Finding magnitude of electrostatic force

  • #1
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We have a question about finding the electrostatic force acting on a particle from another particle. They are colinear since there is only the 2 particles, but not along an axis.

My question is, why does finding the force in the x direction and finding the force in the why direction, then doing

√(Fx2 + Fy2) not give me the same answer as if I did

kq1q2/ [(y2-y1)2 + (x2-x1)2]?

I'm new to typing on here also so I'm not quite sure about complete formatting yet.
 

Answers and Replies

  • #2
TSny
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My question is, why does finding the force in the x direction and finding the force in the why direction, then doing

√(Fx2 + Fy2) not give me the same answer as if I did

kq1q2/ [(y2-y1)2 + (x2-x1)2]?

It should give the the same answer if you found Fx and Fy correctly.

However, it is easier to find the force directly using kq1q2/ [(y2-y1)2 + (x2-x1)2]
 
  • #3
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It should give the the same answer if you found Fx and Fy correctly.

However, it is easier to find the force directly using kq1q2/ [(y2-y1)2 + (x2-x1)2]

My guess would be I didn't find them correctly then. I used the formula for electrostatic force, except for r2, I used (x2-x1)2 and did a similar thing for the y direction
 
  • #4
TSny
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My guess would be I didn't find them correctly then. I used the formula for electrostatic force, except for r2, I used (x2-x1)2 and did a similar thing for the y direction
That's not the correct way to get the components. To convince yourself of that, suppose the two charges were both on the y-axis. What would you get for the x-component of the force if you used (x2-x1)2 for r2?
 
  • #5
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That's not the correct way to get the components. To convince yourself of that, suppose the two charges were both on the y-axis. What would you get for the x-component of the force if you used (x2-x1)2 for r2?

We would end up with 0, making the component undefined, which wouldn't make sense. So why it doesn't work makes sense now, but how would you go about finding the components, despite finding the magnitude first being easier?
 
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  • #6
TSny
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The force is parallel to the line connecting the particles. The angle θ that the force makes to the x axis is the same as the angle that the connecting line makes to the x axis.

So, ##\cos\theta = \frac{x_2-x_1}{r}## and ##F_x = F \cos\theta = \frac{kq_1q_2 (x_2-x_1)}{r^3}##

Similarly for the y component.
 
  • #7
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Okay, I see where that comes from now and that makes sense as well. Thanks for the help!
 

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