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A charge of \(87.6 \mathrm{pC}\) is uniformly distributed on the surface of a thin sheet of insulating material that has a total area of \(29.2 \mathrm{~cm}^{2} .\) A Gaussian surface encloses a portion of the sheet of charge. If the flux through the Gaussian surface is \(5.00 \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C},\) what area of the sheet is enclosed by the Gaussian surface?

Short Answer

Expert verified
Perform the above calculations to find the area of the sheet enclosed by the Gaussian surface. The exact numerical answer depends on the calculated values in the previous steps.

Step by step solution

01

Calculate the total charge density

The total charge \(Q\) is given as \(87.6 \mathrm{pC}\), and the total area \(A_t\) of the sheet is given as \(29.2 \mathrm{cm}^2\). Convert these quantities to SI units (coulombs for charge, meters squared for area), then compute the charge density \(\sigma\) using the formula \(\sigma = Q/A_t\).
02

Use the flux to find the enclosed charge

The flux \(\Phi\) through the Gaussian surface is given as \(5.00 \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}\). As per the Gauss's Law, the flux equals the total charge enclosed \(Q_{enc}\) by the Gaussian surface, divided by the permittivity of free space \(\epsilon_0\). Solve this equation for \(Q_{enc}\), i.e., \(Q_{enc} = \Phi * \epsilon_0\). The value of \(\epsilon_0\) is \(8.85*10^{-12} \mathrm{m}^{2} \mathrm{C}^{-2} \mathrm{~N}^{-1}\)
03

Find the area enclosed by the Gaussian surface

Since we know that the charge density \(\sigma\) equals the charge \(Q_{enc}\) divided by the area \(A_{enc}\), we can solve this equation for \(A_{enc}\), i.e., \(A_{enc} = Q_{enc}/\sigma\). This yields the area enclosed by the Gaussian surface.

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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 concept that describes how an electric field passes through a surface. You can think of it as the number of electric field lines going through a given area.
For instance, if you imagine a flat surface and a constant electric field, the electric flux would depend on both the strength of the electric field and the size of the surface.
The formula to determine electric flux (\( \Phi \)) through a closed surface is given by Gauss's Law:
  • \( \Phi = E \cdot A \cdot \cos(\theta) \)
Here, \( E \) represents the electric field, \( A \) is the area the field lines pass through, and \( \theta \) is the angle between the field lines and the perpendicular to the surface.
In the context of the exercise, the electric flux through the Gaussian surface is given as \(5.00 \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}\), directly helping to find the charge enclosed.
Charge Density
Charge density represents the amount of charge per unit area or volume, depending on whether we deal with surface charge density or volume charge density respectively. In the context of our problem, we are dealing with surface charge density.
Surface charge density (\( \sigma \)) is described by the formula:
  • \( \sigma = \frac{Q}{A_t} \)
where \( Q \) is the total charge, and \( A_t \) is the total area.
In our exercise, the total charge is 87.6 pC and the area is 29.2 cm². By converting these into standard SI units, we find the charge density necessary to determine the area of the sheet enclosed by the Gaussian surface.
This value of charge density helps us relate the enclosed charge in the Gaussian surface to the area we need to find.
Permittivity of Free Space
The permittivity of free space, often denoted as \( \epsilon_0 \), is a constant that characterizes the ability of the vacuum to allow electric field lines to pass through. It plays a fundamental role in electrical and magnetic calculations, particularly in relation to Gauss's Law.
Its value is:
  • \( \epsilon_0 = 8.85 \times 10^{-12} \mathrm{\ m}^{2} \mathrm{ C}^{-2} \mathrm{\ N}^{-1} \)

In our exercise, Gauss's Law states that the electric flux \( (\Phi) \) through any closed surface is equal to the charge enclosed \( (Q_{enc}) \) divided by the permittivity of free space.
Mathematically:
  • \( \Phi = \frac{Q_{enc}}{\epsilon_0} \)
By rearranging this formula, we can calculate the charge enclosed by the Gaussian surface from the given electric flux, further leading us to find the enclosed area. The permittivity of free space is thus crucial in solving problems concerning electromagnetic fields and charge distributions.

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Most popular questions from this chapter

A long coaxial cable consists of an inner cylindrical conductor with radius \(a\) and an outer coaxial cylinder with inner radius \(b\) and outer radius \(c .\) The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length \(\lambda\). Calculate the electric field (a) at any point between the cylinders a distance \(r\) from the axis and (b) at any point outside the outer cylinder. (c) Graph the magnitude of the electric field as a function of the distance \(r\) from the axis of the cable, from \(r=0\) to \(r=2 c .\) (d) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder.

A flat sheet of paper of area \(0.250 \mathrm{~m}^{2}\) is oriented so that the normal to the sheet is at an angle of \(60^{\circ}\) to a uniform electric field of magnitude \(14 \mathrm{~N} / \mathrm{C}\). (a) Find the magnitude of the electric flux through the sheet. (b) Does the answer to part (a) depend on the shape of the sheet? Why or why not? (c) For what angle \(\phi\) between the normal to the sheet and the electric field is the magnitude of the flux through the sheet (i) largest and (ii) smallest? Explain your answers.

A very small object with mass \(8.20 \times 10^{-9} \mathrm{~kg}\) and positive charge \(6.50 \times 10^{-9} \mathrm{C}\) is projected directly toward a very large insulating sheet of positive charge that has uniform surface charge density \(5.90 \times 10^{-8} \mathrm{C} / \mathrm{m}^{2} .\) The object is initially \(0.400 \mathrm{~m}\) from the sheet. What initial speed must the object have in order for its closest distance of approach to the sheet to be \(0.100 \mathrm{~m} ?\)

An electron is released from rest at a distance of \(0.300 \mathrm{~m}\) from a large insulating sheet of charge that has uniform surface charge density \(+2.90 \times 10^{-12} \mathrm{C} / \mathrm{m}^{2}\). (a) How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point \(0.050 \mathrm{~m}\) from the sheet? (b) What is the speed of the electron when it is \(0.050 \mathrm{~m}\) from the sheet?

Electric Fields in an Atom. The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately \(7.4 \times 10^{-15} \mathrm{~m} .\) (a) What is the electric field this nucleus produces just outside its surface? (b) What magnitude of electric field does it produce at the distance of the electrons, which is about \(1.0 \times 10^{-10} \mathrm{~m} ?\) (c) The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?

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