Electro-statics Picture included (got most of it done, numbers not working out)

[mdxod]

Homework Statement

http://classes.uleth.ca/200503/phys2000a/private/ass02sol_files/image002.gif
With $$\theta$$ = 27 degrees, d = .015m, q2 = 6.4 *10^-19, q3 = q4 = -4.8 *10^-19.

Problem states: Find the distance D between the origin and particle 2 if the net force on 1 due to the others is 0.

Homework Equations

F = Kq1q2/ d^2
finding hypot. of a triangle

The Attempt at a Solution

So pretty much...there are no forces in the y direction due to the fact that F13y cancels out F14y. I'm getting really mixed up with the numbers and am not getting to the correct answer. but anyways i will show you what i have done.

(1) F13 + F14 can be = 2* F13 (you can pick F14...it doesn't really matter i just did F13)
so F13 = Kq1q3/ r^2, where r is the dashed lines.

For F12, its = Kq1q2 / (d+D)^2

so Fnet = 0 = 2*F13 + F12

(2) So next, i went to find "r" which is the dashed lines or the distance between q1 and q3...found r = d/(cos 27)

(3) Next is the actual set up..so i have: {2*F13 + F12}

[2Kq1q3] / [(d/cos27)^2] *cos 27 + [Kq1q2] / [d+D]^2 = 0

I know that K and q1 will cancel out.

[q2] / [d+D]^2 = [2q3*cos27] / [d/cos27]^2

from here...when i try to plug in and solve for D...i keep getting negative answers
or ones that just arent correct... but anyways if i were to go further with this problem to just solve for D without any calculations...i would have.

[d+D]^2 = [d*q2] / [2*q3*(cos 27)^3]

then square root both sides...and subtract "d"...

but after doign this and all the calculations, its still not working out to a correct answer. Thanks for any help.

 

[rl.bhat]

Using θ and d, find the distance between 1 and 3. If q is the charge at 1, find the force between q and q3 and q and q4. Find their component along horizontal and vertical direction. vertical components cancel each other. Horizontal components add up.Equate this resultant force with the force between q and q4, which along 4 to 1, and find D.

 

[Doc Al]

Homework Statement

http://classes.uleth.ca/200503/phys2000a/private/ass02sol_files/image002.gif
With $$\theta$$ = 27 degrees, d = .015m, q2 = 6.4 *10^-19, q3 = q4 = -4.8 *10^-19.

Problem states: Find the distance D between the origin and particle 2 if the net force on 1 due to the others is 0.

Homework Equations

F = Kq1q2/ d^2
finding hypot. of a triangle

The Attempt at a Solution

So pretty much...there are no forces in the y direction due to the fact that F13y cancels out F14y. I'm getting really mixed up with the numbers and am not getting to the correct answer. but anyways i will show you what i have done.

(1) F13 + F14 can be = 2* F13 (you can pick F14...it doesn't really matter i just did F13)
so F13 = Kq1q3/ r^2, where r is the dashed lines.

For F12, its = Kq1q2 / (d+D)^2

so Fnet = 0 = 2*F13 + F12

(2) So next, i went to find "r" which is the dashed lines or the distance between q1 and q3...found r = d/(cos 27)

(3) Next is the actual set up..so i have: {2*F13 + F12}

[2Kq1q3] / [(d/cos27)^2] *cos 27 + [Kq1q2] / [d+D]^2 = 0

All good.

I know that K and q1 will cancel out.

[q2] / [d+D]^2 = [2q3*cos27] / [d/cos27]^2

Realize that you snuck in a sign change when you rearranged this equation. That's OK if you only use the magnitudes of the charges.

from here...when i try to plug in and solve for D...i keep getting negative answers
or ones that just arent correct... but anyways if i were to go further with this problem to just solve for D without any calculations...i would have.

[d+D]^2 = [d*q2] / [2*q3*(cos 27)^3]

Looks OK except that the d on the RHS should be d^2.

 

[mdxod]

Looks OK except that the d on the RHS should be d^2.

Thanks Doc Al,

Ok SO...just to clarify i am using |q| magnitude of the charges because i changed that sign...

but rearranging everything i have wrote for D...i should be having (without plugging in numbers yet) that:

D = $$\sqrt{(d^2 * q2)/ (2*q3*(cos27)^3)}$$ - d

where i am using |q2| and |q3| when plugging in...
correct?

but i don't know why i keep getting a negative answer.
I am using q2 and q3 as given in the problem and for d I am using .015.
just don't know what I am doing wrong?

 

[Doc Al]

but i don't know why i keep getting a negative answer.
I am using q2 and q3 as given in the problem and for d I am using .015.
just don't know what I am doing wrong?

I also get a negative answer for D, using the given data. (The distance from the origin would just be the magnitude of D, of course.)

 

[mdxod]

Doc Al,

you are getting -.000438 m or simply taking its magnitude: .000438 meters?

im putting it in, its wrong.

 

[Doc Al]

you are getting -.000438 m or simply taking its magnitude: .000438 meters?

Yes.

The method used is certainly correct. Did you double-check that you copied the data correctly?

 

[mdxod]

checked many times =/ i have no idea what's going on. Have two friends who had similar numbers and their answers came out to be .00XXXXX

this is frustrating lol. is there any other techniques i could use? I mean i tried to keep everything in variable form before plugging in (as this is usually the best way) and then i even
tried plugging in numbers from the start but i got a different answer which was still wrong.

Data copy and pasted:
angle θ = 27.0 ˚ and distance d = 1.50 cm. Particle 2 has charge q2 = 6.40 × 10-19 C; particles 3 and 4 have charges q3 = q4 = -4.80 × 10-19 C. What is the distance D between the origin and particle 2 if the net electrostatic force on particle 1 due to the other particles is zero?

 

[Doc Al]

There's nothing wrong with how you solved the problem--any other method will give you the same result. (Unless you make an error, of course.) I wouldn't get bent out of shape over it.

If this is from a textbook, can you give me the name and problem number? (I've seen plenty of mistakes in textbooks over the years, especially in the problem sets.)

 

[mdxod]

Doc Al,

hope you don't mind, sent you a PM

 

[Doc Al]

I don't mind at all. I assume that this is some sort of online system that generates the questions and accepts the answers? They are notoriously flaky.

In any case, I would not spend any more time on it. If this counts toward your grade, be prepared to defend your answer to your instructor. You should have no problem doing so, since your work is correct. (Perhaps email your instructor that you are doing so. In fact a simple email saying that you think one of the problems is flawed may prompt him or her to check it over.)

Otherwise, spend your time working other problems instead of making yourself nuts over this one.

 

[rl.bhat]

To get the zero force at 1 due to all charge , the distance between 1 and 2 should be 0.01456m which is less than .015 m Hence distance between origin and D is 0.0004378 m.

 

[mdxod]

I don't mind at all. I assume that this is some sort of online system that generates the questions and accepts the answers? They are notoriously flaky.

In any case, I would not spend any more time on it. If this counts toward your grade, be prepared to defend your answer to your instructor. You should have no problem doing so, since your work is correct. (Perhaps email your instructor that you are doing so. In fact a simple email saying that you think one of the problems is flawed may prompt him or her to check it over.)

Otherwise, spend your time working other problems instead of making yourself nuts over this one.

Haha, yeah i mean i finished all of the other problems and only bothered with this because i wanted my 5 points. yes i already did email my professor so let me see what he says, maybe he can give me the points anyways. It just seems ridiculous that it would reject almost every answer i came up with even though there is a +/- %2...oh well.

To get the zero force at 1 due to all charge , the distance between 1 and 2 should be 0.01375m which is less than .015 m Hence distance between origin and D is 0.00125 m.

That would be incorrect (i just checked)

 

[rl.bhat]

After recalculation I got the distance between 1 and 2 equal to 0.01456 Hence D = .0004379m.

 

[mdxod]

After recalculation I got the distance between 1 and 2 equal to 0.01456 Hence D = .0004379m.

Yep, that's why me and Doc Al were stuck...thats the answer we both got, and it is marked as wrong.