Net force is Zero, quadratic answers

Click For Summary
SUMMARY

The discussion centers on the equilibrium of three point charges along the x-axis, specifically analyzing the position of a third charge (q3) that experiences zero net force due to two positive charges, q1 (15.0 µC) at x = 2.00 m and q2 (6.00 µC) at the origin. The solution derived from the quadratic equation 3x² + 8x - 8 = 0 yields two x-coordinates: 0.775 m and -3.44 m. The negative solution is deemed unphysical as it suggests that both forces on q3 would act in the same direction, contradicting the requirement for equilibrium. Thus, only the positive solution is valid for the placement of q3.

PREREQUISITES
  • Understanding of Coulomb's Law and electrostatic forces
  • Familiarity with quadratic equations and their solutions
  • Knowledge of charge interactions and equilibrium conditions
  • Basic principles of physics regarding forces and motion
NEXT STEPS
  • Study the implications of charge placement in electrostatics
  • Learn about the concept of electric field and potential
  • Explore the applications of quadratic equations in physics problems
  • Investigate the conditions for equilibrium in multiple force systems
USEFUL FOR

Students of physics, particularly those studying electrostatics and force interactions, as well as educators seeking to clarify concepts related to charge equilibrium and quadratic solutions.

flyingpig
Messages
2,574
Reaction score
1

Homework Statement



Three point charges lie along the x axis. The positive charge q1 = 15.0uC and is at x = 2.00m (2.00 away from the origin), the positive q2 = 6.00uC is at the origin. The net force on q3 which is between q2 and q1 is 0. What is the x coordinate of q3?


solution

My book (serway and jewett, 8ed, pg 665) says x = 0.775 from 3x^2 + 8x - 8 = 0

the other x = -3.44m

The book says the negative one is (and here I quote) "the magnitude of the forces on q3 are equal, but both forces are in the same direction".


Unfortunately I have no idea what that means, based on my knowledge, isn't the negative one just another point where it satisfies the conditon? So if it was moved to satisfy x = -3.44, then x = 0.755 would be rejected?
 
Physics news on Phys.org
If every charge is positive, would it make sense that the third positive charge could lie on the left-hand side of the first two charges and not feel a force?

The -3.44m is an unphysical answer. It is kind of like the typical ball falling off of a cliff kinematic problem where if you were asked to calculate the time the ball hits the ground, you'll find 2 times due to the quadratic nature of the calculation. One relates to the real situation where the ball hits the ground on the right hand side and the other one is as if time were reversed and the ball went from the top of the hill to the ground in the opposite direction (though the hill is in the way...). I hope that makes sense...
 

Similar threads

Replies
9
Views
3K
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
6
Views
11K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
6K