# Homework Help: Help w/ problem!

1. Jan 8, 2006

I did this problem, yet I found out that its wrong because I didn't plug the correct force in. I have no idea what to do now though. If someone can show me what I did wrong and explain it, I would greatly appreciate that. Thank you!

Two coins with identical charges are placed on a lab table 1.35 m apart.

A.If each coin experiences a force of 2.0 N because of the presence of the other coin, calculate the charge on each coin.

FE=KE (q1q2/r^2)
R^2 = 1.35m
q1q2 = 2.0N
FE=KE (2.0N*2.0N)/(1.35^2m)
FE=2.194 or 2.195 A

B.Would the force be classified as a force of attraction or repulsion? Explain your answer.

Force of attraction because it represents a weight.

2. Jan 8, 2006

### siddharth

This question requires a direct application of Coloumb's law (and the assumption that the two coins are like point charges).

Why have you used q1q2 as 2.0N*2.0N? 2 N is the force which each coin experiences, not the charge.

From the information given in the question, you know the values of $F , r$ and that the fact that the two charges are identical. All you need to find is the charge.

Finally, why would the weight decide if the force is attractive or repulsive?
You should know that it is the nature of the charges(ie, like or unlike).

3. Jan 8, 2006

### Staff: Mentor

(1) 2.0 N is the force, not the charge! The charge is what you are trying to find.
(2) What happened to the constant Ke? (Look it up.)
$$F = k q^2/r^2$$