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

In summary, at a distance of 0.51m from the center of the charge, the error in the electric field magnitude reaches 5% when the approximation is used.
  • #1
Slightly Odd Guy
11
0

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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  • #2
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%".
 
  • #3
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?
 
  • #4
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.
 
  • #5
0.013m2 is ~ (230mm/2)2.
 
  • #6
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.
 
  • #7
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.]
 
  • #8
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?
 
  • #9
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.
 

1. What is meant by "Electric Field along bisector of charged line (w/error)"?

This refers to the electric field strength at points along the bisector of a line segment that has a charge distribution. The "w/error" indicates that there may be some error or uncertainty in the calculations or measurements of the electric field.

2. How is the electric field calculated along the bisector of a charged line?

The electric field along the bisector of a charged line can be calculated by using Coulomb's Law, which states that the electric field strength at a point is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance from the charge.

3. What is the significance of the bisector in this scenario?

The bisector is important because it represents the direction in which the electric field is being measured. The electric field is a vector quantity with both magnitude and direction, and the bisector helps determine the direction of the electric field at different points along the line.

4. Why might there be error in the calculation or measurement of the electric field along the bisector?

There can be error due to various factors such as experimental limitations, inaccurate measurements of distance or charge, or assumptions made in the calculations. Additionally, the electric field may not be constant along the bisector, leading to variations and potential errors.

5. How does the presence of other charges or objects in the vicinity affect the electric field along the bisector?

The presence of other charges or objects can alter the electric field along the bisector. This is because the electric field is affected by all surrounding charges, and the presence of other objects can cause interference or distortion in the electric field. This must be taken into account when calculating or measuring the electric field along the bisector.

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