Electric Field along bisector of charged line (w/error)

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Homework Help Overview

The discussion revolves around determining the electric field magnitude along the perpendicular bisector of a uniformly charged line. The original poster is attempting to find the distance at which the approximation of the line charge as a point charge results in a 5% error in the electric field calculation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the original poster's approach to approximating the line charge as a point charge and question the complexity of the calculations involved. Some suggest a simpler geometric argument to determine when the approximation holds true within the specified error margin.

Discussion Status

There is ongoing exploration of different methods to assess the error in the approximation. Some participants have proposed alternative calculations and comparisons, while others are clarifying the assumptions made in the original approach. No consensus has been reached on the correct method or outcome.

Contextual Notes

Participants note potential confusion regarding the values used in calculations and the implications of using different mathematical forms. There is also mention of imposed homework rules that may limit the approaches considered.

Slightly Odd Guy
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Homework Statement


You wish to determine the electric field magnitude along the perpendicular bisector of a 230-mm line along which35 nC of charge is distributed uniformly. You want to get by with a minimal amount of work, so you need to know when it is sufficient to approximate the line of charge as a charged particle.

At what distance along the perpendicular bisector does your error in E reach 5 % when you use this approximation?

Homework Equations


dE=kλdx/r2
E=kQ/r2
λ=Q/L

The Attempt at a Solution


If I do the approximation part first, I get:

1. Eapprox=kQ/y2
Plugging in the values (minus units at this time):
2. Eapprox=315/y2

So now I work on the Eline-charge.

Because point P at distance d is along the bisector of the line, I know that Ex cancels itself out. All I need to worry about is Ey.

3. dE=((kλdx)/(x2+y2)2)*(y/(x2+y2)1/2)
4. dE=(kλydx)/(x2+y2)3/2)
=>
5. E=kλ∫(ydx/(x2+y2)3/2
I punch that through an integral calculator, and I get
6. E=kλ[x/y(x2+y2)1/2] ---- evaluated from -L/2 to L/2

Now I start to plug in all the values (minus the units for the time being) and I get

7. E=1370[0.23/(y(.013+y2)1/2)]
8. E=315.1/(y(.013+y2)1/2)

Okay.

Now I'm working on finding the approximate error.

error=(|Eapprox-Eexact|)/Eexact

I know the error (5%) and now I want to solve for y at that error.

9. 0.05=|315/y2-315.1/(y(.013+y2)1/2)|/315.1/(y(.013+y2)1/2)

This is pretty nasty, so I plugged it in the Mathway calculator and it promptly informed me there are no solutions.

That sums up what I've done. Where am I going wrong?

Thanks,

SOG
 
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I suspect you are going a much more elaborate path than intended. The objective is to avoid such complications.
Where does the error come from in pretending the charge is a point particle? Can you think of a simple geometric argument for deciding when that error will be less than 5%? I don't think you need to be too precise about it, just able to say "when the distance is more than ... the error will be less than 5%".
 
Well... I could probably compare the radii for both. In which case, I would have:

0.05 = |d2-(d2+0.013)|/d2
0.05=0.013/d2
d=0.51 m

Am I on the right track there?
 
Slightly Odd Guy said:
Well... I could probably compare the radii for both. In which case, I would have:

0.05 = |d2-(d2+0.013)|/d2
0.05=0.013/d2
d=0.51 m

Am I on the right track there?
That's what I had in mind, except I'm not sure how you arrived at 0.013m. That's a lot less than half 230mm.
 
0.013m2 is ~ (230mm/2)2.
 
I switched some numbers around in my second post, so it screw some more things up. I think it's supposed to be:

0.05=|d2-(d2+0.013)|/(d2+0.013)

in which case it gives d=0.501m,

which is wrong.
 
Slightly Odd Guy said:
0.013m2 is ~ (230mm/2)2.
Ah, sorry - didn't notice it was not squared in the equation you wrote. So, yes, I think that's fine.
If you wanted to refine it some, you could show that 1/r2 at the point half way from the middle of the charge to one end is more than the average of the values at the middle and at the end respectively. But that's probably overkill.

[Edit: I've not checked this carefully, but I think this argument might allow you to use 0.013/4.]
 
Slightly Odd Guy said:
I switched some numbers around in my second post, so it screw some more things up. I think it's supposed to be:

0.05=|d2-(d2+0.013)|/(d2+0.013)

in which case it gives d=0.501m,

which is wrong.
Oh - so you know what the answer is supposed to be?
 
haruspex said:
Oh - so you know what the answer is supposed to be?

I don't, unfortunately. My online homework only tells me that it's wrong.
 
  • #10
Slightly Odd Guy said:
I don't, unfortunately. My online homework only tells me that it's wrong.
The refinement I suggested in post #7 gets it down to 0.26m. The exact value is about 0.21m.
 
  • #11
Ok, the correct answer is d=0.36m. I don't know how they come up with that answer.
 
  • #12
Slightly Odd Guy said:
Ok, the correct answer is d=0.36m. I don't know how they come up with that answer.
I just realized what we both forgot, to take only the component in the x direction. So we should use 1/r3, not 1/r2.
 

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