Two oppositely charged wires and their capacitance

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Homework Help Overview

The discussion revolves around calculating the capacitance per unit length of two long parallel wires with opposite charge densities in air. The wires are separated by a distance that is a multiple of their radius, leading to questions about the integration limits for electric field calculations to determine potential.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the integration of the electric field to find potential, questioning the appropriate limits for integration. There is discussion about the existence of the electric field within the wires and how that affects the limits used in calculations.

Discussion Status

Some participants have offered guidance on the limits of integration, while others are still grappling with the implications of their assumptions. There is a mix of interpretations regarding the electric field's behavior and the resulting calculations.

Contextual Notes

Participants note the challenge of undefined logarithmic terms when integrating and the specific condition that η is much greater than 1, which influences the final calculations.

mooncrater
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Homework Statement


Two long wires with the same cross section are arranged in air PARALLEL to each other . Assume both have opposite charge densities +λ and -λ. The distance between the axis of the wire is η times larger than the radius(r) of wires cross section . What would be the capacitance of the wires per unit lenth?

Homework Equations


Electric field of any point between these wires at a distance x from the +λ charged wire is
λ/2πε[1/(x)+1/(ηr-x)]

The Attempt at a Solution


The problem I am facing is about the limits to be used while integrating Electric field to get the potential .
We know that V=-∫Edx
(Between some limits)
So ∫Edx =λ/2πε[ln(x/ηr-x)]
And if I integrate it over 0 to ηr
The I will get stuff like ln 0 and ln∞ which are obviously not defined . Some other limits just don't get the right answer which is πε/ ln η . [/B]
 
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mooncrater said:

Homework Equations


Electric field of any point between these wires at a distance x from the +λ charged wire is λ/2πε[1/(x)+1/(ηr-x)]

1) 'x' gives distance between which two points ?
2) What is the initial point ?
3) Now answer ,what should be the lower limit in the integration ?
 
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O-----------O
<----->is x
Initial point is the centre of the first+λ charged wire (I think).
So what I think is that the lower limit should be 0 and the upper limit should be ηr. BUT , when I put them in the integrated electric field the answer is absurd...
 
mooncrater said:
O-----------O
<----->is x
Initial point is the centre of the first+λ charged wire (I think).

Right .

But , does electric field exist within the wire ? The electric field does work from the surface of wire +λ to the closest point on the surface of other wire -λ.
 
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No. There is no electric field inside the wires ... therefore I should take the limits from r to ηr-r.
After doing the calculation I found that the potential is
V=λ/πε[ln(η-1)]
Since we're talking about a unit length thus λ=Q
So Capacitance=C=Q/V=πε/ln(η-1)
Which is just a little different from the answer(which now seems incorrect to me) So am I right here?
Edit: hey !hey!hey! In the question its given that η>>1 thus η-1≈η so now I got the correct answer ...now I am happy... thank you for your guidance dd.!
 
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