What is the net electric force exerted on the point charge q1?

AI Thread Summary
The discussion focuses on calculating the net electric force on point charge q1, given q = 20 µC and d = 11 cm. Participants work through the formula F(elect) = k(q1,q2)/d^2, adjusting units and values for accuracy. There is confusion regarding the direction of forces and the correct application of the formula, particularly in the denominator for the second force calculation. After troubleshooting, one participant corrects their denominator, leading to a successful calculation of +372 N for the total force. The conversation highlights the importance of precise mathematical representation in physics problems.
Physicsnoob90
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Homework Statement


Given that q = 20 µC and d = 11 cm, find the direction and magnitude of the net electric force exerted on the point charge q1 below.

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Homework Equations


F(elect) = k (q1,q2)/d^2

The Attempt at a Solution



1. after converting each unit to its appropriate form,
F(net 1x) = +(k(q(1)[(2.0)q(2)])/d^2) + -(k(q(1)[(3.0)q(3)])/2d^2)

= +[(9.0e^9 N m^2/C^2)(20e^-6 C)(4e^-5 C)/0.0121 m^2] - [(9.0e^9 N m^2/C^2)(20e^-6)(6e^-5)/0.0484 m^2)

2. F(net1x) = 595.041 - 446.281 = 148.8 N towards q2 (it came back wrong though) [/B]
 
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Check to see if you used the correct charges for your second force.
 
Also, I'm not sure which direction you are taking to be the positive direction.
 
TSny said:
Also, I'm not sure which direction you are taking to be the positive direction.
i'm using +q(1,2) going right and -q(1,3) going left. I still get -446.281 N in the second force
 
Did you use the correct value of q1 in the second force?

(I see that you are using "toward the right" as the positive direction. Good.)
 
It looks like you are plugging the correct numbers into your second force, but I get half the value that you get for the second force.
 
TSny said:
It looks like you are plugging the correct numbers into your second force, but I get half the value that you get for the second force.
Did you use the same number/equation as well?

F(net(q1,3))= k(q1 *(3.0)(q3))/2(d)^2

= - ( (9.0e^9 N m^2/C^2) * 20e^-6 C * (3.0)(20e^-6 C) )/ 2 (0.11 m)^2 = - 446.3 N
 
Physicsnoob90 said:
Did you use the same number/equation as well?

F(net(q1,3))= k(q1 *(3.0)(q3))/2(d)^2

= - ( (9.0e^9 N m^2/C^2) * 20e^-6 C * (3.0)(20e^-6 C) )/ 2 (0.11 m)^2 = - 446.3 N
Here the problem is with your denominator. (You had the correct denominator in your first post.)
 
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TSny said:
Here the problem is with your denominator. (You had the correct denominator in your first post.)
You're right. I got +372 N total as my result (which came back correct!)

Thank You!
 
  • #10
Great! I now see that when you wrote your equation in the first post, the denominator of the second force should have been written (2d)2 rather than 2d2. But you had the correct numerical value for the denominator.
 
  • #11
TSny said:
Great! I now see that when you wrote your equation in the first post, the denominator of the second force should have been written (2d)2 rather than 2d2. But you had the correct numerical value for the denominator.
i was wondering why my initial numbers were looking correct but still getting the wrong result.
 
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