossible to find the exact value of electric flux through each face without knowing the orientation of the field relative to the surfaces. This directional component emphasizes that not all of the field contributes to the flux if it's not perpendicular to the surface.

Enclosed Charge

The enclosed charge inside any given space (such as our cube) is simply the total amount of electric charge within that region. Gauss's Law connects this concept directly to electric flux by stating that the total electric flux through a closed surface is proportional to the charge enclosed by that surface. The mathematical expression of this law is:
\[\begin{equation} \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\epsilon_0} \end{equation}\]
With Gauss's Law, you're able to calculate the charge within a closed surface without knowing the specific details about charge distributions inside, which simplifies analyses in many cases. In our cube exercise, even without knowing the precise direction of the electric field, we could infer some properties of the enclosed charge using the symmetry of the cube and the nature of the electric field.

Vacuum Permittivity

Vacuum permittivity, symbolized by \(\epsilon_0\), is a fundamental constant in electromagnetism that represents how much resistance the free space (vacuum) offers against the formation of an electric field. In more practical terms, the vacuum permittivity shows up in calculations involving electric forces and fields, like in Gauss's Law. The value of vacuum permittivity is approximately \(8.85 \times 10^{-12} \mathrm{C^2/Nm^2}\).
This physical constant is essential in not only describing how an electric field behaves in a vacuum but also in determining the force between charges in free space. For example, when using Gauss's Law to find the total charge within a space – as with our hypothetical cube – the vacuum permittivity value is vital to link the measurable electric flux to the enclosed charge.