PHYSICS F/SCI.+ENGR.,CHAPTERS 1-37
5th Edition
ISBN: 9780134378060
Author: GIANCOLI
Publisher: RENT PEARS
expand_more
expand_more
format_list_bulleted
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
Consider an imaginary box with dimensions 15 cm × 45 cm × 95 cm. There is an E- field with a uniform magnitude 850 N/C going out through the left side of the box and an E-field with a uniform magnitude 550 N/C going in through the right side, as shown in the figure below.
(a) Calculate the total electric flux through the box.
(b) Use Gauss’s law and your previous answer to find the net charge enclosed by the box.
Question 2: Gauss's Law
(a) A uniform electric field of magnitude 1.1 x 104N/C is perpendicular to a square sheet with
sides 2.0 m long. What is the electric flux through the sheet?
(b) Calculate the flux through the sheet of the previous problem if the plane of the sheet is at an
angle of 60° to the field. Find the flux for both directions of the unit normal to the sheet.
(c)
The electric flux through a cubical box 8.0 cm on a side is 1.2 x 10³Nm²/C. What is the total
charge enclosed by the box?
(d) The electric flux through a spherical surface is 4.0 x 104N m²/C. What is the net charge
enclosed by the surface?
A circular plane, with radius of 2.2 m, is immersed in an electric field with a magnitude of 800 N/C. The field makes an angle of 20° with the plane. What is the magnitude of electric flux through the plane?
Calculate the electric flux passing through the surface as shown below.
An infinitely long line of charge carries 0.4 C along each meter of length. Find the electric field 0.3 m from the line of charge.
Chapter 22 Solutions
PHYSICS F/SCI.+ENGR.,CHAPTERS 1-37
Ch. 22.1 - Which of the following would cause a change in the...Ch. 22.2 - A point charge Q is at the center of a spherical...Ch. 22.2 - Three 2.95 C charges are in a small box. What is...Ch. 22.3 - Prob. 1EECh. 22 - If the electric flux through a closed surface is...Ch. 22 - Is the electric field E in Gausss law....Ch. 22 - What can you say about the flux through a closed...Ch. 22 - The electric field E is zero at all points on a...Ch. 22 - Define gravitational flux in analogy to electric...Ch. 22 - Would Gausss law be helpful in determining the...
Ch. 22 - A spherical basketball (a nonconductor) is given a...Ch. 22 - In Example 226, it may seem that the electric...Ch. 22 - Suppose the line of charge in Example 226 extended...Ch. 22 - A point charge Q is surrounded by a spherical...Ch. 22 - A solid conductor carries a net positive charge Q....Ch. 22 - A point charge q is placed at the center of the...Ch. 22 - A small charged ball is inserted into a balloon....Ch. 22 - Prob. 1MCQCh. 22 - Prob. 2MCQCh. 22 - Prob. 3MCQCh. 22 - Prob. 4MCQCh. 22 - Prob. 5MCQCh. 22 - Prob. 6MCQCh. 22 - Prob. 7MCQCh. 22 - Prob. 8MCQCh. 22 - Prob. 9MCQCh. 22 - Prob. 10MCQCh. 22 - Prob. 1PCh. 22 - (I) The Earth possesses an electric field of...Ch. 22 - (II) A cube of side l is placed in a uniform field...Ch. 22 - (II) A uniform field E is parallel to the axis of...Ch. 22 - (I) The total electric flux from a cubical box...Ch. 22 - (I) Figure 2226 shows five closed surfaces that...Ch. 22 - (II) In Fig. 2227, two objects, O1 and O2, have...Ch. 22 - (II) A ring of charge with uniform charge density...Ch. 22 - (II) In a certain region of space, the electric...Ch. 22 - (II) A point charge Q is placed at the center of a...Ch. 22 - Prob. 11PCh. 22 - (I) Draw the electric field lines around a...Ch. 22 - Prob. 13PCh. 22 - (I) Starting from the result of Example 223, show...Ch. 22 - Prob. 15PCh. 22 - (I) A metal globe has l.50 mC of charge put on it...Ch. 22 - Prob. 17PCh. 22 - (II) A solid metal sphere of radius 3.00 m carries...Ch. 22 - (II) A 15.0-cm-diameter nonconducting sphere...Ch. 22 - (II) A flat square sheet of thin aluminum foil,...Ch. 22 - (II) A spherical cavity of radius 4.50 cm is at...Ch. 22 - Prob. 22PCh. 22 - Prob. 23PCh. 22 - (II) Two large, flat metal plates are separated by...Ch. 22 - (II) Suppose the two conducting plates in Problem...Ch. 22 - Prob. 26PCh. 22 - (II) Two thin concentric spherical shells of radii...Ch. 22 - (II) A spherical rubber balloon carries a total...Ch. 22 - (II) Suppose the nonconducting sphere of Example...Ch. 22 - (II) Suppose in Fig. 2232, Problem 29, there is...Ch. 22 - (II) Suppose the thick spherical shell of Problem...Ch. 22 - (II) Suppose that at the center of the cavity...Ch. 22 - (II) A long cylindrical shell of radius R0 and...Ch. 22 - (II) A very long solid nonconducting cylinder of...Ch. 22 - (II) A thin cylindrical shell of radius R1 is...Ch. 22 - (II) A thin cylindrical shell of radius R1 = 6.5...Ch. 22 - (II) (a) If an electron (m = 9.1 1031 kg) escaped...Ch. 22 - (II) A very long solid nonconducting cylinder of...Ch. 22 - (II) A nonconducting sphere of radius r0 is...Ch. 22 - (II) A very long solid nonconducting cylinder of...Ch. 22 - (II) A flat ring (inner radius R0, outer radius...Ch. 22 - (II) An uncharged solid conducting sphere of...Ch. 22 - (III) A very large (i.e., assume infinite) flat...Ch. 22 - (III) Suppose the density of charge between r1 and...Ch. 22 - (III) Suppose two thin flat plates measure 1.0 m ...Ch. 22 - (III) A flat slab of nonconducting material (Fig....Ch. 22 - (III) A flat slab of nonconducting material has...Ch. 22 - (III) An extremely long, solid nonconducting...Ch. 22 - (III) Charge is distributed within a solid sphere...Ch. 22 - Prob. 50GPCh. 22 - Prob. 51GPCh. 22 - The Earth is surrounded by an electric field,...Ch. 22 - Prob. 53GPCh. 22 - Prob. 54GPCh. 22 - Prob. 55GPCh. 22 - Prob. 57GPCh. 22 - Prob. 58GPCh. 22 - Prob. 59GPCh. 22 - Prob. 60GPCh. 22 - Prob. 61GPCh. 22 - Prob. 62GPCh. 22 - Prob. 63GPCh. 22 - Prob. 64GPCh. 22 - Prob. 65GPCh. 22 - Prob. 66GP
Knowledge Booster
Similar questions
- A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge per unit length of , and the cylinder has a net charge per unit length of 2. From this information, use Gausss law to find (a) the charge per unit length on the inner surface of the cylinder, (b) the charge per unit length on the outer surface of the cylinder, and (c) the electric field outside the cylinder a distance r from the axis.arrow_forwardGauss's Law problem: An infinite, uniform, line of charge is on the x-axis. The linear charge density is (lambda), with units of C/m. Find an expression for the electric field at a particular y-value on the y-axis at x=0, using Gauss's Law. Do this problem as if (lambda) is positive -- the answer is valid regardless of the sign. Suppose you moved the problem left or right, or rotated the problem about the x-axis. How does the problem change?arrow_forwardGauss's Law problem: An infinite, uniform, line of charge is on the x-axis. The linear charge density is (lambda), with units of C/m. Find an expression for the electric field at a particular y-value on the y-axis at x=0, using Gauss's Law. Do this problem as if (lambda) is positive -- the answer is valid regardless of the sign. Pick a shape for your Gaussian surface. Since you don't know the value (or expression) for E, you must pick a surface where the electric flux is either EA or zero. (E must be uniform over the surface for EA.)arrow_forward
- If the electric field of a point charge were proportional to 1/r3instead of 1/r2, would Gauss’s law still be valid? Explain your reasoning.(Hint: Consider a spherical Gaussian surface centered on a singlepoint charge.)arrow_forwardConsider Gauss's law: f Ē - d.Ã=q/€0 . Which of the following is true? E must be the electric field due to the enclosed charge If q = 0, then E = 0 everywhere on the Gaussian surface If the three particles inside have charges of +q, +q, and -2q, then the integral is zero If a charge is placed outside the surface, then it cannot affect E at any point on the surface on the surface E is everywhere parallel to dAarrow_forwardThe figure below shows a closed cylinder with cross-sectional area A = 2.00m2. The constant electric field E has magnitude 3.50 X 103 N/C and is directed vertically upward, perpendicular to the cylinder's top and bottom surfaces so that no field lines pass through the curved surface. Calculate the electric flux through the cylinder's (a) top and (b) bottom surfaces. (c) Determine the amount of charge inside the cylinderarrow_forward
- Gauss’ Law has an important practical application. Imagine a solid conducting object which may be in a strong electric field. If all excess charge is on the surface of the conductor and if there is no electric field inside the conductor, then it should be possible to remove any conducting material from the interior without changing the location of the charges or the electric field inside. This will leave a hollow metallic shell which has no electric field inside it. There is no electric field inside even if there is a strong field outside: the conducting shell acts as a “shield” against electric fields. Furthermore, the hollow shell can have holes in it or even be made of a wire mesh and still retain its shielding property. Such a space surrounded by a conductor is called a “Faraday cage”. Materials : • A metal cup • A hard plastic rod and fur • A metallic ball suspended by a thread A. Hold the rod in its position outside the cup as near as possible to the ball inside the cup.…arrow_forwardGauss’ Law has an important practical application. Imagine a solid conducting object which may be in a strong electric field. If all excess charge is on the surface of the conductor and if there is no electric field inside the conductor, then it should be possible to remove any conducting material from the interior without changing the location of the charges or the electric field inside. This will leave a hollow metallic shell which has no electric field inside it. There is no electric field inside even if there is a strong field outside: the conducting shell acts as a “shield” against electric fields. Furthermore, the hollow shell can have holes in it or even be made of a wire mesh and still retain its shielding property. Such a space surrounded by a conductor is called a “Faraday cage”. Materials : • A metal cup • A hard plastic rod and fur • A metallic ball suspended by a thread Now try the experiment without touching the can with your hand or any other electrical conductor.…arrow_forwardA solid, insulating sphere of radius a has a uniform charge density throughout its volume and a total charge Q. Concentric with this sphere is an uncharged, conducting, hollow sphere whose inner and outer radii are b and c as shown. We wish to understand completely the charges and electric fields at all locations. (a) Find the charge contained within a sphere of radius r < a. (b) From this value, find the magnitude of the electric field for r < a. (c) What charge is contained within a sphere of radius r when a < r < b? (d) From this value, find the magnitude of the electric field for r when a < r < b. (e) Now consider r when b < r < c. What is the magnitude of the electric field for this range of values of r ? (f) From this value, what must be the charge on the inner surface of the hollow sphere? (g) From part (f), what must be the charge on the outer surface of the hollow sphere? (h) Consider the three spherical surfaces of radii a, b, and c. Which of these…arrow_forward
- (a) Define uniform and non-uniform electric flux passing through a closed surface. (a) Explain Gauss's Law for Point Charges Inside Spherical and Nonspherical Surfaces, followed by the general form of Gauss's law.arrow_forwardConsider two parallel copper plates both 1m? in area. The lower plate is charged with 4µC of charge and the upper plate with -4µC of charge. (a) Determine the charge density o on each plate. (b) Neglecting edge effects (assuming a homogeneous field between the plates), show that the electric field can be expressed in the form E = Eoê using Gauss' Law. Determine the value of Eo. (c) Consider now the addition of an open hemispherical surface (no base) of radius R between the two plates in the orientation shown in Fig. 1. What is the expression for electric flux through this surface? Given R = 3cm, what is the value of the flux? R FIGURE 1 (d) Consider now a rotation of the hemispherical surface of 90° about the x-axis. What is the the new flux through the surface? (e) Following (d), a point charge q is placed at the origin of the hemispherical surface. Assume that small enough that it does not affect the charge density on the plates. What is the new flux through the hemisphere? isarrow_forwardA rod with linear charge density λ is bent into a quarter circle with radius R. (a) Find expressions for the electric field (x and y direction) at the point P shown due to a small piece of charge at some angle θ. (b) Integrate your expression to find the total electric field due to the entire rod of charge. (c) Without integrating again, determine the electric field (at point P) for the case shown of two semi-circular rods of equal and opposite charge density as shown.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning