I need some help with an electricity problem

1. Feb 9, 2005

benji

So I thought I understood this electricity stuff pretty well--I've read through the chapter once in full and skimmed over it a couple times. I've studied all the examples and still my problems aren't coming out correct. Here is one problem in particular that is giving me a lot of trouble:

What I did with this problem is draw a free-body diagram for the forces that act upon the sphere attatched to the string. I found the force of gravity by simply using F=mg, so I have the left side of the right triangle. Then I solved for the electrical force between the two spheres by using F=(k*q1*q2)/(r^2). That gives me the bottom of the right triangle. So shouldn't I just be able to use simple trig to find the hypotenuse of the triangle, thus finding tension on the string? I did this and my answer came out to be 0.615 N--the correct answer is 0.813 N. Where did I go wrong? I've checked and re-checked my math and I still can't figure out why. So that was part (b), for part (a) of the problem I just used trig--inverse tangent multiplied by the bottom side of the trianle over the tension. For this answer I'm WAY off--I get -2.3X10^-21 degrees when the correct answer is 15.4 degrees.

So I know I wrote a little novel here, but I could really use the help!

Thanks.

2. Feb 9, 2005

StatusX

I don't know if this is the problem, but did you keep in mind that the electrical force will increase as the ball swings closer to the charge?

3. Feb 9, 2005

MathStudent

Have you tried using newton's second law for the horizontal and vertical components of force on the sphere?

$$\Sigma F_x = ma_x$$

$$\Sigma F_y = ma_y$$

its hard to tell what went wrong because we can't see your equations, post any work you've done so far

Last edited: Feb 9, 2005
4. Feb 10, 2005

benji

I think I'll just ask about this one tomorrow in class, thanks anyways though. I should have just scanned my paper ;) .