How to Find R from Electric Charge Problem

[kidi3]

#30
Well.. would it make any difference?
#29
yeah i may have understood your previous post..
Now is the result
1.89577

http://snag.gy/E71gm.jpg

 

[TSny]

#30
Well.. would it make any difference?
#29

Yes. The big fraction in the square root is upside down.

 

[kidi3]

...'
I still get an incorrect answer..

http://snag.gy/veJcm.jpg

 

[TSny]

It's not the d/cosθ that should be flipped. When you got down to your last expression with the square root, the main fraction inside the square root appears to be upside down (not the d/cosθ).

For the moment, let's let r = d/cosθ and R = d+D

Then your original equation can be written as

2q₃cosθ/r²= q₂/R²

What does this become if you multiply both sides by r²R²?

 

[kidi3]

I am getting confused.. wasn't F_x i shoud have used..

And the error, I really don't understand how there can be a error in the equation i wrote..

 

[TSny]

I am getting confused.. wasn't F_x i shoud have used..

Exactly, F_x is what you should use. The force that q₃ exerts on q₁ is

F = kq₁q₃/r²

The x-component of this force is F_x = Fcosθ. So,

F_x = kq₁q₃cosθ/r².

q₄ creates the same x-component of force on q₁. So, the total force on q₁ from q₃ and q₄ together is

2kq₁q₃cosθ/r² $\;\;\;\;$ (in the positive x direction)

q₂ creates a force on q₁ in the negative x direction of magnitude kq₁q₂/R².

So, the net force will be zero if

2kq₁q₃cosθ/r² = kq₁q₂/R²

This leads to

2q₃cosθ/r² = q₂/R²

And the error, I really don't understand how there can be a error in the equation i wrote..

Right. That's what we need to determine. So, let's go through the steps for solving the last equation above for R. I suggested what I thought was a good first step; namely, to multiply the equation through by the least common denominator r²R². If that's not how you want to do it, that's ok. But, can you please show your next step or two in solving for R?