Magnitude of Force of two point charges on a third (electrostatics)

  • #1
Two point charges are placed on the x-axis as follows: charge q1 = 4.01nC is located at x= 0.201m , and charge q2 = 5.00nC is at x= -0.301m.

What is the magnitude of the total force exerted by these two charges on a negative point charge q3 = -6.03nC that is placed at the origin?

I know I need to use:
F = (k * q_1 * q_2)/d^2, where k = 9E9


I've tried taking the sum of magnitudes of F(1 on 3) and F(2 on 3), but have been incorrect. I set it up like below:

F(1 on 3) = (k * 4.01E-9 * -6.03E-9) / 0.201^2 = -0.000005 N

F(2 on 3) = (k * 5.00E-9 * -6.03E-9) / 0.301^2 = -0.000003 N


This would give the answer as 0.000008 N, or more precisely 8.37E-6 N. Could anyone tell me where I'm going wrong?

Homework Statement





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The Attempt at a Solution

 

Answers and Replies

  • #2
rude man
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check the sign of F(2 on 3). Which direction is it?
 
  • #3
Yeah, it should be positive, my bad. However I've also tried 0.000002 as an answer and that was incorrect:/
 
  • #4
gneill
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Lose the fixed point notation and stick to scientific notation so that significant figures are clear. What's the value of the net force to three significant figures?
 
  • #5
F(1 on 3) = -5.387E-6
F(2 on 3) = -2.995E-6

This is a magnitude of 8.37E-6 N, which was incorrect
 
  • #6
gneill
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F(1 on 3) = -5.387E-6
F(2 on 3) = -2.995E-6

This is a magnitude of 8.37E-6 N, which was incorrect
The first force is wrong (direction).
The second force is okay.
 
  • #7
The magnitude appears to be correct though, right? For some reason it's not
 
  • #8
rude man
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The magnitude appears to be correct though, right? For some reason it's not
Probably because I mixed you up. I suggested F(2 on 3 ) should be +, but it's -. Similarly, F(1 on 3) should be +, not -. But the magnitude of the difference is the same.

Do it gneill's way which is right.
 
  • #9
Probably because I mixed you up. I suggested F(2 on 3 ) should be +, but it's -. Similarly, F(1 on 3) should be +, not -. But the magnitude of the difference is the same.

Do it gneill's way which is right.
In that case, F(1 on 3) is = 5.387E-6
F(2 on 3) is = -2.995E-6

Magnitude still appears to be 8.37E-6 N like you said. Am I using the wrong equation or something? This doesn't make sense to me
 
  • #10
gneill
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Add the forces, then take the magnitude.
 
  • #11
Ohhh I gotcha, so 5.387E-6 + (-2.995E-6), which = 2.392E-6. So the magnitude is 2.392E-6N?
 
  • #12
gneill
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Yup. Round to 3 significant figures for your final result.
 
  • #13
Perfect. Thank you all so, so much
 

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