Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 1563

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.

Start your free trial

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

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

Electric potential at any point is $\mathrm{V}=-5 \mathrm{x}+3 \mathrm{y}+\sqrt{(15 \mathrm{z})}$, then the magnitude of the electric field is \(\ldots \ldots \ldots \mathrm{N} / \mathrm{C}\). (A) \(3 \sqrt{2}\) (B) \(4 \sqrt{2}\) (C) 7 (D) \(5 \sqrt{2}\)

The capacitors of capacitance \(4 \mu \mathrm{F}, 6 \mu \mathrm{F}\) and $12 \mu \mathrm{F}$ are connected first in series and then in parallel. What is the ratio of equivalent capacitance in the two cases? (A) \(2: 3\) (B) \(11: 1\) (C) \(1: 11\) (D) \(1: 3\)

Charges \(+q\) and \(-q\) are placed at point \(A\) and \(B\) respectively which are a distance \(2 \mathrm{~L}\) apart, \(\mathrm{C}\) is the midpoint between \(\mathrm{A}\) and \(\mathrm{B}\). The work done in moving a charge \(+Q\) along the semicircle \(C R D\) is \(\ldots \ldots\) (A) $\left[(\mathrm{qQ}) /\left(2 \pi \mathrm{\epsilon}_{0} \mathrm{~L}\right)\right]$ (B) \(\left[(-q Q) /\left(6 \pi \in_{0} L\right)\right]\) (C) $\left[(\mathrm{qQ}) /\left(6 \pi \mathrm{e}_{0} \mathrm{~L}\right)\right]$ (D) \(\left[(q Q) /\left(4 \pi \in_{0} L\right)\right]\)

Three charges, each of value \(Q\), are placed at the vertex of an equilateral triangle. A fourth charge \(q\) is placed at the centre of the triangle. If the charges remains stationery then, \(q=\ldots \ldots \ldots\) (A) \((\mathrm{Q} / \sqrt{2})\) (B) \(-(\mathrm{Q} / \sqrt{3})\) (C) \(-(Q / \sqrt{2})\) (D) \((\mathrm{Q} / \sqrt{3})\)

A charged particle of mass \(\mathrm{m}\) and charge \(q\) is released from rest in a uniform electric field \(E\). Neglecting the effect of gravity, the kinetic energy of the charged particle after 't' second is ...... (A) $\left[\left(\mathrm{Eq}^{2} \mathrm{~m}\right) /\left(2 \mathrm{t}^{2}\right)\right]$ (B) $\left[\left(\mathrm{E}^{2} \mathrm{q}^{2} \mathrm{t}^{2}\right) /(2 \mathrm{~m})\right]$ (C) \(\left[\left(2 \mathrm{E}^{2} \mathrm{t}^{2}\right) /(\mathrm{qm})\right]\) (D) \([(\mathrm{Eqm}) / \mathrm{t}]\)
See all solutions

Recommended explanations on English Textbooks

Cues and Conventions

Read Explanation

Syntax

Read Explanation

Free Response Essay

Read Explanation

English Grammar

Read Explanation

Single Paragraph Essay

Read Explanation

History of English Language

Read Explanation
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