How to Find R from Electric Charge Problem

[kidi3]

Homework Statement

http://snag.gy/Zg3SO.jpg
At the moment is it question A

The Attempt at a Solution

A)
I've realized that Force applied by F_31 + F_41 = F_21

SO I've writte this equation up but Mathmathica, says there is an errror, and gives me an incorrect result.

http://snag.gy/rQ2kz.jpg

Mathmatica says it should {{D -> -0.225008}, {D -> 0.185008}}

The correct answer should be 2,27cm.

 

[TSny]

Did you take into account the directions of the forces F_31 and F_41? They have both x and y components.

 

[kidi3]

No..
y -component must be zero since it is zero for q2
and the sum of q3 and q4 would negate the y component.

 

[TSny]

No..
y -component must be zero since it is zero for q2
and the sum of q3 and q4 would negate the y component.

Correct. So, you only need to work with the x components of the forces.

 

[kidi3]

But aren't I doing it already?

 

[TSny]

But aren't I doing it already?

I don't think so. It appears to me that on the left hand side of http://snag.gy/rQ2kz.jpg you are using the full magnitude of the forces F_31 and F_41, rather than their x-components.

 

[kidi3]

Ahh.. so you want me to skip
d/cos(\theta)
and just use d.. ?

 

[TSny]

Ahh.. so you want me to skip
d/cos(\theta)
and just use d.. ?

No, that part is correct, r = d/cosθ is the correct distance between charges 4 and 1. Draw a sketch showing the direction of the force F₄₁. How would you express the x-component of F₄₁ in terms of F₄₁ and θ?

 

[kidi3]

well the distance it moves is d.. so I have to write an expression for d..

So it would be rcosθ = d??

 

[TSny]

Sorry, but I don't understand. What is moving a distance d? I thought you were trying to find the distance D so that the net force on q1 is zero.

Your first equation which you derived looks good except it doesn't take into account that you want the x-components of the forces to add to zero. (You already know the y-components add to zero by symmetry.)

 

[kidi3]

well q1 lies on a distance d from q3 and q4...

and isn't that distance given by r cos(θ)= d

where r is the Hypotenuse of the triangle`..

 

[TSny]

well q1 lies on a distance d from q3 and q4...

and isn't that distance given by r cos(θ)= d

where r is the Hypotenuse of the triangle`..

Yes. All of that is correct. So, your expression (d/cosθ)² in the denominator of your equation is correct.

Force is a vector quantity. The equation F = q₁q₂/(4∏ε_or²) gives the magnitude of the force. You need to determine the x-component of the force vector. Again, a sketch of the forces will help.

 

[kidi3]

... hmm would it then be
F_x = q1q2/(4∏εo(d))

Since both lies in a distance d?... or how..

 

[TSny]

... hmm would it then be
F_x = q1q2/(4∏εo(d))

Since both lies in a distance d?

No. Consider the x-component of F₃₁. Can you use trigonometry or similar triangles to find F_x in the attached picture?

 

[kidi3]

hmmm.. is it then cos(\theta)=r/x <=> x = cos(theta)/r

 

[TSny]

hmmm.. is it then cos(\theta)=r/x <=> x = cos(theta)/r

Not sure what you are writing here. You need to find F_x, rather than x.

 

[kidi3]

Ahh.. I see what you mean now..
My drawings weren't correct..
But by using youre drawing I see that Cos(theta) = F_x/F <=> F_x = Fcos(theta)

So the expression for F_X = (q1q2/(4∏εo(d^2))) cos(theta)

 

[TSny]

Yes. Good.

[EDIT: oops, you did mean r instead of d in the denominator, right? And the charges are q₁ and q₃ ]

 

[kidi3]

Hmm.. but i still get an incorrect answer.. :(

http://snag.gy/LDwAe.jpg

When i Tries to calculate D i get complex number..

 

[TSny]

It seems to work out ok. Note that you don't want to plug in a negative value for q₃ since you have set up your equation to state that the magnitudes of the forces balance out.

 

[kidi3]

D has to be 2,266 cm
But i get 1471,4 m...

 

[TSny]

Well, I get D = 1.92 cm.

If you go to your original equation, put in the cosθ factor on the left to get the x-component, and then simplify, see if you get

2q₃cos³θ/d² = q₂/R²

where I let R be the total distance between 1 and 2.

If so, what do you get if you solve for R?

 

[kidi3]

How do you get this ?
2q3cos3θ/d2 = q2/R2

 

[TSny]

Well, what do you get if you cancel out everything you can between the left and right hand sides?

 

[kidi3]

But isn't what i have already done before..

 

[TSny]

I don't know. You really haven't shown any steps involved in solving your equation. So, I can't tell you were you are going wrong.

 

[TSny]

Start with your original equation and put in a factor of cosθ on the left side to take care of getting the x-component.

Then, do you see that q₁ cancels out as well as 4πε_o?

For now, just let r stand for the distance between #3 and #1 charge. So, instead of writing d/cosθ on the left, just write r. And on the right side, let R be the total distance between #2 and #1. So, instead of writing D+d just write R. Can you state what the equation simplifies to?

 

[kidi3]

These are my steps...

http://snag.gy/3oGJ5.jpg

 

[TSny]

The first equation is incorrect. I see you have the cosθ factor to get the x-component. Good. But for some reason you forgot to write (d/cosθ)² in the denominator to represent r².

 

[TSny]

Also looks like you didn't go from the second to the third equation correctly. [Shouldn't the fraction inside the square root be flipped over?]

 

[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?