Four Charged Particles in a Square

Click For Summary
SUMMARY

The problem involves calculating the ratio of charges q/Q for a system of four charged particles arranged in a square. The charges are defined as q1 = +Q, q2 = q3 = q, and q4 = -5.25Q. The net electrostatic force on particle 1 is zero when the resultant forces from particles 2 and 3 balance the force from particle 4. The correct ratio is determined to be q/Q = 1.31 after correctly calculating the vector components of the forces acting on particle 1.

PREREQUISITES
  • Understanding of Coulomb's Law for electrostatic forces
  • Knowledge of vector addition and resultant forces
  • Familiarity with the Pythagorean theorem
  • Ability to draw and interpret free body diagrams
NEXT STEPS
  • Study vector addition in physics to enhance understanding of resultant forces
  • Review Coulomb's Law and its applications in electrostatics
  • Practice problems involving multiple charges and net force calculations
  • Explore free body diagram techniques for complex systems
USEFUL FOR

Students studying electrostatics, physics educators, and anyone interested in solving problems related to charged particle interactions.

doogles8654
Messages
11
Reaction score
0

Homework Statement


In the figure below, four particles form a square. The charges are q1 = +Q, q2 = q3 = q, and q4 = -5.25Q. What is q/Q if the net electrostatic force on particle 1 is zero?
hrw7_21-22.gif



Homework Equations


F=k|q1||q2|/r2
a2+b2=c2


The Attempt at a Solution


I first drew a free body diagram of particle one with the forces of particles 2 and 3. Their resultant force needs to balance exactly the force of particle 4 in order for the net force on particle 1 to be zero.

Forces 2 and 3 both equal kQq/a2 so their resultant is 2kQq/a2.

The distance between particle 1 and 4 i called c. a2+b2=c2. In this case a and b are both a(according to the diagram) so: a2+a2=c2. Thus c=a\sqrt{2}.

I used this to get the force of 4 on 1: kQ|-5.25Q|/a\sqrt{2}2, which when simplified becomes 5.25q2k/2a2.

In setting this equal to the resultant of forces 2 and 3, I end up with 5.25/4=1.31=q/Q. This is not coming up as the right answer.
 
Physics news on Phys.org
Check your resultant for forces 2 and 3. They form vector components and so should be added accordingly to find the magnitude of the resultant.
 
Okay wow, thanks! Hehe, totally forgot how to find the resultant of two vectors I guess. :P Good call, I got it now!
 
F2+F3=\sqrt{2}kQq/a2
 

Similar threads

Replies
23
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
14
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
5
Views
7K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K