/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q7P A particle of charge 1.8μ°ä is ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A particle of charge 1.8μ°äis at the center of a Gaussian cube55cmon edge. What is the net electric flux through the surface?

Short Answer

Expert verified

The net electric flux through the surface is 2.0×105N.m2/C.

Step by step solution

01

The given data

  1. Charge of the particle,q=1.8μ°ä
  2. Edge length of cube,a=55cm1m100cm=0.55m
02

Understanding the concept of Gauss law-planar symmetry

Using the concept of Gauss law and the planar symmetry, we can get the required values of the electric field at the left of the plates, right of the plates, and between the plates.

Formula:

The total flux through any surface, ϕ=qε0 (1)

03

Calculation of the net flux through any cube surface

As the cube has six surfaces, thus, the net flux through each surface of the cube is given using equation (1) such that,

ϕ=1.8×10-6C6×8.85×10-12C2/N.m2=2.0×105N.m2/C

Hence, the value of the required flux is2.0×105N.m2/C

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 91Ó°ÊÓ!

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

In Fig. 23-44, two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge densities of opposite signs and magnitude7.0×10-22C/m2. In unit-vector notation, what is the electric field at points (a) to the left of the plates, (b) to the right of them, and (c) between them?

A uniformly charged conducting sphere of1.2 mdiameter has surface charge density 8.1 mC/m2 . Find (a) the net charge on the sphere and (b) the total electric flux leaving the surface.

Figure 23-58 shows, in cross-section, two solid spheres with uniformly distributed charges throughout their volumes. Each has radius R. Point Plies on a line connecting the centers of the spheres, at radial distance from the center of sphere 1. If the net electric field at point Pis zero, what is the ratio of the total charges?

Flux and conducting shells. A charged particle is held at the center of two concentric conducting spherical shells. Figure 23-39ashows a cross section. Figure 23-39b gives the net flux ϕthrough a Gaussian sphere centered on the particle, as a function of the radius rof the sphere. The scale of the vertical axis is set byϕ=5.0×105m2/C.What are (a) the charge of the central particle and the net charges of (b) shell A and (c) shell B?

Equation 23-11 (E=σ/ε0) gives the electric field at points near a charged conducting surface. Apply this equation to a conducting sphere of radius rand charge q, and show that the electric field outside the sphere is the same as the field of a charged particle located at the center of the sphere.

See all solutions

Recommended explanations on Physics 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.