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?]