Calculate electric field applied along a wire

In summary, the previous poster was correct and they had conductedivity of gold wrong, it is actually 4.1*10^7. Thanks for all the help!
  • #1
jendrix
122
4
Hi, I'm doing some revision and I'm having a problem with the following:

Conductivity of gold is 4.1*10^-8 , what is the electric field along the wire? The current is 13.2A and the radius is 3.8mm

I've used 1/conductivity to find resistivity of (2.4*10^-8) and used this along with area to find the resistant for 1m

(2.4*10^-8) *1m/pi*(0.0038)^2

This gives 5.3*10^-4 which when multiplied by the current gives 0.007 V/m, however I'm told this is the wrong answer, can anyone see where I might have gone wrong?

Thanks
 
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  • #2
jendrix said:
Hi, I'm doing some revision and I'm having a problem with the following:

Conductivity of gold is 4.1*10^-8 , what is the electric field along the wire? The current is 13.2A and the radius is 3.8mm

I've used 1/conductivity to find resistivity of (2.4*10^-8) and used this along with area to find the resistant for 1m

(2.4*10^-8) *1m/pi*(0.0038)^2

This gives 5.3*10^-4 which when multiplied by the current gives 0.007 V/m, however I'm told this is the wrong answer, can anyone see where I might have gone wrong?

Thanks

I think you've shown an incorrect value for the exponent for the conductivity of gold; probably a typo, since the value for resistivity looks okay.

Your steps to find the electric potential across a meter length is okay. What's probably "doing you in" on your result is the significant figures. All the values you've worked with in the problem have at least 2 significant figures, and the result you've presented has only one.
 
  • #3
Use j = σE
j = current density, amp/sq. m
σ = conductivity, S/m
E = electric field, V/m

Vectors in bold.
 
  • #4
rude man said:
Use j = σE
j = current density, amp/sq. m
σ = conductivity, S/m
E = electric field, V/m

Vectors in bold.

The previous poster was right, I did have conductivity of gold wrong, it is actually 4.1*10^7

So using j=σE

I found j to be 2.91*10^5 and dividing this by σ 4.1*10^7 my answer is 7.1*10^-3 V/m to 3sf, does this look correct?

Thanks
 
  • #5
jendrix said:
The previous poster was right, I did have conductivity of gold wrong, it is actually 4.1*10^7

So using j=σE

I found j to be 2.91*10^5 and dividing this by σ 4.1*10^7 my answer is 7.1*10^-3 V/m to 3sf, does this look correct?

Thanks

Yes, that looks okay (but you've specified two sig figs, not three, which is fine for the given values).

Your first method would give you the same result (and mathematically amounts to the same thing, but "takes the scenic route").
 
  • #6
jendrix said:
The previous poster was right, I did have conductivity of gold wrong, it is actually 4.1*10^7

So using j=σE

I found j to be 2.91*10^5 and dividing this by σ 4.1*10^7 my answer is 7.1*10^-3 V/m to 3sf, does this look correct?

Thanks

j is right, the rest I leave to you & gneill. I would suggest including units for each number. It's a powerful checking tool and really makes no sense without them.
 
  • #7
Thanks for all the help guys :)
 

1. What is the formula for calculating electric field along a wire?

The formula for calculating electric field along a wire is E = V/d, where E represents electric field, V represents voltage, and d represents distance along the wire. This formula assumes a uniform electric field and a straight wire.

2. How do you calculate the electric field at a specific point on a wire?

To calculate the electric field at a specific point on a wire, you can use the formula E = kQ/r^2, where k is the Coulomb's constant, Q is the charge of the wire, and r is the distance between the point and the wire. This formula applies to a point charge, so if the wire has a distribution of charge, you will need to use integration to calculate the electric field.

3. What factors affect the electric field along a wire?

The electric field along a wire is affected by the voltage applied, the distance along the wire, and the distribution of charge on the wire. Additionally, the presence of other nearby charges or conductors can also affect the electric field.

4. How does the direction of the electric field along a wire change?

The direction of the electric field along a wire depends on the direction of the voltage applied. The electric field will always be directed away from the positive terminal and towards the negative terminal of the voltage source.

5. Can the electric field along a wire be negative?

Yes, the electric field along a wire can be negative. This would occur if the voltage applied is negative or if the distribution of charge on the wire results in a net negative charge in the direction of the electric field. In this case, the electric field would be directed towards the positive terminal of the voltage source.

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