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MateoConLechuga
Minor Calculator Deity
Joined: 14 Jul 2013 Posts: 1025

Posted: 26 Feb 2015 10:28:25 pm Post subject: Physics Question [SOLVED] 


Okay, I've been thinking about this problem for a little bit, and I think I've got it worked out, but I am a little unsure of how to work out the final part. So, here goes:
Code: Consider two long, parallel and oppositely charged wires of radius 'r' with their centers separated by a distance D (D>>r). The charges are uniformly distributed strictly on the surface of each wire, with a line charge density λ.
 () Wire a
↑
D
↓
 (+) Wire b
Question: So, here's the question:
a) Using Gauss's Law, derive an expression for the total electric field in the space between the two wires.
b) What is the potential difference between the two wires?
c) Show that the capacitance per unit length of this pair of wires is: C/(length of wire)= (πε₀)/(ln((Dr)/r)
So far, here is my interpretation:
a) I determined that the electric field of one wire is E=λ/(2πε₀*R), which I then figured that the total field is E=σ/(ε₀)...
b) To get the potential difference, I did the following:
V=∫Eds (Integrated from a to b)
V=E∫ds
V=Ed
c) For c, I am not really sure where to begin...
I could probably spent some more time on it and work it out fully; I just wanted to see how far off I am, and how I could go about thinking about it. Any input and ideas are much appreciated! Thank you so much in advance! _________________ Portal Returns
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Last edited by MateoConLechuga on 27 Feb 2015 04:08:38 pm; edited 1 time in total 

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Tari
Systems Integrator
Joined: 03 Jul 2006 Posts: 2377 Location: Mormontown

Posted: 26 Feb 2015 11:04:41 pm Post subject: 


Without thinking through it too much, you should be able to derive the expression in (c) from C = Q/V. You've computed V, and Q is a function of length and charge density which are arbitrary and known, respectively. _________________
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MateoConLechuga
Minor Calculator Deity
Joined: 14 Jul 2013 Posts: 1025

Posted: 26 Feb 2015 11:07:26 pm Post subject: 


Ah, that makes sense, but I am a little unsure of where the ln((Dr)/r) part is derived from.... Or is that just individual sections of the length? The πε₀ part I assume comes from the charge itself. _________________ Portal Returns
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Tari
Systems Integrator
Joined: 03 Jul 2006 Posts: 2377 Location: Mormontown

Posted: 26 Feb 2015 11:14:57 pm Post subject: 


I'd guess the logarithmic term comes out of the integral, suggesting your approach to (b) is incorrect. It looks like you factored E out of the integral as if it were a constant factor, but E is itself a function of s, the distance from the starting wire. Thus your calculus is incorrect. _________________
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charlessprinkle
New Member
Joined: 27 May 2013 Posts: 87

Posted: 26 Feb 2015 11:17:14 pm Post subject: 


You have part (a) correct about the expression for a single wire if you look at a distance R away from the wire. Both electric fields point in the same direction, so they add constructively. Notice the dependency on R from each wire. If I am not an equal distance away from both wires in between, i.e. R_{1} = R_{2}, then they are not equal strength electric fields. So, declare one as R_{1} and the remaining distance to the other as R_{2}. The summation of those electric fields should be the answer. You can try expressing it in terms of D if you wish, knowing that R_{1}+R_{2}=D.
I agree with Tari. For the electric potential of part (b), you can't pull out the electric field from the integral. Note:
V = ∫E▪dR, where E is a function of the distance from the wire. So, you have a straight field, but it's variable according to how far away from the wire you are. Since you have a 1/R relationship, integration should give you some natural logarithm form for the answer.
Once you find the potential for each wire individually, we understand that potentials are scalars, which add numerically as opposed to vectorially (as in Coulomb's law or electric fields). So having both potentials, you can straight up add them.
For part (c), notice that the capacitance definition is dependent on the total charge and total potential. Now, "long" wires we can assume to be infinite, and that means there would be an infinite amount of charge. So, it's not sensible to be talking about the charge total on the entire rod. What's more sensible to be talking about is the charge per unit length. In doing so, we will likewise be talking about capacitance per unit length. Divide both sides of the equation by the length L to yield that C/L = Q/(L*V). Q/L, we define as the charge per unit length, λ. So, the capacitance per unit length will be given now by C/L = λ/V. Given the 1dimensional charge density, and the potential difference for that single wire you found, you should be able to calculate this out. Note the natural logarithm in the answer, as I've drawn your attention to the fact that the electric field falls off with 1/R and that the potential will then be some natural logarithm. If you are close to the answer but the logarithm argument doesn't appear right, you may also want to consider using properties of logarithms to arrange the terms inside the argument. 

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MateoConLechuga
Minor Calculator Deity
Joined: 14 Jul 2013 Posts: 1025

Posted: 26 Feb 2015 11:35:58 pm Post subject: 


Okay, so if I do (b) like this:
V=∫(E/r)dr ; r represents the radius from the wire, then the integral gives
V=E*(ln(Dr)  ln(r))
which can be simpilfied to
V=E*(ln((Dr)/r)).
Does this seem to be more along the right lines, or can I still not pull the E from the integral, as the field varies? _________________ Portal Returns
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charlessprinkle
New Member
Joined: 27 May 2013 Posts: 87

Posted: 26 Feb 2015 11:54:17 pm Post subject: 


Where did the random E/r come from? What I meant was that E is a function of r, which has a 1/r dependence according to what you derived in part (a):
E = λ/2πεr
If we want to make it more apparent what is constant and what is variable, I can write it like this:
E = (λ/2πε)*1/r
You simply drop it into E.
V = ∫E▪dr = ∫(λ/2πεr)dr
If you want to call the variable radius away from the wire r, then notice r will go from 0 to R_{1}, some arbitrary distance away from wire 1. We go from 0 to R_{1} because of the fact we will be comparing the potential difference between being at the rod, and being a distance R_{1} away from it.
Note you can never pull a variable out of an integral because it varies. You're doing a summation over all the values of that variable.
Remember how we found the electric field anywhere in space between the wires, and called the distance away R_{1} and R_{2} for wires 1 and 2 respectively, such that they were not necessarily equal? Specifically, we stated that if we're one distance R_{1} away from wire 1, and a different distance from wire 2, R_{2}, then the only general relationship could be expressed as R_{1}+R_{2}=D, the total distance between the wires.
The only difference in the problem is they choose to keep using r for the distance from 1 rod, and express the other distance as Dr. It's exactly the same thing.
Keep this idea altogether in mind as we will need it.
Work from there and see what you get for wire 1. Do likewise for wire 2 to get a similar expression. Once you get both expressions, compare them for a total potential difference. 

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MateoConLechuga
Minor Calculator Deity
Joined: 14 Jul 2013 Posts: 1025

Posted: 27 Feb 2015 12:19:41 am Post subject: 


Thank you so much for your responses! I apologize for seeing somewhat silly at times; going to try to give it a better shot now.
One quick question: Is it possible to assume the field is 2*(λ/2πεr), which is (λ/πεr) because the wires have the field lines in the same direction and are identical? If so, can I use that instead in the integral? _________________ Portal Returns
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charlessprinkle
New Member
Joined: 27 May 2013 Posts: 87

Posted: 27 Feb 2015 12:29:15 am Post subject: 


Hopefully this visual geometry will clear up the confusion that they're not the same distance to each wire, and hence will be different field strengths:
(right click and view the image to see it in the correct orientation) 

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MateoConLechuga
Minor Calculator Deity
Joined: 14 Jul 2013 Posts: 1025

Posted: 27 Feb 2015 12:42:51 am Post subject: 


Thank you for drawing that! I do understand that the fields are definitely going to be different at different points, but as the question is asking for the total field, then wouldn't the distances that are opposite add up to the same value? For instance:
These two points would have the same field, and thus the field is simply twice the field one one... I apologize for continuously bringing this up; but it is just odd to me. Thank you again! _________________ Portal Returns
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charlessprinkle
New Member
Joined: 27 May 2013 Posts: 87

Posted: 27 Feb 2015 01:56:24 am Post subject: 


Remember that you're trying to find the total electric field at a single point, so you want to use the same point for both wires. You're sticking to one coordinate system such that the electric field at the point I've drawn, the same for both wires, is given by E_{1} and E_{2} I've written. You're doing something I'm unsure of where you're making a mirror image for the other wire. Remember that you want to find the electric field at the same point in space (fixed, constant). If you want to find the electric field at a point in space due to both wires of charge acting at that point, you must use the same point. Hence, the r and Dr are values that cannot be changed or flipped.
When you calculate the electric potentials, you should do them separately, and add the final results. Use E_{1} to get V_{1} via integration (again, because E is variable with distance from the wire). Because you want to find the potential due to the wire at some distance r, so you can compare it with the other wire's potential, you will want to integrate from 0 to r. For the other wire under the same distancing of the problem, you will want to go from 0 to Dr for V_{2}. 

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