How do I solve for the correct value of x in this electric charge problem?

[Zythyr]

Homework Statement

http://img210.imageshack.us/img210/3239/q1ae6.jpg

The attempt at a solution

Kq1q3/(L+x)2 = Kq2q3/(x)2
q1/(L+x)2 = q2/(x)2

I came up with the question above. But when I plug in the numbers I get 2 different answers. For x, I got -7.363450837 and -21.73332335. I tired both answers in the system and I got it wrong. I am not sure why this is happening.

Do I have to add +11 to the x to get the real answer? If so then to which one do I add 11 to?

 

[Doc Al]

Hint: First figure out in what region along the x-axis must q3 be placed? (x < 0, x > L, or 0 < x < L).

(If you can't figure out which region, try them all!)

Also: Be careful with your signs. (You can't just use magnitudes.)

 

[Zythyr]

Hint: First figure out in what region along the x-axis must q3 be placed? (x < 0, x > L, or 0 < x < L).

(If you can't figure out which region, try them all!)

I can't figure out which region it is in. I have only have 1 more attempt left to get the answer correct.

 

[Doc Al]

I can't figure out which region it is in. I have only have 1 more attempt left to get the answer correct.

Then try each region (one at a time) and set up the equation such that the net field is zero. Of the three regions, only one will give a sensible answer. Try it!

Note: The direction of the field counts.

 

[Zythyr]

I think its on the left of the q1. But then I am not sure about the formual.

If it is to the left of q1, that means that x is the distance between q1 and q3. And L+x is the distance between q2 and q3. I am guessing I would use this question

q1/(x)2 = q2/(L+x)2

Am I right? Also since q2 is a negitive, when I try to solve in get complex number which doesn't let me solve for x.

 

[Doc Al]

You are almost there.

Let q1 and q2 stand for just the magnitudes of the charges. (Put in the signs by hand.) The field from q1 (which is the positive charge) points to the left, which we'll call negative. So: E1 = -q1/(x)^2. Similarly, E2 = +q2/(L+x)^2.

Find x when E1 + E2 = 0.

 

[Zythyr]

I am so lost now. I don't get it at all. Can you please give me the answer and I will be sure to study the soulution when it is posted.

 

[Doc Al]

Don't stop now. You are practically done.

I am guessing I would use this question

q1/(x)2 = q2/(L+x)2

That's the exact equation you need. Just use the magnitudes of the charges.

 

[Doc Al]

So: E1 = -q1/(x)^2. Similarly, E2 = +q2/(L+x)^2.

Find x when E1 + E2 = 0.

Note that -q1/(x)^2 + q2/(L+x)^2 = 0 is the same as:
q1/(x)^2 = q2/(L+x)^2

(q1 and q2 are both positive numbers.)

 

[Zythyr]

Okay I took the absoulute value of the charges and used the question. I got two values for X. I got -.0363654916 and .1073332336.

I tried putting in the answer -.0363654916 and I got it wrong. I have one attempt left. What do I do?

 

[hage567]

Are your answers expressed in meters? It looks like it. I think the homework program is asking for the answer in cm. Check your units.

 

[Doc Al]

Okay I took the absoulute value of the charges and used the question. I got two values for X. I got -.0363654916 and .1073332336.

I tried putting in the answer -.0363654916 and I got it wrong. I have one attempt left. What do I do?

I thought we had agreed that the charge must be placed to the left of q1. And we used "x" to represent the distance to the left of q1. So "x" must be positive. (A negative value of "x" means to the right of q1--which means it's not a solution that satisfies our initial conditions.)

It probably would have been a bit smarter if we called the distance to the left of q1 by the letter "d", so we wouldn't confuse it with the x-coordinates. Oh well! (This is probably the source of your confusion!)

So you solved for "x" and chose the only answer that makes sense (the positive value). Now you must translate our "x" to the coordinate along the x-axis. A positive value for "x" means a negative x-coordinate. (Since q1 is at the origin.)

I hope this makes sense.

Also: As hage567 advised, make sure your answer is in cm, not meters.