Component Force on a 1.0 nC Charge

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SUMMARY

The discussion centers on calculating the net force acting on a 1.0 nC charge due to three other charges: a +2.0 nC charge, a -6.0 nC charge, and a -2.0 nC charge, positioned at specific angles and distances. The total force calculated was 3.6 x 10-5 N, with the y-component determined to be approximately 2.16 x 10-5 N. The x-component, however, was incorrectly calculated initially, leading to confusion. The correct approach involves vector addition of forces rather than merely summing magnitudes.

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Physistory

Homework Statement


What is the force F on the 1.0 nC charge at the bottom in the figure? Give your answer in component form.

In the figure, the +!.0 nC charge is adjacent to a +2.0 nC charge to the left at a 45° angle, a -6.0 nC charge directly above, and a -2.0 nC charge to the right at a 45° angle. Both distances between the first and third charges adjacent to the 1.0 nC charge are 5.0 centimeters in length.

Homework Equations



F = F2on1 + F3on1 + F4on1 = (K)(q1)(q2)/(r2)

x2 + y2 = z2; x2 = z2 - y2

C = (pi)*r2 = (pi)*(5)2 = 78.5 cm
C/4 = (19.6 cm)/2 = 9.81 cm

The Attempt at a Solution


I foremost determined the direction of the net force to be somewhere between east and northeast of the 1.0 nC charge (hereafter denoted q1 and the other charges so forth). I have assumed this because q2 would repel q1, while q3 and q4 would attract it. Thus, q1 would be pulled more up and eastward than it would be pulled southeastward. In contrast, q1 would only be pulled directly upward, in the direction of q3 if q4 were positive, because q2 and q4 would then be canceled out.

Using the first equation listed above, I added the forces and obtained a total force of 3.6*10-5 N. The three values I added to obtain this number were 7.2*10-6, 2.16*10-5, and 7.2*10-6 N again.

On my first attempt, I thought that to solve for the x-component I merely had to multiply each force by their respective cosines. In doing so, I obtained the values 5.09*10-6, 2.16*10-5 and 5.09*10-6 N which added to 3.17*10-5 N. This was not correct.

The value 2.16*10-5 was also obtained by plugging q1, q3 and 5.0 cm into the first equation. My textbook lists this (or precisely, 2.2, which is very close to 2.16) as the correct value for the y-component of the force. I believe that I understand why (that is, the direction from q1 to q3 is up and can therefore be found just by plugging in the two charges and the distance between them).

I am as yet unsure how to solve for the x-component. I have tried using Pythagoras' equation with multiple value combinations, but have not yet obtained anything close to the correct value, which my textbook lists to be 1.02*10-5 N. I would be very grateful for any hints as to whether there is something I am overlooking or am failing to realize. Thank you in advance!
 
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Physistory said:
In the figure, the +!.0 nC charge is adjacent to a +2.0 nC charge to the left at a 45° angle, a -6.0 nC charge directly above, and a -2.0 nC charge to the right at a 45° angle. Both distances between the first and third charges adjacent to the 1.0 nC charge are 5.0 centimeters in length
Just stating the angle is ambiguous. Are the ±2nC charges both also "above"?
Physistory said:
I added the forces and obtained a total force of 3.6*10-5 N.
Please show your working, and keep it all symbolic. Avoid plugging in numbers until the final step.
Force is a vector. Forces in different directions must be added accordingly. Without seeing your working or going through the calculation myself I cannot tell if you did that correctly.
 
I'm not sure that I understand the placement of the particles. Based on your numbers that your are calculating I'm guessing it is as shown.
upload_2017-8-6_17-4-9.png

As haruspex noted, you should show more detail of your calculations.

Physistory said:
Using the first equation listed above, I added the forces and obtained a total force of 3.6*10-5 N. The three values I added to obtain this number were 7.2*10-6, 2.16*10-5, and 7.2*10-6 N again.
You don't want to add the magnitudes of the three forces since they don't point in the same direction.

On my first attempt, I thought that to solve for the x-component I merely had to multiply each force by their respective cosines. In doing so, I obtained the values 5.09*10-6, 2.16*10-5 and 5.09*10-6 N which added to 3.17*10-5 N. This was not correct.
What is the x component of the force between the 1 nC and the -6 nC charge?

The value 2.16*10-5 was also obtained by plugging q1, q3 and 5.0 cm into the first equation. My textbook lists this (or precisely, 2.2, which is very close to 2.16) as the correct value for the y-component of the force. I believe that I understand why (that is, the direction from q1 to q3 is up and can therefore be found just by plugging in the two charges and the distance between them).
That looks right if the particles are arranged as shown above. But, it would be helpful to us if you showed more details of your calculations.
 

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