Help with Coulomb's law: Net electrostatic force

  • #1
mousey
9
2
Homework Statement:
Given the arrangement of charged particles in the figure below, find the net electrostatic force on the
q1 = 5.35-µC charged particle. (Assume q2 = 14.33 µC and q3 = −18.12 µC. Express your answer in vector form.)

The three charges (q1, q2, and q3) are located at the following points:
q1: (-2.00cm, 0cm)
q2: (1.00cm, 1.00cm)
q3: (0cm, -1.00cm)

I converted uC into C and cm into m, found the distance between q1 and q2 (r_12) and between q1 and q3 (r_13), and I know I have to compare q2 to q1 and q3 to q1. I can figure out the force just by plugging numbers into Coulomb's law, but I'm not sure how to calculate the vector components.

Thank you!
Relevant Equations:
F =(k{q_1}(q_2))\ d^{2)
(Coulomb's Law)
I tried just calculating the force with Coulomb's law, then calculating the forces for each vector individually and adding, but I got it wrong both ways
 

Answers and Replies

  • #2
Gordianus
458
116
We need your solution to find what went wrong.
 
  • #3
mousey
9
2
We need your solution to find what went wrong.
Thank you! I tried uploading a picture of my work but I couldn't figure out how to initially. Also here's the original problem in all it's glory.
 

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  • #4
Gordianus
458
116
The modulus of F12 Is 690 N/C. However its components are larger than that value. There's something wrong there.
 
  • #5
mousey
9
2
The modulus of F12 Is 690 N/C. However its components are larger than that value. There's something wrong there.
Right, I tried that answer and it was marked wrong, so I tried finding the force for the individual vector components (table in the lower left), by using just the distance in the x and y directions, but in meters in place of "d" for d^2. Then I added the components (i of q2 + i of q3, j of q2 +j of q3), but that answer was also wrong.
 
  • #6
haruspex
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Haven’t checked the details of arriving at F12 and F13, but they look about right. After that I am lost. How did you get those component force values in the table?
 
  • #7
mousey
9
2
Haven’t checked the details of arriving at F12 and F13, but they look about right. After that I am lost. How did you get those component force values in the table?
these are my equations. Basically in terms of a right triangle, I used the "legs" to calculate for each vector component instead of the hypotenuse as the whole force.
 

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  • #8
Gordianus
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I checked the modulus in a hurry and they look OK. I don't understand the table
 
  • #9
mousey
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I checked the modulus in a hurry and they look OK. I don't understand the table
I tried to explain it further in my reply to haruspex.
 
  • #10
haruspex
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these are my equations. Basically in terms of a right triangle, I used the "legs" to calculate for each vector component instead of the hypotenuse as the whole force.
You have a basic misunderstanding in how to find the components. You should not be dividing by e.g. 0.03^2.
 
  • #11
haruspex
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Having found the overall force magnitude, multiply it by the cosine of the angle (x/r) to find the i component and sine (y/r) to find the j component.
Watch the signs carefully.
 
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  • #12
mousey
9
2
Having found the overall force magnitude, multiply it by the cosine of the angle (x/r) to find the i component and sine (y/r) to find the j component.
Watch the signs carefully.
That makes perfect sense. Thank you.
 

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