Magnitude of an electric force problem with vectors

AI Thread Summary
The discussion revolves around calculating the electric force on a charge Q due to two equal positive charges. Participants focus on resolving the forces into their x and y components using trigonometric functions. There is clarification needed on the correct use of "cos" instead of "sin" in the equations. The balancing of vertical components and the addition of horizontal components are highlighted as key steps in simplifying the problem. The overall approach involves using the Pythagorean theorem to find the net force after calculating the components.
AladdinSane
Messages
8
Reaction score
0

Homework Statement


Two equal positive charges, both q = 2.0 x 10^ -6 C, interact with a third charge Q = 4.0 x 10^ -6 C, as shown above. What is the magnitude of the electric force on Q?[/B][/B]



Homework Equations


Electric force = Kc (q1/r^2)


The Attempt at a Solution


Basically, I believe that my problem is in solving the vectors.
I attempted to break down the charges(it is hard to show this work on here but...) I broke it down into...
E(y) = E(1) x sin Θ(1) + E(2) x sin Θ(2)
E(x) = E(1) x cox Θ(1) - E(2) sin Θ(2)
I'm not sure if that makes sense, but it was all I could think of...Any help would be greatly appreciated. Thanks a lot!
 

Attachments

  • scan.jpg
    scan.jpg
    30.2 KB · Views: 451
Physics news on Phys.org
AladdinSane said:
E(x) = E(1) x cox Θ(1) - E(2) sin Θ(2)
I'm sure you meant "cos", not "sin" there right?

While waiting for the image, it sounds like your reasoning is correct. You are summing the x and y components of the separate forces on the thrid charge. Next you need to determine the net force.

Hey, how did you type those thetas in there?
 
Last edited:
thanks chi,
so i believe my second set should actually look like...
E(x) = E(1) x cosx Θ(1) - E(2) cos Θ(2)
then once i plug in my values and simplify, i should get my x component any my y component, right? then i just use the pythagorean theorem to solve?

oh, and for the thetas i just happened to come across one on a wensite and i copied and pasted. .
 
Θ

wow, must be some ASCII trick.Anyway, now that I see the drawing, please notice that the two vertical components are balancing out while the two horizontal components are working together. Simpler now, isn't it?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top