Force on q2 from two charges at equal distances

[PattyCake]

1. Given that q=+12uc distance=19cm
Find Magnitude and direction exerted on point q2.
What happens to the force if the distance is tripled?q____________ q2 ____________q3
q1=+12
q2=-2.0q
q3=+3.0q
distances between point are equidistant.

The Attempt at a Solution

F=k Q Q/ d^2
F=F+F
F=k/d^2(QQ+QQ)

F=k(q)(q)/d^2

(9*10^9)(12*10^-6)(2*10^-6)
(.19)^2

=5.98

(9*10^9)(-2*10^-6)(3*10^-6)
(.19)^2

=-1.50

thus 5.98-(-1.50)=7.48

 

[BiGyElLoWhAt]

You're sure you don't have a value for q1?

 

[PattyCake]

You're sure you don't have a value for q1?

I think q1 is 12uc

 

[BiGyElLoWhAt]

Ok, so how do you treat separate forces? You have coloumbs law correct. I guess I don't see where you're stuck, unless you don't understand the formula. If that's the case, please clarify.

 

[PattyCake]

Ok, so how do you treat separate forces? You have coloumbs law correct. I guess I don't see where you're stuck, unless you don't understand the formula. If that's the case, please clarify.

I think I'm just confused because the book does it completely differently and my answer isn't the same. I think there Is error rounding. and I didn't initially understand the formula. The more I looked at it I had a urikea moment.

 

[PattyCake]

I think I'm just confused because the book does it completely differently and my answer isn't the same. I think there Is error rounding. and I didn't initially understand the formula. The more I looked at it I had a urikea moment.

How do you remove this from the feed?

 

[BiGyElLoWhAt]

Well I hope you did. As to removing it, this thread will stay here in case someone else can get some use out of it at a later time.

Would you humor me? Walk me through this, so I know what's going on.

 

[BiGyElLoWhAt]

By the way, welcome to Physics Forums. :)

 

[PattyCake]

Would you humor me? Walk me through this, so I know what's going on.

So I used Coulombs equation.
F=kQQ/d^2
I knew that F=F+F
So I just wrote the equation
F=k/d^2(QQ+QQ)
I plugged the constant in for k. converted 19cm to meters so it changed to .19m.
Then plugged in ((-12)(2)+(2)(3))
but then I didn't change them to coulombs.
So that through me off.

So instead I plugged everything into the original format of the equation and did it in parts because it just was easier on the eyes.

I hope your humored.

 

[BiGyElLoWhAt]

=]
I am.

 

[PattyCake]

but are you sure I did it right?

 

[BiGyElLoWhAt]

Also, if you feel so inclined:
https://www.physicsforums.com/threads/introducing-latex-math-typesetting.8997/
One of the best things in life, IMO, second only to corndogs.
$F_{net\ on\ q_2} = F_{q_1\ on\ q_2}+F_{q_3\ on\ q_2}$

 

[BiGyElLoWhAt]

but are you sure I did it right?

Assuming there are subscripts below your q's, and also assuming that (QQ+QQ) is in the numerator, then yes.

 

[BiGyElLoWhAt]

Also assuming that you ultimately converted your charges correctly. Neglected that initially, because you, yourself, mentioned it. I figured I might as well include it.