Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The inward and outward electric flux for a closed surface in units of \(\mathrm{Nm}^{2} / \mathrm{C}\) are respectively \(8 \times 10^{3}\) and \(4 \times 10^{3}\). Then the total charge inside the surface is \(\ldots \ldots \ldots \ldots . . \mathrm{c}\). (A) \(\left[\left(-4 \times 10^{3}\right) / \epsilon_{0}\right]\) (B) \(-4 \times 10^{3}\) (C) \(4 \times 10^{3}\) (D) \(-4 \times 10^{3} \mathrm{E}_{0}\)

Short Answer

Expert verified
The total charge inside the surface is approximately \(\left[-4 \times 10^{3}/\epsilon_{0}\right]\), which is closest to option (A). However, the exact value of the total charge enclosed is \(-3.54 \times 10^{-8} \ \mathrm{C}\).

Step by step solution

01

Given information

We are given the inward and outward electric fluxes for a closed surface. The inward electric flux is \(8 \times 10^3 \ \mathrm{Nm}^2/\mathrm{C}\), and the outward electric flux is \(4 \times 10^3 \ \mathrm{Nm}^2/\mathrm{C}\).
02

Applying Gauss's Law

Gauss's Law states that the total electric flux through a closed surface is equal to the total charge enclosed by the surface divided by the permittivity of free space (\(\epsilon_0\)). Mathematically, it can be written as: \[\Phi_E = \frac{Q_{enclosed}}{\epsilon_0}\] Where \(\Phi_E\) is the total electric flux through the closed surface, \(Q_{enclosed}\) is the total charge enclosed by the surface, and \(\epsilon_0\) is the permittivity of free space (\(\epsilon_0 \approx 8.85 \times 10^{-12} \ \mathrm{C}^2/\mathrm{Nm}^2\)).
03

Find the net electric flux

We need to find the net electric flux through the closed surface. Since the inward flux is negative and the outward flux is positive, we subtract the inward flux from the outward flux: \[\Phi_{net} = \Phi_{outward} - \Phi_{inward}\] \[\Phi_{net} = 4\times 10^3 - 8\times 10^3 = - 4\times 10^3 \ \mathrm{Nm}^2/\mathrm{C}\]
04

Calculate the total charge enclosed

Now, we can use Gauss's Law to find the total charge enclosed by the surface: \[Q_{enclosed} = \Phi_{net} \cdot \epsilon_0\] \[Q_{enclosed} = (-4 \times 10^3 \ \mathrm{Nm}^2/\mathrm{C}) \cdot (8.85 \times 10^{-12} \ \mathrm{C}^2/\mathrm{Nm}^2)\] \[Q_{enclosed} = -3.54 \times 10^{-8} \ \mathrm{C}\] We can see that the total charge enclosed by the surface is closest to option (A): (A) \(\left[-4 \times 10^{3}/\epsilon_{0}\right]\) However, the \(Q_{enclosed}\approx -3.54 \times 10^{-8}\), so the correct answer is closest to, but not exactly equal to option (A).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Electric Flux
Electric flux is an essential concept in electromagnetism, particularly when working with Gauss's Law. It measures the number of electric field lines passing through a given area. Electric flux can be visualized as how much of the electric field "flows" through a surface.
The formula for electric flux through a surface is given by:
  • \(\Phi_E = \mathbf{E} \cdot \mathbf{A} = E \cdot A \cdot \cos\theta\)
Here, \(\mathbf{E}\) is the electric field, \(\mathbf{A}\) is the area vector perpendicular to the surface, \(E\) is the magnitude of the electric field, \(A\) is the area of the surface, and \(\theta\) is the angle between \(\mathbf{E}\) and \(\mathbf{A}\).
If the field lines are perpendicular to the surface, flux is maximized, whereas if they run parallel, flux is zero. This idea helps in calculating how much electric charge is enclosed within a specific region, as part of Gauss's Law. In our exercise, the net electric flux is calculated by considering both inward and outward fluxes passing through a closed surface.
Closed Surface
A closed surface serves an important role in physics, particularly in understanding electrostatic principles through Gauss's Law. Imagine a surface that completely encloses a volume, like a sphere or a cube, with no openings. This is what we refer to as a closed surface in the context of electric flux.
The significance of a closed surface comes from its ability to illustrate the flow of electric flux around charges:
  • Within a closed surface, electric field lines can enter and exit, illustrating inward and outward fluxes.
  • Gauss's Law states that the net electric flux through a closed surface is directly related to the total charge contained within that surface.
In the provided exercise, a closed surface is used to determine the net electric flux by accounting for all field lines crossing the boundary. Through this method, we can calculate the total charge enclosed by understanding both the positive (outward) and negative (inward) contributions to electric flux.
Permittivity of Free Space
Permittivity of free space, often symbolized as \(\epsilon_0\), is a fundamental constant that plays a key role in electromagnetism. It serves as a measure of how much electric field (or force) is "permitted" through the vacuum free of electric charges. The value of permittivity of free space is approximately \(8.85 \times 10^{-12} \, \mathrm{C}^2/\mathrm{Nm}^2\).
Permittivity is crucial because:
  • It appears in the denominator of Coulomb's Law, influencing the force between two charges in vacuum.
  • In Gauss's Law, it relates the total charge inside a closed surface to the resulting electric flux.
In the context of the exercise, we use \(\epsilon_0\) to calculate the total charge within a closed surface from the net electric flux. The relationship is given by the formula \(Q_{enclosed} = \Phi_{net} \cdot \epsilon_0\), demonstrating how permittivity directly connects electric flux and charge in a closed system. Understanding this constant helps decipher the relationship between electric charges and fields.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A parallel plate capacitor has the space between its plates filled by two slabs of thickness \((\mathrm{d} / 2)\) each and dielectric constant \(\mathrm{K}_{1}\) and \(\mathrm{K}_{2}\) If \(\mathrm{d}\) is the plate separation of the capacitor, then capacity of the capacitor is .......... (A) \(\left[\left(2 \mathrm{~d} \in_{0}\right) / \mathrm{A}\right]\left[\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right) /\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right)\right]\) (B) \(\left[\left(2 \mathrm{~A} \in_{0}\right) / \mathrm{d}\right]\left[\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right) /\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right)\right]\) (C) \(\left[\left(2 \mathrm{Ad} \epsilon_{0}\right) / \mathrm{d}\right]\left[\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right) /\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right)\right]\) d] \(\left(K_{1}+K_{2}\right)\) (D) \(\left[\left(2 \mathrm{~A} \in_{0}\right) /\right.\)

Two Points \(P\) and \(Q\) are maintained at the Potentials of \(10 \mathrm{v}\) and \(-4 \mathrm{v}\), respectively. The work done in moving 100 electrons from \(\mathrm{P}\) to \(\mathrm{Q}\) is \(\ldots \ldots \ldots\) (A) \(2.24 \times 10^{-16} \mathrm{~J}\) (B) \(-9.60 \times 10^{-17} \mathrm{~J}\) (C) \(-2.24 \times 10^{-16} \mathrm{~J}\) (D) \(9.60 \times 10^{-17} \mathrm{~J}\)

A ball of mass \(1 \mathrm{gm}\) and charge \(10^{-8} \mathrm{c}\) moves from a point \(\mathrm{A}\), where the potential is 600 volt to the point \(B\) where the potential is zero. Velocity of the ball of the point \(\mathrm{B}\) is \(20 \mathrm{~cm} / \mathrm{s}\). The velocity of the ball at the point \(\mathrm{A}\) will be \(\ldots \ldots\) (A) \(16.8(\mathrm{~m} / \mathrm{s})\) (B) \(22.8(\mathrm{~cm} / \mathrm{s})\) (C) \(228(\mathrm{~cm} / \mathrm{s})\) (D) \(168(\mathrm{~m} / \mathrm{s})\)

The electric flux for gaussian surface \(\mathrm{A}\) that enclose the \(\ldots \ldots\) charged particles in free space is (given \(\left.q_{1}=-14 n c, q_{2}=78.85 \mathrm{nc}, q_{3}=-56 n c\right)\) (A) \(10^{4} \mathrm{Nm}^{2} / \mathrm{C}\) (B) \(10^{3} \mathrm{Nm}^{2} / \mathrm{C}\) (C) \(6.2 \times 10^{3} \mathrm{Nm}^{2} / \mathrm{C}\) (D) \(6.3 \times 10^{4} \mathrm{Nm}^{2} / \mathrm{C}\)

If an electron moves from rest from a point at which potential is 50 volt, to another point at which potential is 70 volt, then its kinetic energy in the final state will be \(\ldots .\) (A) \(1 \mathrm{~N}\) (B) \(3.2 \times 10^{-18} \mathrm{~J}\) (C) \(3.2 \times 10^{-10} \mathrm{~J}\) (D) 1 dyne

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free