h) In symbols, what is the total flux through the sphere? (Use equation from introduction) 1) Notice that in this case the electric field lines intersect the sphere perpendicular to its surface and that the electric field strength is uniform over the surface. How do we know that the electric field strength does not vary over the surface? j) Because of the simplifying conditions discussed in part (f), we can apply Gauss' Law to find the electric field due to +Q. Find the electric field for a point charge +Q. (Recall that the surface area of a sphere is 4tr²) k) Graph E versus r, using proper scales and units to axis

h) In symbols, what is the total flux through the sphere? (Use equation from introduction) 1) Notice that in this case the electric field lines intersect the sphere perpendicular to its surface and that the electric field strength is uniform over the surface. How do we know that the electric field strength does not vary over the surface? j) Because of the simplifying conditions discussed in part (f), we can apply Gauss' Law to find the electric field due to +Q. Find the electric field for a point charge +Q. (Recall that the surface area of a sphere is 4tr²) k) Graph E versus r, using proper scales and units to axis

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter24: Gauss’s Law

Section: Chapter Questions

Problem 24.10OQ: A cubical gaussian surface is bisected by a large sheet of charge, parallel to its top and bottom...

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A cubical gaussian surface is bisected by a large sheet of charge, parallel to its top and bottom faces. No other charges are nearby. (i) Over how many of the cubes faces is the electric field zero? (a) 0 (b) 2 (c) 4 (d) 6 (ii) Through how many of the cubes faces is the electric flux zero? Choose from the same possibilities as in part (i).
1. An thin rod bent into the shape of an arc of a circle of radius R carries a uniform charge of linear density (i.e. charge on per unit length) 1. The arc subtends a total angle 20,, symmetric about the x axis. Determine the electric field at the origin O. See Fig.1 below. R R Fig.1 for problem 1.
PHYSICS. RATE WILL BE GIVEN! NO LONG EXPLANATION NEEDED. MULTIPLE CHOICE: CHOOSE THE CORRECT ANSWER Consider a cubic surface C, a spherical surface S, and right cylindrical surface R. All of the mentioned surfaces enclose the same amount of volume and their geometric centers coincides. If a negative point charge is placed at the geometric center, which of the surfaces will have the greatest electric flux through it? Note: The right cylinder has a radius to height ratio of 1/2. O spherical surface S O cubic surface C O right cylindrical surface R O all of the mentioned surfaces will have the same flux.
Problem 1 We have said that Gauss's law is always true, but only useful for calculating the electric field created by source charge distributions that are spheres, infinite straight cylinders, and infinite flat sheets, and even those cases have additional restrictions. a) Explain why we are limited to those distributions. Discuss what additional restrictions apply. For example, can we use Gauss's law to find the field of a sphere whose density depends on distance r from the center? Can we do it for a sphere whose density depends on angle (e.g. different at the poles from at the equator)? Why or why not? b) Can we use it for a uniformly charged ring? a uniformly charged cube? Why or why not? c) Summarize the key features for each of the possible distributions when the charge density is uniform throughout the sphere, cylinder or plate by filling out the table below answering the following questions: What is the shape of the test surface? What variable do we (or the text) usually use to…
is not a multiple of 3, press ENG or SHIFT+ENG, whichever the case.) Example: 0.06N or 6.0x10-2 N must be expressed to 60x10-3 N or 60mN 1. A hemispherical surface with radius r = 12 cm in a region of uniform electric field E = 10 V/m has its axis aligned parallel to the direction of the field. Calculate the flux through the surface. 2. In the fiqure shown, the magnitude of the uniform electric
1) Assume we have a thin conducting ring (radius R) that is uniformly charged with net charge Q. We're going to derive an equation for the electric field along the axis going through the center of the ring. Choose the z axis to coincide with the axis extending vertically through the middle of the ring. R ith "bit" Charged ring (Image from textbook by T.A. Moore) a) Explain why the net electric field at the center of the ring (in the plane of the ring) is zero b) We want to slice the ring into tiny charged "bits". Explain why (show mathematically) that the charge in the "ith bit", q; is: qi =,A0 2n c) Find an equation for |E (the magnitude of the electric field) at point P, that is caused by q: in terms of Q, 8o, A¢, R, and z (the z coordinate of point P). d) Now you've calculated the magnitude of the electric field contributed by one bit of charge. Explain why when you add in all the bits of charge from the entire ring, the total electric field must point straight up along the z-axis.…
QUESTION 6 Problem Using the method of integration, what is the electric field of a uniformly charged thin circular plate (with radius Rand total charge Q) at xo distance from its center? (Consider that the surface of the plate lies in the yz plane) Solution A perfect approach to this is to first obtain the E-field produced by an infinitesimal charge component of the charge Q. There will be several approaches to do this, but the most familiar to us is to obtain a very small shape that could easily represent our circular plane. That shape would be a ring. So for a ring whose charge is q. we recall that the electric field it produces at distance x0 is given by E= (1/ Mx0qV 242) Since, the actual ring (whose charge is dg) we will be dealing with is an infinitesimal part of the circular plane, then, its infinitesimal electric field contribution is expressed as = (1/ Xx0 2. We wish to obtain the complete electric field contribution from the above equation, so we integrate it from 0 toR to…
As shown in the figure, inner radius R1 outer radius R2homogeneously distributed over the metal sphere shell, total positiveelectric charge is given as + 4?. If in the center of the sphere shellThere is a point electrical charge with a total amount of 2?.Answer the following questions using the Gaussian flux law. a) Find the electric field strength in the region r <1. b) Find the electric field strength in the region 1 <r <?2. ) c) Find the electric field strength in the region r> ?2. d) Express the electric field strengths you find as vectors.
Time left C For which of the following charge distributions would Gauss's law not be useful for calculating the electric field? a spherical shell of radius R with charge uniformly distributed over its surface O a. O b. an infinitely long cylinder of radius R with charge uniformly distributed over its surface Oc. a uniformly charged sphere of radius R an infinite planar sheet having constant surface charge density Oe a cylinder of radius R and height h with charge uniformly distributed over its surface NEXT PAGE
I Review The figures show cross sections of three-dimensional closed surfaces. They have a flat top and bottom surface above and below the plane of the page. However, the electric field is everywhere parallel to the page, so there is no flux through the top or bottom surface. The electric field is uniform over each face of the surface. Submit Previous Answers v Correct Part C For (Figure 3), does the surface enclose a net positive charge, a net negative charge, or no net charge? Match the words in the left column to the appropriate blanks in the sentences on the right. Rese Figure « 1 of 3 > an outward There is flux through each side of the surface, which indicates that there is an inward enclosed in this surface. zero no net charge 10 N/C 10 Ν/C a net negative charge 10 N/C 10 N/C a net positive charge Submit Request Answer
Consider an infinitely long slab with thickness d and some density of charge p. What is the electric field at all points in space? Plot the field as a function of the distance from the slab. How is the field different from the infinite plane with uniform charge o? po
A rod of length e has a uniform positive charge per unit length A and a total charge Q. Calculate the electric field at a point P that is located along the long axis of the rod and a distance a from one end (see figure). The electric field at P due to a uniformly charged rod lying along the x-axis. dx P a SOLUTION Conceptualize The field dE at P due to each segment of charge on the rod is in the --Select-- v because point P is farther from the charge distribution. v x-direction because every segment carries a positive charge. The figure shows the appropriate geometry. In our result, we expect the electric field to become smaller as the distance a becomes larger Categorize Because the rod is continuous, we are evaluating the field due to a continuous charge distribution rather than a group of individual charges. Because every segment of the rod produces an electric field in the -Select--- vy x-direction, the sum of their contributions can be handled without the need to add vectors.…