Example Problems 1. An electric charge q,-10 μC remains located at the origin of a coordinate system. A second electrical charge q2 20 uC is moved from point to point. (a) How much work is required by an external force to move the second charge along the x-axis from point (1 m, 0) to (2m, 0)? (b) What is the electric potential at the point (1 m, 0) due to the charge at the origin? (c) What is the potential difference between the points (1 m, 0) and (2 m, 0)? (d) What is the potential difference between points (1 m, 0) and (0, 1 m)? (e) How much work is required by an external force to move the second charge from point (1 m, 0) to (0, 1 m)? () What is the integral SE di along a circle from point (I m, 0) to (0, 1 m)? Solution: Given: q1 = 1.00 x 10-5 C Eo-8.85x10-12 C2/Nm? q2 = 2.00 x 10-5 C Equations Substitution: 1 9192 4TnEo r2 (a) Felec_4πεο r Fext elec =-F 2m dx 2 1m r=x 9192 1 1 4πεο (2 m-1 = |-0.90 J Vb = 9.0x104 V Ve 4.5x104 v Vbc = 4.5x104 V Va = 9.0x104 V 1 91 Vc = 41160 (d) Vd Edl 0 The same result is obtained by noticing that E is always perenclar to di along the circular path of integration, so Edl-0 and di-0.
Example Problems 1. An electric charge q,-10 μC remains located at the origin of a coordinate system. A second electrical charge q2 20 uC is moved from point to point. (a) How much work is required by an external force to move the second charge along the x-axis from point (1 m, 0) to (2m, 0)? (b) What is the electric potential at the point (1 m, 0) due to the charge at the origin? (c) What is the potential difference between the points (1 m, 0) and (2 m, 0)? (d) What is the potential difference between points (1 m, 0) and (0, 1 m)? (e) How much work is required by an external force to move the second charge from point (1 m, 0) to (0, 1 m)? () What is the integral SE di along a circle from point (I m, 0) to (0, 1 m)? Solution: Given: q1 = 1.00 x 10-5 C Eo-8.85x10-12 C2/Nm? q2 = 2.00 x 10-5 C Equations Substitution: 1 9192 4TnEo r2 (a) Felec_4πεο r Fext elec =-F 2m dx 2 1m r=x 9192 1 1 4πεο (2 m-1 = |-0.90 J Vb = 9.0x104 V Ve 4.5x104 v Vbc = 4.5x104 V Va = 9.0x104 V 1 91 Vc = 41160 (d) Vd Edl 0 The same result is obtained by noticing that E is always perenclar to di along the circular path of integration, so Edl-0 and di-0.
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
Chapter25: Electric Potential
Section: Chapter Questions
Problem 25.12OQ: A particle with charge -40.0 nC is on the x axis at the point with coordinate x = 0. A second...
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Transcribed Image Text:Example Problems
1. An electric charge q,-10 μC remains located at the origin of a coordinate system. A second electrical
charge q2 20 uC is moved from point to point. (a) How much work is required by an external force to
move the second charge along the x-axis from point (1 m, 0) to (2m, 0)? (b) What is the electric
potential at the point (1 m, 0) due to the charge at the origin? (c) What is the potential difference
between the points (1 m, 0) and (2 m, 0)? (d) What is the potential difference between points (1 m, 0)
and (0, 1 m)? (e) How much work is required by an external force to move the second charge from point
(1 m, 0) to (0, 1 m)? () What is the integral SE di along a circle from point (I m, 0) to (0, 1 m)?
Solution:
Given:
q1 = 1.00 x 10-5 C
Eo-8.85x10-12 C2/Nm?
q2 = 2.00 x 10-5 C
Equations
Substitution:
1 9192
4TnEo r2
(a) Felec_4πεο r
Fext elec
=-F
2m
dx
2
1m
r=x
9192 1 1
4πεο (2 m-1
= |-0.90 J
Vb = 9.0x104 V
Ve 4.5x104 v
Vbc = 4.5x104 V
Va = 9.0x104 V
1 91
Vc = 41160
(d) Vd
Edl 0
The same result is obtained by noticing that E is always perenclar to di along the circular path of
integration, so Edl-0 and di-0.
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