Attraction of -Q to +Q in x and y directions

  • Thread starter Thread starter Violagirl
  • Start date Start date
  • Tags Tags
    Attraction
Click For Summary
SUMMARY

The discussion focuses on calculating the force experienced by a charge -Q due to two other charges, Q1 and Q2, in both x and y directions. The participant correctly identifies the forces as Fx = k(Q1)(-Q)/(a) x-hat and Fy = (k)(Q2)(-Q)/(a) y-hat, leading to the resultant force F = Fx + Fy. The participant seeks confirmation on the correctness of their calculations and the direction of the forces, emphasizing the importance of understanding vector components in electrostatic force calculations.

PREREQUISITES
  • Understanding of Coulomb's Law and electrostatic forces
  • Familiarity with vector addition in physics
  • Knowledge of the concepts of charge interaction and directionality
  • Basic proficiency in using the formula F = k(q)(Q)/r²
NEXT STEPS
  • Review vector addition in electrostatics
  • Study the implications of charge polarity on force direction
  • Learn about the superposition principle in electrostatic forces
  • Explore advanced applications of Coulomb's Law in multi-charge systems
USEFUL FOR

Students studying physics, particularly those focusing on electrostatics, as well as educators looking to enhance their understanding of charge interactions and vector forces.

Violagirl
Messages
112
Reaction score
0

Homework Statement



Find the magnitude and direction of the force on the charge -Q on the figure in the attached document.

Homework Equations



F = k(q)(Q)/r2r-hat


The Attempt at a Solution



So I have a force in the x direction and a force in the y direction.

Fx = k(Q1)(-Q)/(a) x-hat

Fy = (k)(Q2)(-Q)/(a) y-hat

F = Fx + Fy = -kQ22/(a)2 x-hat + -kQ12/a2 y-hat

Is this right? Or can more be done to solve it?
 

Attachments

  • Attraction of -Q to +Q in +x and +y directions.jpg
    Attraction of -Q to +Q in +x and +y directions.jpg
    3.4 KB · Views: 471
Physics news on Phys.org
Almost there - check the directions of the forces.
Your formula has the -Q charge being pushed in the -x direction.
Is that what would happen?

In general - the force experienced by charge q due to charge Q is:
$$\vec{F}=\frac{kQ}{r^3}q\vec{r}$$ Here r points from Q to q.
 

Similar threads

Replies
12
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
1
Views
3K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
5K