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\(\bullet\) (a) A closed surface encloses a net charge of 2.50\(\mu \mathrm{C}\) . What is the net electric flux through the surface? (b) If the electric flux through a closed surface is determined to be \(1.40 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C},\) how much charge is enclosed by the surface?

Short Answer

Expert verified
(a) Flux = \( 2.82 \times 10^{5} \, \text{N} \cdot \text{m}^2/\text{C} \), (b) Charge = \( 1.24 \times 10^{-11} \, \text{C} \).

Step by step solution

01

Understanding Electric Flux and Gauss's Law

Electric flux \( \Phi \) through a closed surface is related to the electric field and the area through which it passes. Gauss's Law states that the net electric flux through a closed surface is equal to the net charge \( Q \) enclosed divided by the permittivity of free space \( \varepsilon_0 \). The formula is \( \Phi = \frac{Q}{\varepsilon_0} \), where \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \).
02

Calculating the Electric Flux (Part a)

Given that the enclosed charge \( Q \) is 2.50 \( \mu \)C, we convert this to Coulombs: \( 2.50 \times 10^{-6} \) C. Using Gauss's Law, \( \Phi = \frac{2.50 \times 10^{-6}}{8.85 \times 10^{-12}} \, \text{N} \cdot \text{m}^2/\text{C} \). Calculate this to find \( \Phi \).
03

Solving for Part a

Calculate \( \frac{2.50 \times 10^{-6}}{8.85 \times 10^{-12}} \): \( \Phi = 2.82 \times 10^{5} \, \text{N} \cdot \text{m}^2/\text{C} \). So, the net electric flux through the surface for part (a) is \( 2.82 \times 10^{5} \, \text{N} \cdot \text{m}^2/\text{C} \).
04

Calculating Enclosed Charge (Part b)

Given the electric flux \( \Phi = 1.40 \, \text{N} \cdot \text{m}^2/\text{C} \), we use Gauss's Law rearranged to find the charge: \( Q = \Phi \times \varepsilon_0 \). Substitute the given flux and \( \varepsilon_0 \) to get \( Q = 1.40 \times 8.85 \times 10^{-12} \).
05

Solving for Part b

Calculate \( 1.40 \times 8.85 \times 10^{-12} \, \text{C} \): \( Q = 1.24 \times 10^{-11} \, \text{C} \). Therefore, the enclosed charge for part (b) is \( 1.24 \times 10^{-11} \, \text{C} \).

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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 used to describe the flow of the electric field through a given area. Imagine the electric field lines as water flowing through a net. The amount of water passing through the net is akin to the electric flux. In formal terms, electric flux is the product of the electric field and the perpendicular area it penetrates. A key point is that if more field lines pass through the surface, the electric flux increases. For a closed surface, according to Gauss's Law, the net electric flux is determined not by the field alone but by the charge enclosed within.
  • The electric flux is represented by the symbol \( \Phi \).
  • It is measured in Newton-meter squared per Coulomb \( \text{N} \cdot \text{m}^2/\text{C} \).
  • Gauss's Law links electric flux with the enclosed charge.
Enclosed Charge
The enclosed charge within a closed surface is the amount of electric charge present inside that surface. According to Gauss's Law, the electric flux through a closed surface is directly proportional to the charge enclosed by the surface. This means, if the enclosed charge increases, the electric flux through the surface increases, and vice versa. To calculate the net electric flux through the surface, we use:\[\Phi = \frac{Q}{\varepsilon_0}\]where \( Q \) is the enclosed charge. In practice, to solve problems, determining the enclosed charge involves applying this formula by rearranging it as needed to find either \( \Phi \) or \( Q \). Remember, Gauss's Law only considers the net charge inside the surface for such calculations.
  • An increase in enclosed charge means higher electric flux.
  • It is a foundation for understanding how electric fields behave in closed systems.
Permittivity of Free Space
The permittivity of free space, denoted as \( \varepsilon_0 \), is a fundamental physical constant in electromagnetism. It characterizes how electric fields interact in a vacuum - essentially the "resistance" of the vacuum to electric field lines. This constant is crucial for calculations involving electric fields, especially in Gauss's Law.Given by: \[\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2\]This value affects how much electric field is generated per unit charge. When we evaluate scenarios like those in Gauss's Law, \( \varepsilon_0 \) plays an essential role in relating the enclosed charge to electric flux.
  • It allows us to calculate how electric fields interact with charges in vacuum.
  • Acts as a key link between charge, electric field, and flux.

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Most popular questions from this chapter

\(\bullet\) A proton is traveling horizontally to the right at \(4.50 \times 10^{6} \mathrm{m} / \mathrm{s}\) (a) Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.20 \(\mathrm{cm} .\) (b) How much time does it take the proton to stop after entering the field? (c) What mini- mum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)?

\(\bullet\)(a) How many excess elec- trons must be distributed uni- formly within the volume of an isolated plastic sphere 30.0 \(\mathrm{cm}\) in diameter to produce an elec- tric field of 1150 \(\mathrm{N} / \mathrm{C}\) just out- side the surface of the sphere? (b) What is the electric field at a point 10.0 cm outside the surface of the sphere?

\(\bullet\) The electric field caused by a certain point charge has a mag- nitude of \(6.50 \times 10^{3} \mathrm{N} / \mathrm{C}\) at a distance of 0.100 \(\mathrm{m}\) from the charge. What is the magnitude of the charge?

\(\bullet\) A particle has a charge of \(-3.00 \mathrm{nC}\) . (a) Find the magnitude and direction of the electric field due to this particle at a point 0.250 m directly above it. (b) At what distance from the parti- cle does its electric field have a magnitude of 12.0 \(\mathrm{N} / \mathrm{C}\) ?

\(\bullet$$\bullet\) Two unequal charges repel each other with a force \(F .\) If both charges are doubled in magnitude, what will be the new force in terms of \(F ?\)

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