Force acting on a charge

  • Thread starter stobbz
  • Start date
  • #1
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Homework Statement



A charge +2nC is located at (0, 2 m). Another +2 nC charge is
located at (0, -2 m). A third charge, +3 nC, is located at (2 m, 0).
Calculate the total force (in vector form) acting on charge +3nC
from both +2nC charges.

Homework Equations



Coulomb's law in vector form

The Attempt at a Solution



I worked out the force on q3 due to q1 to be:

9.54 nano Newton in the positive x direction and
-9.54 nano Newton in the negative y direction

Next, I worked out the force on q3 due to q2 and got:

9.54 nano Newton in the positive x direction and in the positive y direction.

Adding these together, the component would be zero and the total force would be 19.08 nano Newton in the x direction. This would be my answer. However, the answer given is 9.537 nano Newton in the x direction.

Any help is appreciated!
Thanks
 
Last edited:

Answers and Replies

  • #2
gneill
Mentor
20,816
2,792
What is the difference between "in the x direction" and "in the positive x direction"?:confused:
 
  • #3
13
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What is the difference between "in the x direction" and "in the positive x direction"?:confused:
Nothing, I've changed it to avoid any more confusion :smile:
 
  • #4
43
1

I worked out the force on q3 due to q1 to be:

9.54 nano Newton in the positive x direction and
-9.54 nano Newton in the negative y direction

So, a force in the positive y-direction?

19.08 nano Newton in the x direction. This would be my answer. However, the answer given is 19.08 nano Newton in the positive x direction.
You got the right answer?

Please correct and I'll see if I can help...
 
  • #5
13
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So, a force in the positive y-direction?


You got the right answer?

Please correct and I'll see if I can help...
Ahh sorry, the given answer is 9.537 nN.
 
  • #6
gneill
Mentor
20,816
2,792
So it looks like you've mucked up a factor of two somewhere. Can you show your calculations? What value did you use for the distance in Coulomb's law?
 
  • #7
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So it looks like you've mucked up a factor of two somewhere. Can you show your calculations? What value did you use for the distance in Coulomb's law?
I used 4 (two squared) for the position vector and root 2/2 for the unit vector. This is for the x component for the force on q3 exerted by q1.
 
  • #8
gneill
Mentor
20,816
2,792
Let's take the case of q1 and q3. q1 is at (0,2), while q3 is at (2,0). The distance betwixt is:

[tex] \sqrt{(2 - 0)^2 + (0 - 2)^2} = \sqrt{4 + 4} = 2 \sqrt{2}[/tex] (meters)
 
  • #9
13
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r13 = (2, -2), unit vector r = root 2 / 2 in the x direction - root 2 / 2 in the y direction,

F13 = Q1Q3 * unit vector / 4 PI Epsilon nought r^2

(with r = 2)
 
  • #10
13
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Ahh ok, now I see where I was going wrong!

Thanks for walking me through that gneill, massively appreciate it.
 

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