Potential difference between wire and ground

  • #1
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Homework Statement


A long, straight power line is made from a wire with radius ra = 1.0 cm and carries a line charge density λ = 2.6 μC/m. Assuming there are no other charges present, calculate the potential difference between the surface of the wire and the ground, a distance of rb = 22 m below.

Homework Equations


ΔV= -∫E⋅ds
E due to infinite line of charge: 2kλ/r

The Attempt at a Solution


what I did was -2kλ∫1/r dr, with limits .01m to 22m. So -2kλ(ln(22/.01). I feel like this might be wrong because I am only taking into account the bottom of the wire aren't I?
 

Answers and Replies

  • #2
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I'm pretty sure that's correct. With a sphere, for instance, Gauss's Law means that just outside the surface of the sphere the field is identical to that of a point charge located at the center of the sphere. This problem is the 2D version of that. Instead of a sphere you can think of it as a circle, the cross section of the wire. Just outside of the wire the field should be identical to that of an ideal infinitely-thin wire located at the center of the finite wire.
 
  • #3
haruspex
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Homework Statement


A long, straight power line is made from a wire with radius ra = 1.0 cm and carries a line charge density λ = 2.6 μC/m. Assuming there are no other charges present, calculate the potential difference between the surface of the wire and the ground, a distance of rb = 22 m below.

Homework Equations


ΔV= -∫E⋅ds
E due to infinite line of charge: 2kλ/r

The Attempt at a Solution


what I did was -2kλ∫1/r dr, with limits .01m to 22m. So -2kλ(ln(22/.01). I feel like this might be wrong because I am only taking into account the bottom of the wire aren't I?
Aren't you using the field as it would be if no earth were present?
I get a slightly different result. I used the method of images to replace the ground by another charged object and considered the sum of the potentials.
 
  • #4
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Aren't you using the field as it would be if no earth were present?
I get a slightly different result. I used the method of images to replace the ground by another charged object and considered the sum of the potentials.
I'm not sure what the method of images is, we have not done that. Am I supposed to assume the Earth has a charge? We have not done any examples involving the Earth yet either.
 
  • #5
haruspex
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I'm not sure what the method of images is, we have not done that. Am I supposed to assume the Earth has a charge? We have not done any examples involving the Earth yet either.
OK. I consider the earth as an infinite flat conducting plate. It has no net charge, but there will be an induced charge distribution.
In the method of images you seek to replace such an infinite sheet by some other body with a known charge distribution. The replacement is valid (above the sheet) provided it produces the same potential everywhere in the plane of the sheet. In this case, a potential of zero.
Given you have this charged wire above the ground, what charged body below the ground would combine with it to result in a net zero potential everywhere at ground level?
 
  • #6
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OK. I consider the earth as an infinite flat conducting plate. It has no net charge, but there will be an induced charge distribution.
In the method of images you seek to replace such an infinite sheet by some other body with a known charge distribution. The replacement is valid (above the sheet) provided it produces the same potential everywhere in the plane of the sheet. In this case, a potential of zero.
Given you have this charged wire above the ground, what charged body below the ground would combine with it to result in a net zero potential everywhere at ground level?
Would a wire of opposite linear charge density work, or am I misunderstanding this?
 
  • #7
haruspex
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Would a wire of opposite linear charge density work, or am I misunderstanding this?
Exactly.
So now you can figure out the potential due to each at any point you like and sum them.
 

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