Two oppositely charged wires and their capacitance

In summary, this conversation discusses the problem of finding the capacitance of two long wires with opposite charge densities arranged parallel to each other. The electric field between the wires is given by a formula involving ln(x/ηr-x), but there is confusion about the limits to be used while integrating to find the potential. After some discussion, it is determined that the lower limit should be 0 and the upper limit should be ηr-r, and the resulting capacitance is found to be πε/ln(η-1).
  • #1
mooncrater
217
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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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  • #2
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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  • #3
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...
 
  • #4
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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  • #5
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.!
 
Last edited:

1. What is capacitance?

Capacitance is a measure of an object's ability to store electrical charge. It is typically measured in units of farads (F).

2. How do two oppositely charged wires affect each other's capacitance?

When two oppositely charged wires are placed near each other, they form a capacitor. This increases the overall capacitance of the system as the electric field between the wires is strengthened.

3. How does the distance between the wires affect their capacitance?

The closer the wires are to each other, the higher the capacitance will be. This is because the electric field between the wires is stronger when they are closer together, allowing for more charge to be stored.

4. How does the size of the wires affect their capacitance?

The larger the surface area of the wires, the higher the capacitance will be. This is because a larger area allows for more charge to be stored on the wires.

5. What is the relationship between capacitance and voltage for two oppositely charged wires?

The capacitance of two oppositely charged wires is directly proportional to the voltage between them. This means that as the voltage increases, the capacitance also increases, and vice versa.

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