Sample Problems 1.4 1. Three point charges are located along the x-axis. Point charge q,=+3.5×10-6 C is at x=0, point charge q,=+8.5×10-6 C is at x=2.0 m, and point charge q3=-5.0×10-6 C is at x= 3.0 m. Find the resultant electric force acting on q,. 92 2.0 m 3.0 m Given: 9,=+3.5×10-6 C 92=+8.5×10- C 93=-5.0×10-6 C r2 on 1=2.0 m-0=2.0 m r3 on 1=3.0 m–0=3.0 m Solution: Solve first for the individual forces F2on 1 (force exerted by q, on q,) and F3 on 1 (force exerted by q3 on q,). Using Eq. (1.1), Fm1 = k- (r, on 1) 2 on 1 (+3.5×10“CX+8.5×10-“C)| =[9×10° N-m²/C²|| (2.0 m) = 0.0669 N 0.07 N. Point charge q, repels q1. Thus, F is directed to the left. 2 on 1 F, m 1 = k- 3 on (+3.5×10“CX-5.0×10-“C)| = [9×10° N-m²/C²]| (3.0 m)? = 0.0175 N20.02 N. Point charge q, attracts q,. Therefore, F3 on 1 is directed to the right. Setting the left direction to be positive and the right direction to be negative, the magnitude F of the resultant force is F=0.0669 N – 0.0175 N=0.0494 N, directed to the left.
Sample Problems 1.4 1. Three point charges are located along the x-axis. Point charge q,=+3.5×10-6 C is at x=0, point charge q,=+8.5×10-6 C is at x=2.0 m, and point charge q3=-5.0×10-6 C is at x= 3.0 m. Find the resultant electric force acting on q,. 92 2.0 m 3.0 m Given: 9,=+3.5×10-6 C 92=+8.5×10- C 93=-5.0×10-6 C r2 on 1=2.0 m-0=2.0 m r3 on 1=3.0 m–0=3.0 m Solution: Solve first for the individual forces F2on 1 (force exerted by q, on q,) and F3 on 1 (force exerted by q3 on q,). Using Eq. (1.1), Fm1 = k- (r, on 1) 2 on 1 (+3.5×10“CX+8.5×10-“C)| =[9×10° N-m²/C²|| (2.0 m) = 0.0669 N 0.07 N. Point charge q, repels q1. Thus, F is directed to the left. 2 on 1 F, m 1 = k- 3 on (+3.5×10“CX-5.0×10-“C)| = [9×10° N-m²/C²]| (3.0 m)? = 0.0175 N20.02 N. Point charge q, attracts q,. Therefore, F3 on 1 is directed to the right. Setting the left direction to be positive and the right direction to be negative, the magnitude F of the resultant force is F=0.0669 N – 0.0175 N=0.0494 N, directed to the left.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Question
Using the same charge configuration in Sample Problems 1.4 item number 1, find the resultant electric force in q2.
![Sample Problems 1.4
1. Three point charges are located along the x-axis. Point charge q, =+3.5×10-6 C is at
x=0, point charge q,=+8.5×10-6 C is at x=2.0 m, and point charge q;=-5.0×10-“ C is at
x= 3.0 m. Find the resultant electric force acting on qi.
92
93
2.0 m
3.0 m
Given: 9,=+3.5×10-6 C
92=+8.5×10-6 C
93=-5.0×10-6 C
r2 on 1=2.0 m-0=2.0 m
r3 on 1=3.0 m-0=3.0 m
Solution:
Solve first for the individual forces F2 on 1 (force exerted by q, on q,) and F,
3 on 1 (force exerted
by
93 on q1). Using Eq. (1.1),
F. om 1 = k-
(r, on 1)°
[[(+3.5×10-“CX+8.5×10-“C)|
= [9x10° N-m³/C']|
(2.0 m)
= 0.0669 N 0.07 N.
Point charge q, repels q,. Thus, F2 on 1 is directed to the left.
F, m1 = k
(r, on 1)
3 on
(+3.5×10“CX-5.0×10-“C)|
=[9×10° N-m²/C}|
(3.0 m)?
= 0.0175 N 20.02 N.
Point charge q; attracts q,. Therefore, F, on 1 is directed to the right. Setting the left
direction to be positive and the right direction to be negative, the magnitude F of the resultant
force is F=0.0669 N – 0.0175 N=0.0494 N, directed to the left.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdfbfb0f6-c041-4d62-b974-c733a0bdfd24%2F819fde67-04f0-4f2a-8407-5de497a24feb%2Fr8ynbn_processed.png&w=3840&q=75)
Transcribed Image Text:Sample Problems 1.4
1. Three point charges are located along the x-axis. Point charge q, =+3.5×10-6 C is at
x=0, point charge q,=+8.5×10-6 C is at x=2.0 m, and point charge q;=-5.0×10-“ C is at
x= 3.0 m. Find the resultant electric force acting on qi.
92
93
2.0 m
3.0 m
Given: 9,=+3.5×10-6 C
92=+8.5×10-6 C
93=-5.0×10-6 C
r2 on 1=2.0 m-0=2.0 m
r3 on 1=3.0 m-0=3.0 m
Solution:
Solve first for the individual forces F2 on 1 (force exerted by q, on q,) and F,
3 on 1 (force exerted
by
93 on q1). Using Eq. (1.1),
F. om 1 = k-
(r, on 1)°
[[(+3.5×10-“CX+8.5×10-“C)|
= [9x10° N-m³/C']|
(2.0 m)
= 0.0669 N 0.07 N.
Point charge q, repels q,. Thus, F2 on 1 is directed to the left.
F, m1 = k
(r, on 1)
3 on
(+3.5×10“CX-5.0×10-“C)|
=[9×10° N-m²/C}|
(3.0 m)?
= 0.0175 N 20.02 N.
Point charge q; attracts q,. Therefore, F, on 1 is directed to the right. Setting the left
direction to be positive and the right direction to be negative, the magnitude F of the resultant
force is F=0.0669 N – 0.0175 N=0.0494 N, directed to the left.
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