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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Using the same charge configuration in Sample Problems 1.4 item number 2, find the resultant electric force on the charge at vertex B.
![Three identical point charges with charge q=+3.0×10-"C are placed
at each vertex of an equilateral triangle ABC as shown. If the side of the
equilateral triangle is 0.01 m, find the resultant electric force on the
charge at vertex A.
2.
A
r
Given: q,=q, = 9c =+3.0×10-6 C
r = 0.01 m
Solution:
Fs on A
= k
=[9x10° N m²/C²|(+3.0×10""CX+3.0×10-°)]
(0.01 m)
-9-
%3D
= 810 N
Point charge q, repels q.. Thus, F on A is directed to the right up at 60° (since the triangle
is equilateral) with the positive x-axis. Force Fc on A is computed using the same equation.
Fe on A
= k
|(+3.0×10 “CX+3.0×10 °C)
= 19x10' N m²/C²|
(0.01 m)²
=810 N
810 N
Point charge qc repels q,. Thus, Fcon A is directed upward to
the left at an angle of 60° with the negative x-axis.
810 N
To visualize FBon A and Fc on A» Ia is placed at the origin of the
Cartesian coordinate system as shown. The component method is
used to determine the resultant electric force F.
60°
60°
Force
Horizontal Component
Vertical Component
+810 N (cos 60°) = +405 N
-810 N (cos 60°)
+810 N (sin 60°) = 702 N
+810 N (sin 60°) = 702 N
Fs on A = 810 N
Fcon A = 810 N
-405 N
Resultant electric force F
EF, = 0
EF, = 1404 N
Therefore, the resultant electric force is equal to 1404 N~1400 N acting vertically up.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdfbfb0f6-c041-4d62-b974-c733a0bdfd24%2F6331bdd7-270d-440e-bec1-504954530500%2Fpdhapjb_processed.png&w=3840&q=75)
Transcribed Image Text:Three identical point charges with charge q=+3.0×10-"C are placed
at each vertex of an equilateral triangle ABC as shown. If the side of the
equilateral triangle is 0.01 m, find the resultant electric force on the
charge at vertex A.
2.
A
r
Given: q,=q, = 9c =+3.0×10-6 C
r = 0.01 m
Solution:
Fs on A
= k
=[9x10° N m²/C²|(+3.0×10""CX+3.0×10-°)]
(0.01 m)
-9-
%3D
= 810 N
Point charge q, repels q.. Thus, F on A is directed to the right up at 60° (since the triangle
is equilateral) with the positive x-axis. Force Fc on A is computed using the same equation.
Fe on A
= k
|(+3.0×10 “CX+3.0×10 °C)
= 19x10' N m²/C²|
(0.01 m)²
=810 N
810 N
Point charge qc repels q,. Thus, Fcon A is directed upward to
the left at an angle of 60° with the negative x-axis.
810 N
To visualize FBon A and Fc on A» Ia is placed at the origin of the
Cartesian coordinate system as shown. The component method is
used to determine the resultant electric force F.
60°
60°
Force
Horizontal Component
Vertical Component
+810 N (cos 60°) = +405 N
-810 N (cos 60°)
+810 N (sin 60°) = 702 N
+810 N (sin 60°) = 702 N
Fs on A = 810 N
Fcon A = 810 N
-405 N
Resultant electric force F
EF, = 0
EF, = 1404 N
Therefore, the resultant electric force is equal to 1404 N~1400 N acting vertically up.
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