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posted by  siriusangel on 9/4/2008 12:54:12 AM  |  status: Live  

Electrical forces

Course Textbook Chapter Problem
General Physics College Physics (6th) by Buffa, Lou, Wilson 15 N/A
Question Details:
 Three charges are located at the corners of an equilateral triangle, what are the magnitude and the direction of the force on q1?
What is the electric field at the center of the triangle ?
 
 

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posted by Antony08 on 9/4/2008 11:50:23 AM  |  status: Live
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siriusangel's comment:
"Thanks for the detailed step by step explanation makes much more sense! :- )"
Response Details:
The force on charge q1 due to charge q2 is
F1 = (1/4πεo) * (q1q2/r2)
or F1 = (9 * 109) * [(4 * 10-6 * 4 * 10-6)/(20 * 10-2)2]
or F1 = 3.8 N
 
The force on charge q1 due to charge q3 is
F2 = (1/4πεo) * (q1q3/r2)
or F2 = (9 * 109) * [(4 * 10-6 * -4 * 10-6)/(20 * 10-2)2]
or F2 = -3.8 N
 
The net force on the charge q1 is
Fnet = [F12 + F22 + 2F1F2cosθ]1/2
or Fnet = [(3.8)2 + (-3.8)2 + 2 * 3.8 * -3.8 * cos(60o)]1/2
or Fnet = [14.44]1/2 = 3.8 N
 
The center of the triangle is at a distance of 8.66 cm from the vertices of the triangle.
The electric field at the center of the triangle is
E = (1/4πεo) * (2q/r2)
or E = 9 * 109 * [(2 * 4 * 10-6)/(20 * 10-2)2]
or E = 18 * 105 N/C = 1800 * 103 N/C
or E = 1800 kN/C
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