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

  • Thread starter Violagirl
  • Start date
  • #1
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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?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Attachments

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
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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.
 

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