# Coulombs law stuff I dont know what they are asking

1. Jan 30, 2006

### red ink

Two charged particles, with charges q_1=q and q_2=4q, are located a distance d apart on the x axis. A third charged particle, with charge q_3=q, is placed on the x axis such that the magnitude of the force that charge 1 exerts on charge 3 is equal to the force that charge 2 exerts on charge 3.
Find the position of charge 3. Assume that all three charges are positive.

So charge 3 could be inbetween charge 1 and 2, or to the left of charge 1. The problem wants me to find "2 possible VALUES" of charge 3. How can i find a value when no values were given to me in the word problem. THey want the X_3,1, and X_3,2 in terms of Q, D, and K (coulomb constant).

2. Jan 30, 2006

### cepheid

Staff Emeritus
Okay, so it sounds like the problem is asking for you to find a location for charge 3 on the x-axis such that the electric forces exerted on it by each of the other two charges are equal (in magnitude).

No, the value of charge 3 is given, q3 = q. You must mean that the problem wants you to find two possible positions for charge 3.

You answered that question yourself right here:

This problem is a fairly straightforward application of Coulomb's law, which allows you to calculate the electric force between two charges given the magnitude of the charges and the distance between them. It shouldn't be too hard considering you're only working in one dimension. For simplicity, why not put q1 at the origin and q2 a distance d to the right? If q3 is at some position x3, just write an expression for the force exerted by each of the other two charges on it according to Coulomb's law.

e.g

$$F_{\textrm{2 \,on \,3}} = k\frac{q_2 q_3}{d-x_3} = k\frac{4q^2}{d-x_3}$$

Can you write a similar expression for the force due to q1 on q3? If the required condition is that these two forces must be equal, equate the expressions! Can you solve for x3?

3. Jan 31, 2006

### red ink

Thank you in advance for your reply...forgive me for my poor formating in my posts. This LaTeX Code stuff is kinda confusing!!

So by equate the expressions, you mean Force 1 on 3 will look just like the expression you gave me for Force 2 on 3, with the only difference being the numerator is "q^2" instead of 4q^2 since Charge 3=q?
$$F_{\textrm{1 \,on \,3}} = k\frac{q_1 q_3}{d-x_3} = k\frac{q^2}{d-x_3}$$
to solve for x_3, set $$F_{\textrm{1 \,on \,3}} = F_{\textrm{2 \,on \,3}}$$. This should give me a quadratic equation with 2 real value solutions?
x_3,1, x_3,2 = these values?

Last edited: Jan 31, 2006
4. Jan 31, 2006

### cepheid

Staff Emeritus
You've almost got it, except that the expression for F 1 on 3 due to Coulomb's law should have in the denominator the *distance between charge 1 and charge 3.* What is this distance?

5. Feb 1, 2006

### red ink

thats what stumpped me...they gave no distance. Only the varible "d". I already failed it online, but i'd like to know what i shoulda put. I had no clue what they were asking for.