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CALC The electric potential V in a region of space is

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 86P Chapter 23

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 86P

CALC? The electric potential V in a region of space is given by where A is a constant. (a) Derive an expression for the electric field at any point in this region. (b) The work done by the field when a 1.50-µC test charge moves from the point (x, y, z) = (0, 0, 0.250 m) to the origin is measured to be 6.00 X 10-5 J. Determine A. (c) Determine the electric field at the point (0, 0, 0.250 m). (d) Show that in every plane parallel to the xz-plane the equipotential contours are circles. (e) What is the radius of the equipotential contour corresponding to V = 1280 V and y = 2.00 m?

Step-by-Step Solution:

Introduction First we have to calculate the electric field as a function of position. Then we have to calculate the constant A from the given work done. Then we have to calculate the electric field at the given point. Then we have to show that the equipotential contour parallel to xz plane are circle. And then we have to find out the radius of the equipotential couture at xz plane at y = 2.00 m Step 1 The potential is given by V (x,y,z) = A(x 3y + z )2 Now the electric field is given by 2 2 2 E(x,y,z) = V (x,y,z) = A(x 3y + z ) ˆ ˆ ˆ 2 2 2 = ( xi + y j + zk )(x 3y + z ) = 2A(xi 3yj + zk) ˆ ………………….(1) So the above equation represents the electric field as a function of x,y,z . Step 2 Now the work done in the electric field is given by W = q(V (x ,y ,z ) V (x ,y ,z ) = qA[(x 2 3y 2+ z ) (x 2 3y 2 + z )] 2 2 2 1 1 1 2 2 2 1 1 1 Here we have (x1,y 1z 1 = (0,0,0.250 m) and (x 2y 2z 2 = (0,0,0) 6 also q = 1.50 C = 1.50 × 10 C and W = 6.00 × 10 5 J So using the above values we have 5 6 2 6 × 10 J = (1.50 × 10 C)A( 0.25 m) 6×105J 2 A = (1.50×106C)(0.25 m) = 640 J/C.m ……………(2) Step 3 The electric field at point (0,0,0.250 m) is given by ˆ ˆ ˆ 2 ˆ ˆ E(0,0,0.25 m) = 2A(xi 3yj + zk) = 2(640 J/C.m )(0.250 m)k = ( 320 V /m)k Hence the electric field at the point (0,0,0.250 m) is ( 320 V /m)k .

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Chapter 23, Problem 86P is Solved
Step 5 of 5

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

The full step-by-step solution to problem: 86P from chapter: 23 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. University Physics was written by and is associated to the ISBN: 9780321675460. The answer to “CALC? The electric potential V in a region of space is given by where A is a constant. (a) Derive an expression for the electric field at any point in this region. (b) The work done by the field when a 1.50-µC test charge moves from the point (x, y, z) = (0, 0, 0.250 m) to the origin is measured to be 6.00 X 10-5 J. Determine A. (c) Determine the electric field at the point (0, 0, 0.250 m). (d) Show that in every plane parallel to the xz-plane the equipotential contours are circles. (e) What is the radius of the equipotential contour corresponding to V = 1280 V and y = 2.00 m?” is broken down into a number of easy to follow steps, and 116 words. This textbook survival guide was created for the textbook: University Physics, edition: 13. Since the solution to 86P from 23 chapter was answered, more than 553 students have viewed the full step-by-step answer. This full solution covers the following key subjects: Field, electric, point, Plane, equipotential. This expansive textbook survival guide covers 26 chapters, and 2929 solutions.

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