# Circular motion and charge

1. Jan 17, 2010

### nothingatall

1. The problem statement, all variables and given/known data
A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.776 g, q = 5.04 µC is located on the x axis at x = 18.7 cm, moving with a speed of 39.8 m/s in the positive y direction. For what value of Q (in μC) will the moving particle execute circular motion? (Neglect the gravitational force on the particle.)

2. Relevant equations
F=mv^2/r- centripetal force
F=k*q1*Q/r^2

3. The attempt at a solution
i'm thinking of starting by setting the two forces together and find the other Q. Just making sure if its the correct way to find it. If its not can someone outline the steps i need to follow? thanks.

2. Jan 17, 2010

### tiny-tim

Hi nothingatall!

(try using the X2 tag just above the Reply box )

Yes, that's exactly the way to do it.

3. Jan 17, 2010

### nothingatall

ok so i tried it and apparently my answer is wrong. Here's my work:

Fc= (7.76e-4kg)(39.8m/s)^2/18.7e-2m
6.573=(8.99e9)(5.04e-6C)(Q)/(18.7e-2m)^2
=5.07uC.

it asks for the answer in microC so I don't know where i messed up. I'm at my last attempt before it get marked wrong and my professor is no help.

4. Jan 17, 2010

### tiny-tim

hmm … using your figures of 7.76*(.398)2*0.187/8990*5.04, I get 5.07 10-6 also.

Does the sign matter?

5. Jan 17, 2010

### nothingatall

I'm not sure but i'm afraid to try it on account of its my final attempt :(

6. Jan 17, 2010

### ideasrule

I'll take a gamble and say that yes, the sign does matter, so the answer is -5.07 10^-6 C.