# Fields and Waves HELP

1. Jan 28, 2005

### Deviousfred

Fields and Waves HELP!!!!

I'm stuck on a problem on my homework assignment.

A ring-shaped conductor with a radius a=2.90 cm has a total positive charge q1=0.126 nanoCoulombs uniformly distributed around it. The center of the ring is at the origin of coordinates O.

What is the magnitude of the electric field at point P, which is on the positive x-axis at x=45.0 cm?

I found that to be 5.56 Newtons/ Coulombs

What is the direction of the electric field at point P?

Positve x

A particle with a charge of -2.30 microCoulombs is placed at the point P described in part (a). What is the magnitude of the force exerted by the particle on the ring?

Anyone know of an equation that I can use to solve this?

I tried F=(1/4*pi*epsilon_0)((q1q2/r^2) I used both x=45cm and the hypotenuse by taking the sqrt(.029^2+.45^2) and both give the same answer and its wrong. I also used E=F/q where I used 5.56 N/C and 2.3*10^-9 C and it gave me the same answer.

2. Jan 28, 2005

### MathStudent

You already know the electric field value at the point of interest...
What is electric field?
It is the amount of force a unit charge would feel at a given point in space..
so you can just use
$$\vec{F} = q \vec{E}$$

You already did the hard stuff... just need to know how to make use of it

-MS

edit: Ah, I see you already tried that at the bottom of your first post, sorry

Last edited: Jan 28, 2005
3. Jan 28, 2005

### Deviousfred

where q would be 2.3*10^-9 C and E would be 5.56 N/C?

I used it and it gave me the same thing as if I used the other equations mentioned.

4. Jan 28, 2005

### MathStudent

Well I havent checked that the value of E you calculated is correct... but for one, micro stands for 10^-6, nano is 10^-9

5. Jan 28, 2005

### Deviousfred

you got me!!!! that's where I'm wrong. Let me check.

6. Jan 28, 2005

### Deviousfred

thank you thank you thank you!!!!

You caught my dumb mistake.

7. Jan 28, 2005

### Deviousfred

awesome!!!! I have a feeling I'm gonna be spending a lot of time on these forums.

Hopefully I can help some other people like you did for me today.