College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Related questions
Concept explainers
Question
Problem 13: Each plate of a parallel-plate capacitor has an area of A = 0.61 m2. The plate separation is 3.0 mm, and one of these plates is shown in the figure. It carries a charge of Q = 2.9 μC, which is concentrated on its inner surface. Imagine a cylindrical Gaussian surface, of radius r = 0.012 m, whose axis is perpendicular to the plates. One end of the cylinder is inside the plate shown and the other end is located between the plates. Both ends are parallel to the plates.
Part (a) What is the flux through surface 1 Φ1, in newton meters squared per coulomb?
Part (b) What is the flux through surface 2 (the outside of the cylinder not including the ends), in newton meters squared per coulomb?
Part (c) What is the flux through surface 3 Φ3, in newton meters squared per coulomb?
Part (d) Using all of your results, input an expression for the field within the capacitor E, in terms of the quantities given in the problem.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution
Trending nowThis is a popular solution!
Step by stepSolved in 6 steps with 5 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- 8arrow_forwardOne way to make a uniform electric field is to use what is known as a parallel-plate capacitor. To make one, you take two metal cookie sheets from your kitchen, and place them parallel to one another. Then, you connect the positive terminal of a 9-volt battery to the top cookie sheet, so that it becomes positively charged, and you connect the negative terminal of the same battery to the bottom cookie sheet, so that the bottom cookie sheet becomes negatively charged. In the picture, the top cookie sheet is drawn in red (positive) and the bottom cookie sheet is drawn in blue (negative). Note that the electric field between the cookie sheets points from the positive sheet to the negative sheet, and has a magnitude of the battery voltage (9 V) divided by the distance between the cookie sheets Part (c) The distance between the cookie sheets is 24.0 cm. An electron (in blue) is given an initial velocity horizontally. The initial velocity of the electron is directed along the line…arrow_forwardTwo parallel plates, each charged equally and oppositely to the other, are separated by 2.3500 cm. A proton is let go from rest at the positive plate's surface and, at the same time, an electron is let go from rest at the negative plate's surface. What is the distance between the negative plate and the point where the proton and the electron go by each other? Note: unlike most questions, this one will need your answer correct to 5 significant digitsarrow_forward
- The image shows a coaxial cable (two nested cylinders) of length l, inner radius a and outer radius b. Note that l >> a and l >> b. The inner cylinder is charged to +Q and the outer cylinder is charged to −Q. a) Use Gauss’s Law to calculate the electric field E at r < a, a < r < b, and r > b. b) Calculate the electric potential difference between a and b that is ∆Vabarrow_forwardQuestion 6: The plane y = 0 distinguishes two lossless dielectric zones with dielectric constants ɛr1 = 2 and ɛr2 = 3. 1. If the region is known as E ****1 = x2 - ŷ3 + bilin5, calculate the electric field and displacement vectors E****2 and D****2 in region 2. (There is no load density on the interface.arrow_forwardA parallel-plate capacitor is formed from two 5.8 cm -diameter electrodes spaced 1.9 mm apart. The electric field strength inside the capacitor is 1.0 ×106N/C. What is the charge (in nC) on each electrode?arrow_forward
- Case: Observation of two point charges revealed the following data. For referencing the position of the point charges in vacuum, an empty region which will be called point zero will the origin of the angle which will expressed in radians. Values: Point charge A Charge = +133 μC Distance from point zero = 376 cm Angular position = π rad Point charge B Charge = –293 μC Distance from point zero = 217 cm Angular position = 0.20943951023932 rad a) Magnitude of electric field due to point charge A at point zero b) Magnitude of electric field due to point charge B at point zero c) Magnitude of electric field due to point charge A and point charge B at point zero[1] d) Direction of the net electric field in proximity with point zero e) Acceleration of a proton if placed at point zero[2] Notes: [1] direction of the vector components of the electric field based on standard test charge polarityarrow_forwardThe drawing shows an electron entering the lower left side of a parallel plate capacitor and exiting at the upper right side. The initial speed of the electron is 3.13 × 106 m/s. The capacitor is 2.00 cm long, and its plates are separated by 0.150 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude. 15.30 + A + 2.00 cm Number i + + + 0.150 cm Unitsarrow_forwardA 1.4-µC point charge is placed between the plates of a parallel plate capacitor. The charge experiences a force of 0.89 N. What is the magnitude o of the charge density on either plate of the capacitor? 0 = iarrow_forward
- A spherical capacitor is composed of two concentric conducting spheres, one of radius a and the other of radius c (c > a). In addition, between the two conductors there is a spherical shell of dielectric material (relative permittivity/relative dielectric constant ) with inner radius b (c > b > a) and outer radius c. The charge on the inner conductor is +Q. The charge on the outer conductor is -Q. (a) Make a sketch of the situation, indicating the relevant dimensions. (b) Determine the magnitude of the electric field E at radius r for a < r < b. (c) Determine the magnitude of the electric field E at radius r for b < r < c. (d) What is the (induced) surface charge density on the inner surface of the dielectric. (e) Sketch the radial component of the electric field versus r . (f) Sketch the electrostatic potential versus r . (g)Calculate the potential difference between the conductor at r = a and that at r = c. (h) What is the capacitance of this capacitor?arrow_forwardu.li)!ransactloh=LMS.EDU.StudêntEvaluation. TakeStudentEvaluation/1228666 Va 2d P + Q v. A particle of charge +q and mass m is ejected from the point P with the initial velocity = 40î + 30f(), traveling in a parabolic orbit with a maximum height d = 2cm in the uniform electric field and hitting the point Q at a distance of L 10cm. Since the distance between the plates is 2d and the potential difference is 1000V. find the ratio of the particle's charge to its mass (q/m). (Ignore the gravitational force). a) 16. 10-2 (C) edu tr/upload/ytu/CourseScan/2c8c1008-1e08-4748-b296-a55f1284f4e5.png mak için buraya yazın e 明 KI Coparrow_forwardSuppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rb has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = αe-r/a0 + β/r + b0 where alpha (α), beta (β), a0 and b0 are constants. Find an expression for its capacitance. First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: Calculating the antiderivative or indefinite integral , Vab = (-αa0e-r/a0 + β + b0 ) By definition, the capacitance C is related to the charge and potential difference by: C = / Evaluating with the upper and lower limits of integration for Vab, then simplifying: C = Q / ( (e-rb/a0 - e-ra/a0) + β ln() + b0 () )arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
University Physics (14th Edition)PhysicsISBN:9780133969290Author:Hugh D. Young, Roger A. FreedmanPublisher:PEARSON
Introduction To Quantum MechanicsPhysicsISBN:9781107189638Author:Griffiths, David J., Schroeter, Darrell F.Publisher:Cambridge University Press 
Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Lecture- Tutorials for Introductory AstronomyPhysicsISBN:9780321820464Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina BrissendenPublisher:Addison-Wesley
College Physics: A Strategic Approach (4th Editio...PhysicsISBN:9780134609034Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart FieldPublisher:PEARSON
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON