
#1
Jan2311, 02:01 PM

P: 17

1. The problem statement, all variables and given/known data
Three point charges are fixed in place in the right triangle shown below, in which q1 = 0.71 µC and q2 = 0.67 µC. What is the electric force on the +1.0µC (let's call this q3) charge due to the other two charges? image: http://www.webassign.net/grr/p1615alt.gif magnitude ____ N direction ______°, measured counterclockwise from the +xaxis 2. Relevant equations kq/r^2 3. The attempt at a solution The only thing I don't know how to do is calculate the x and y components of the charge on q3 due to q1. Here's what I did: First I found the missing side, 6 cm. Then I converted all the cm to m by multiplying it by .1 m and then I converted all the uC to C by multiplying it by X10^6. q1: Fx= k(q1)(q3)/r^2 = 8.99e9 (0.71e6) (1.0e6)/(0.10)^2 * 6/10 Fy=k(q1)(q3)/r^2 = 8.99e9 (0.71e6) (1.0e6)/(0.10)^2 * 8/10 q2: Fx=k(q2)(q3)/r^2 =8.99e9 (0.67e6) (1.0e6)/(0.10)^2 Total force= Fx added together. And then sqrt. (Fx^2 + Fy^2) As for the direction, tan theta= (Fy/Fx) and then add 180 to it? Please help me, it's due tonight, thank you so much in advance. 



#2
Jan2311, 02:15 PM

Mentor
P: 39,648

You should convert the polar form of the vectors to rectangular form, add the forces in rectangular form, and then convert back to polar form. Otherwise, it's too easy to make mistakes. The polar > rectangular conversion for the force due to q2 is easy, since all of the force is in the x direction, right? What about the conversion for the force due to q2? Can you show us the x and y components of that force? And then add the forces componentwise, and convert back to the polar answer format that they are asking for... 



#3
Jan2311, 04:27 PM

P: 17

I don't know what polar or rectangular form mean, I'm sorry, I'm in Intro physics. I think the xcomponent of the force q2 is (0.71e6 cos 30) and the ycomponent is (0.71e6 sin 30)? Can you please let me know if I'm on the right track? 



#4
Jan2311, 05:06 PM

Mentor
P: 39,648

Force problem! What is the electric force on q3 due to the other 2 charges?Once you have the x and y components of the forces (making sure the signs are right), add the x components to get the resultant x force, and add the y components. Then covert back to the polar form that the question asks for. 



#5
Jan2311, 07:40 PM

P: 17

So I have: q2: Fx=  k(0.71e6)(1e6)/(0.71e6 cos30)^2 Fy=  k(0.71e6)(1e6)/(0.71e6 sin30)^2 (both components are negative, because the force points in the negative x direction, since they're repulsive to the positive charge of q3.) q3: Fx= k(0.67e6)(1e6)/(0.6)^2 So, Fx= 532160742267 + 0.0167313889 = 5.32160742e11 Fy= 1.29705876e10 What does "polar form" mean? 



#6
Jan2311, 09:25 PM

Mentor
P: 39,648

Polar/Rectangular component conversions: http://en.wikipedia.org/wiki/Polar_coordinate_system . 


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