# 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 ;) .