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A charge \(Q\) is located inside a rectangular box. The electric flux through each of the six surfaces of the box is: \(\Phi_{1}=+1500 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C},\) \(\Phi_{2}=+2200 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}, \Phi_{3}=+4600 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}, \Phi_{4}=-1800 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}\) \(\Phi_{5}=-3500 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C},\) and \(\Phi_{6}=-5400 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C} .\) What is \(Q ?\)

Short Answer

Expert verified
The charge \( Q \) is \( -2.125 \times 10^{-8} \, \mathrm{C} \).

Step by step solution

01

Identify Gauss's Law

Gauss's Law states that the net electric flux \( \Phi_{\text{net}} \) through a closed surface is equal to the charge \( Q \) enclosed divided by the electric constant \( \varepsilon_0 \). The formula is: \[ \Phi_{\text{net}} = \frac{Q}{\varepsilon_0} \]
02

Calculate Net Electric Flux

To find \( \Phi_{\text{net}} \), sum the electric fluxes of all six surfaces: \[ \Phi_{\text{net}} = \Phi_1 + \Phi_2 + \Phi_3 + \Phi_4 + \Phi_5 + \Phi_6 \] Substituting the given values, we get: \[ \Phi_{\text{net}} = 1500 + 2200 + 4600 - 1800 - 3500 - 5400 \] \[ \Phi_{\text{net}} = -2400 \, \mathrm{N} \cdot \mathrm{m}^2 / \mathrm{C} \]
03

Apply Gauss's Law to Solve for the Charge Q

Solve for \( Q \) using Gauss's Law formula: \[ Q = \Phi_{\text{net}} \times \varepsilon_0 \] The value of \( \varepsilon_0 \) is approximately \( 8.854 \times 10^{-12} \, \mathrm{C}^2 / \mathrm{N} \cdot \mathrm{m}^2 \). Therefore, \[ Q = (-2400) \times (8.854 \times 10^{-12}) \] \[ Q = -2.125 \times 10^{-8} \, \mathrm{C} \]
04

Conclusion

The charge \( Q \) enclosed within the rectangular box is \( -2.125 \times 10^{-8} \, \mathrm{C} \) indicating it is a negative charge.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Flux
Electric flux is a crucial concept in understanding the behavior of electric fields. It refers to the number of electric field lines passing through a given area. The electric flux (\( \Phi \) ) through a surface is the product of the electric field (\( E \) ), the area (\( A \) ) of the surface, and the cosine of the angle (\( \theta \) ) between the field lines and the normal (perpendicular line) to the surface. Mathematically, it can be expressed as:
  • \( \Phi = E \cdot A \cdot \cos(\theta) \)
In closed surfaces, the sign of electric flux indicates whether the field lines are leaving or entering the surface. Positive flux means that field lines are leaving the surface, while negative flux indicates they are entering. Therefore, calculating the net electric flux involves summing the contributions from each surface of a closed shape, like a box or sphere, thereby enabling us to understand how the electric field interacts with objects.
Enclosed Charge
Enclosed charge refers to the net amount of electric charge that is contained within a closed surface. According to Gauss's Law, the total electric flux through a closed surface is directly related to the charge enclosed within that surface. This relationship is what allows us to determine enclosed charge (\( Q \) ) using the formula:
  • \( \Phi_{\text{net}} = \frac{Q}{\varepsilon_0} \)
Understanding enclosed charge is essential in calculations involving electrostatics, as it implies that any charge that is not enclosed by the surface will not factor into the net electric flux. Hence, it is critical in applications such as calculating the charge distribution on conductors or within cavities. Besides, the concept assists in visualizing how charges influence electric field lines and how those lines traverse surfaces.
Electric Constant (\varepsilon_0)
The electric constant, also known as the permittivity of free space (\( \varepsilon_0 \) ), is a fundamental constant in electromagnetism. It characterizes the ability of a vacuum to allow electric fields to establish. The electric constant is instrumental in equations like Gauss's Law, where it appears in the relationship between electric flux, enclosed charge, and Gauss's surface:
  • \( Q = \Phi_{\text{net}} \times \varepsilon_0 \)
The value of \( \varepsilon_0 \) is approximately \( 8.854 \times 10^{-12} \text{C}^2 / \text{N} \cdot \text{m}^2 \), and it is derived from the units of charge, distance, and force. It is a key element of the relationship between electric force and electric fields, illustrating how strongly two charges interact in a vacuum compared to when other materials influence the interaction. Understanding and applying \( \varepsilon_0 \) properly enables accurate calculations in electric and electromagnetic fields, making it fundamental for physics students and engineers alike.

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Most popular questions from this chapter

Conceptual Example 13 deals with the hollow spherical conductor in Figure \(18.30 .\) The conductor is initially electrically neutral, and then a charge \(+q\) is placed at the center of the hollow space. Suppose the conductor initially has a net charge of \(+2 q\) instead of being neutral. What is the total charge on the interior and on the exterior surface when the \(+q\) charge is placed at the center?

The membrane surrounding a living cell consists of an inner and an outer wall that are separated by a small space. Assume that the membrane acts like a parallel plate capacitor in which the effective charge density on the inner and outer walls has a magnitude of \(7.1 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} .\) (a) What is the magnitude of the electric field within the cell membrane? (b) Find the magnitude of the electric force that would be exerted on a potassium ion \(\left(\mathrm{K}^{+} ;\right.\) charge \(\left.=+e\right)\) placed inside the membrane.

ssm mmh Two small objects, A and B, are fixed in place and separated by \(3.00 \mathrm{cm}\) in a vacuum. Object \(\mathrm{A}\) has a charge of \(+2.00 \mu \mathrm{C}\), and object \(\mathrm{B}\) has a charge of \(-2.00 \mu \mathrm{C}\). How many electrons must be removed from object A and put onto object \(\mathrm{B}\) to make the electrostatic force that acts on each object an attractive force whose magnitude is \(68.0 \mathrm{N} ?\)

ssm Two spherical objects are separated by a distance that is \(1.80 \times 10^{-3} \mathrm{m} .\) The objects are initially electrically neutral and are very small compared to the distance between them. Each object acquires the same negative charge due to the addition of electrons. As a result, each object experiences an electrostatic force that has a magnitude of \(4.55 \times 10^{-21} \mathrm{N} .\) How many electrons did it take to produce the charge on one of the objects?

A small object, which has a charge \(q=7.5 \mu \mathrm{C}\) and mass \(m=9.0 \times 10^{-5} \mathrm{kg},\) is placed in a constant electric field. Starting from rest, the object accelerates to a speed of \(2.0 \times 10^{3} \mathrm{m} / \mathrm{s}\) in a time of \(0.96 \mathrm{s}\). Determine the magnitude of the electric field.

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