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

[Physicsnoob90]

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.

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]

 

[TSny]

Check to see if you used the correct charges for your second force.

 

[TSny]

Also, I'm not sure which direction you are taking to be the positive direction.

 

[Physicsnoob90]

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

 

[TSny]

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.)

 

[TSny]

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.

 

[Physicsnoob90]

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

 

[TSny]

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.)

 

[Physicsnoob90]

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!

 

[TSny]

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.

 

[Physicsnoob90]

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.