Physics for Science and Engineering With Modern Physics, VI - Student Study Guide

4th Edition

ISBN: 9780132273244

Author: Doug Giancoli

Publisher: PEARSON

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Textbook Question

Chapter 21, Problem 21P

(I) What are the magnitude and direction of the electric force on an electron in a uniform electric field of strength 1920 N/C that points due east?

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Chapter 21, Problem 20P

Chapter 21, Problem 22P

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4.) The diagram shows the electric field lines of a positively charged conducting sphere of radius R and charge Q. A B Points A and B are located on the same field line. A proton is placed at A and released from rest. The magnitude of the work done by the electric field in moving the proton from A to B is 1.7×10-16 J. Point A is at a distance of 5.0×10-2m from the centre of the sphere. Point B is at a distance of 1.0×10-1 m from the centre of the sphere. (a) Explain why the electric potential decreases from A to B. [2] (b) Draw, on the axes, the variation of electric potential V with distance r from the centre of the sphere. R [2] (c(i)) Calculate the electric potential difference between points A and B. [1] (c(ii)) Determine the charge Q of the sphere. [2] (d) The concept of potential is also used in the context of gravitational fields. Suggest why scientists developed a common terminology to describe different types of fields. [1]

3.) The graph shows how current I varies with potential difference V across a component X. 904 80- 70- 60- 50- I/MA 40- 30- 20- 10- 0+ 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 VIV Component X and a cell of negligible internal resistance are placed in a circuit. A variable resistor R is connected in series with component X. The ammeter reads 20mA. 4.0V 4.0V Component X and the cell are now placed in a potential divider circuit. (a) Outline why component X is considered non-ohmic. [1] (b(i)) Determine the resistance of the variable resistor. [3] (b(ii)) Calculate the power dissipated in the circuit. [1] (c(i)) State the range of current that the ammeter can measure as the slider S of the potential divider is moved from Q to P. [1] (c(ii)) Describe, by reference to your answer for (c)(i), the advantage of the potential divider arrangement over the arrangement in (b).

1.) Two long parallel current-carrying wires P and Q are separated by 0.10 m. The current in wire P is 5.0 A. The magnetic force on a length of 0.50 m of wire P due to the current in wire Q is 2.0 × 10-s N. (a) State and explain the magnitude of the force on a length of 0.50 m of wire Q due to the current in P. [2] (b) Calculate the current in wire Q. [2] (c) Another current-carrying wire R is placed parallel to wires P and Q and halfway between them as shown. wire P wire R wire Q 0.05 m 0.05 m The net magnetic force on wire Q is now zero. (c.i) State the direction of the current in R, relative to the current in P.[1] (c.ii) Deduce the current in R. [2]

Chapter 21 Solutions

Physics for Science and Engineering With Modern Physics, VI - Student Study Guide

Ch. 21.5 - Return to the Chapter-Opening Question, page 559,...Ch. 21.5 - What is the magnitude of F12 (and F21) in Example...Ch. 21.5 - Determine the magnitude and direction of the net...Ch. 21.5 - (a) Consider two point charges of the same...Ch. 21.6 - Four charges of equal magnitude, but possibly...Ch. 21 - If you charge a pocket comb by rubbing it with a...Ch. 21 - Why does a shirt or blouse taken from a clothes...Ch. 21 - Explain why fog or rain droplets tend to form...Ch. 21 - A positively charged rod is brought close to a...Ch. 21 - Why does a plastic ruler that has been rubbed with...

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