Question

The six faces of a cubical box each measure \(20.0 \mathrm{~cm}\) by \(20.0 \mathrm{~cm},\) and the faces are numbered such that faces 1 and 6 are opposite to each other, as are faces 2 and \(5,\) and faces 3 and \(4 .\) The flux through each face is given in the table. Find the net charge inside the cube. $$ \begin{array}{|c|c|} \hline \text { Face } & \text { Flux }\left(\mathrm{N} \mathrm{m}^{2} / \mathrm{C}\right) \\ \hline 1 & -70.0 \\ \hline 2 & -300.0 \\ \hline 3 & -300.0 \\ \hline 4 & +300.0 \\ \hline 5 & -400.0 \\ \hline 6 & -500.0 \\ \hline \end{array} $$

Short answer

Answer: The net charge inside the cube is approximately \(-1.12 \times 10^{-8} \mathrm{~C}\).

Step-by-step solution

Calculate the total electric flux through the cube

First, we need to find the total electric flux through the cube. According to the problem, the fluxes through the six faces are: $$\Phi_1 = -70 \mathrm{~Nm^2/C} \quad \Phi_2 = -300 \mathrm{~Nm^2/C} \quad \Phi_3 = -300 \mathrm{~Nm^2/C}$$ $$\Phi_4 = 300 \mathrm{~Nm^2/C} \quad \Phi_5 = -400 \mathrm{~Nm^2/C} \quad \Phi_6 = -500 \mathrm{~Nm^2/C}$$ Now, we'll sum these fluxes to get the total electric flux through the cube: $$\oint \vec{E} \cdot d\vec{A} = \Phi_1 + \Phi_2 + \Phi_3 + \Phi_4 + \Phi_5 + \Phi_6$$ $$\oint \vec{E} \cdot d\vec{A} = -70 - 300 - 300 + 300 - 400 - 500 = -1270 \mathrm{~Nm^2/C}$$

Find the net charge inside the cube

Now that we have the total electric flux, we can use Gauss's law to find the net charge inside the cube: $$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{encl}}{\epsilon_0}$$ Rearranging the equation to solve for the net charge, we get: $$Q_{encl} = \epsilon_0 \oint \vec{E} \cdot d\vec{A}$$ The vacuum permittivity constant \(\epsilon_0\) is approximately \(8.85 \times 10^{-12} \mathrm{~C^2/Nm^2}\). Plugging in the value of the total electric flux and the vacuum permittivity constant, we find the net charge inside the cube: $$Q_{encl} = (8.85 \times 10^{-12} \mathrm{~C^2/Nm^2}) \times (-1270 \mathrm{~Nm^2/C}) = -1.12 \times 10^{-8} \mathrm{~C}$$ So, the net charge inside the cube is approximately \(-1.12 \times 10^{-8} \mathrm{~C}\).

Key concepts

Electric Flux

Imagine electric field lines flowing through a surface like water through a net. The strength of this 'flow' is what we call electric flux. Technically, it's a measure of how much electric field penetrates through a given area. It's calculated by taking the product of the electric field strength and the area it's passing through, and you take into account the angle between the field lines and the normal (perpendicular) direction of the area. If the field lines are perpendicular to the surface, the flux is maximized. When the lines are parallel, the flux is zero because the field doesn't 'penetrate' the surface.

In the cube exercise, the total electric flux is found by adding together the flux through each face. Negative and positive signs represent the direction of the field – whether it's entering or leaving the cube. In the given problem, there's a mix of both, which suggests that the field lines are entering some faces and exiting others, conveying the presence of a net charge enclosed within the cube.

Net Charge

The concept of 'net charge' is fundamental. It tells us whether there's an imbalance in the number of protons and electrons in an object. If there are more electrons, the object has a negative charge, and if there are more protons, it has a positive charge. Net charge is what causes electric forces and fields to arise.

In the context of Gauss's Law, which you apply in the cube problem, the net charge is what you're solving for. What's interesting about Gauss's Law is that it doesn't matter how the charge is distributed within the surface; the electric flux only depends on the total net charge inside. That's why, with the flux calculated, you can directly find the net charge enclosed without any knowledge of how the charge is arranged inside the cube.

Vacuum Permittivity

Now let's talk about a less intuitive concept: vacuum permittivity, often denoted by \(\epsilon_0\). This is a constant that appears in equations of electromagnetism, representing the ability of a vacuum to permit electric field lines to flow through it. Think of it as how 'thin' or 'thick' the vacuum is to electric fields. The standard value for vacuum permittivity is approximately \(8.85 \times 10^{-12} \mathrm{\~C^2/Nm^2}\).

In our cube problem, you used this constant along with the electric flux to find the net charge. By multiplying the negative total electric flux with the vacuum permittivity, you computed the amount of net charge within the cube. Since vacuum permittivity is a universally agreed-upon value, the calculation gives a definite result for the net charge, providing a direct link between observable electric fields and the amount of charge causing those fields.

Electromagnetism

Last but definitely not least, electromagnetism is one of the four fundamental forces of nature and is the framework connecting electricity and magnetism. It encompasses many phenomena, including how electric charges create electric fields and how charges react to being in those fields. Gauss’s law, which was crucial in solving the cube exercise, is a cornerstone of electromagnetism. This law gives us a method to link the net charge inside a closed surface to the electric field outside that surface.

In simpler terms, Gauss's law bridges the gap between invisible charges and their visible effects, like electrical forces. Understanding electromagnetism, starting with concepts like Gauss's law, not only explains the behavior of charges and fields but also underpins much of modern technology – from the electricity that powers our homes to the transmission of information via mobile devices.