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Electric field using Gauss's Law, but in open cylinder given only r?

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ktw

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Homework Statement
Hello, this is my first post on the physics forums. Please forgive me if I am not following any post structure well. The question is supposed to use Gauss's Law, but I cannot understand how it would apply to a cylinder that you cannot obtain the surface area of (we are not given L), let alone an object that is not closed. How do you go about this?
Homework Equations
(Difficult to write without access to integral and vector symbols.) I am using Gauss's Law and General superposition of an electric field.
I have no idea how to approach the problem using Gauss's Law.
I found the electric field using superposition, and it was incorrect.

I am assuming you treat the wire as a continuous electric field, and then also treat the pipe as a continuous electric field. I solved for this using superposition, but it was incorrect, so then I tried multiplying it by 2πr (circumference) and it was incorrect as well. Maybe I input the wrong values? Any help is appreciated. Thank you.
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collinsmark

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Homework Statement: Hello, this is my first post on the physics forums. Please forgive me if I am not following any post structure well. The question is supposed to use Gauss's Law, but I cannot understand how it would apply to a cylinder that you cannot obtain the surface area of (we are not given L), let alone an object that is not closed. How do you go about this?
Homework Equations: (Difficult to write without access to integral and vector symbols.) I am using Gauss's Law and General superposition of an electric field.

I have no idea how to approach the problem using Gauss's Law.
I found the electric field using superposition, and it was incorrect.

I am assuming you treat the wire as a continuous electric field, and then also treat the pipe as a continuous electric field. I solved for this using superposition, but it was incorrect, so then I tried multiplying it by 2πr (circumference) and it was incorrect as well. Maybe I input the wrong values? Any help is appreciated. Thank you.
View attachment 249321→→
Hello @ktw,

Welcome to PF! :welcome:

Per the forum rules, you are going to have to show your work before you can get any help. We should be able to help show you where something went wrong or point you in the right direction if you get stuck.

---------
Regarding using Gauss's Law for finding the electric field strength:

Gauss's Law is only useful for finding the electric field in certain situations:
  • Objects with spherical symmetry such as point particles, spherical shells, etc.
  • Objects with cylindrical symmetry such as infinitely long wires or cylinders (you can use this to approximate long, yet finite length objects)
  • Infinitely long planes (although you can use this to approximate large, yet finite planes).

Don't get the wrong idea though, Gauss's Law always holds true. Always. It's just that it's only useful for finding the electric field in certain situations. This is one of those situations.

So that fact that you are not given the length is no surprise. Just treat the wire and tube as being infinitely long, but make your Gaussian surface with finite length [itex] \ell [/itex].
 

Delta2

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Hi @ktw and welcome at PF forums.

If you don't know Latex ( a way to write mathematical text using commands ) you still can type various mathematical symbols with the help of the toolbar. Above the textbox you are writing your reply there is a toolbar, and in that toolbar there is a button "insert symbol" which has the square root icon √x. Use that and other options of toolbar to write your equations.

Other than that i have not much to add to what @collinsmark said above.
 

ktw

6
2
Thank you for the welcome. Say I created a Gaussian cylinder with a radius of 1.4 cm, centered at the infinite wire, with a finite length of 10 cm. Would this provide me with the answer I am seeking? Or, would I then superposition that field value with the given charge of the outer cylinder?
 

collinsmark

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Thank you for the welcome. Say I created a Gaussian cylinder with a radius of 1.4 cm, centered at the infinite wire, with a finite length of 10 cm. Would this provide me with the answer I am seeking? Or, would I then superposition that field value with the given charge of the outer cylinder?
Yes, that approach would work.

But may I suggest that instead of using a finite length of "10 cm," just denote the length of your Gaussian surface as the variable [itex] \ell [/itex]. Keeping it as a variable has an advantage that should become obvious later, as you proceed.

[Edit: Of course, you can use whatever variable name you choose. I suggest the variable name [itex] \ell [/itex] because it is a common name to use for such a purpose. But the symbol/name doesn't matter.]
 

ktw

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2
I imposed my own Gaussian surface using a cylinder centered at the wire with a radius of 1.2 cm. I used the formula Φ = qenclosed0. I ended up with an electric field of 711.86 N/C. To clarify, should I superposition this value with the outer tube's electric field? Or, when the question is asking for the field, do you think it does not include the outer value? I am unsure of how to go about using superposition here.
 

ktw

6
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At this point I want to clarify that obtaining the Gaussian surface is not where I am struggling- rather, it is what to do with said Gaussian surface whenever there is another, oddly shaped electric field acting upon it. How do I generate a net field from that? I have attached a rudimentary drawing to explain my dilemma. The outer sphere is a positively charged sphere of negligible diameter. Could someone elaborate on this conceptually for me? I'm sorry if I am not being descriptive enough. I am having trouble expressing my issue.
Untitled.png
 

collinsmark

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I imposed my own Gaussian surface using a cylinder centered at the wire with a radius of 1.2 cm. I used the formula Φ = qenclosed0. I ended up with an electric field of 711.86 N/C.
I came up with a different answer. Show your work and we'll figure out what went wrong.

To clarify, should I superposition this value with the outer tube's electric field? Or, when the question is asking for the field, do you think it does not include the outer value? I am unsure of how to go about using superposition here.
Gauss's Law states:

[tex] \oint \vec E \cdot \vec {dA} = \frac{Q_{\mathrm{enc}}}{\varepsilon_0} [/tex]

In other words, if you integrate the dot product of the electric field with the corresponding surface vector over the entire surface of the Gaussian surface, the result is the total charge enclosed within the Gaussian surface, divided by [itex] \varepsilon_0 [/itex].

So, in this situation, is outer pipe enclosed within the Gaussian surface, or is the pipe outside of the Gaussian surface?
 

collinsmark

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At this point I want to clarify that obtaining the Gaussian surface is not where I am struggling- rather, it is what to do with said Gaussian surface whenever there is another, oddly shaped electric field acting upon it. How do I generate a net field from that? I have attached a rudimentary drawing to explain my dilemma. The outer sphere is a positively charged sphere of negligible diameter. Could someone elaborate on this conceptually for me? I'm sorry if I am not being descriptive enough. I am having trouble expressing my issue.View attachment 249328
Generally speaking, Gauss's Law only applies to the total charge within the closed, Gaussian surface. Any charge outside the closed, Gaussian surface doesn't affect the end result. That idea is the crux of Gauss's Law.
 

ktw

6
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Φ=qenclosed0
qenclosed=6.3x10-9 C/m (Just realized this is C/m instead of only C. What does this mean?)
ε0=8.85x10-12 F/m.

[6.3x10-9 C/m ] / [ 8.85x10-12 F/m = 711.86 C/F]. What is C/F? Coulombs per Farad? I think the unit I am looking for here is N/C, but please correct me if I am wrong.

So, I understand that the Gaussian surface does not contain the outer pipe. However, is the Electric field I obtain from that the answer this problem is looking for? I feel as though the outer field would affect the inner field, as it makes the inner field stronger.
 

collinsmark

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Φ=qenclosed0
qenclosed=6.3x10-9 C/m (Just realized this is C/m instead of only C. What does this mean?)
That's the linear charge density. It means that every meter of wire contains [itex] 6.3 \times 10^{-9} [/itex] Coulombs of charge. Typically, this is denoted by the variable "lambda", [itex] \lambda [/itex]. So here, [itex] \lambda = 6.3 \ \mathrm{\frac{nC}{m}} [/itex].

If you have strip of wire of length [itex] \ell [/itex] with a linear charge density of [itex] \lambda [/itex], then the total charge on the wire is [itex] q = \lambda \ell [/itex].

Now apply that to this problem. If you have a Gaussian surface of length [itex] \ell [/itex] surrounding a wire with linear charge density [itex] \lambda [/itex] (and nothing else inside the surface), then what is the total charge enclosed within the Gaussian surface?

ε0=8.85x10-12 F/m.

[6.3x10-9 C/m ] / [ 8.85x10-12 F/m = 711.86 C/F]. What is C/F? Coulombs per Farad? I think the unit I am looking for here is N/C, but please correct me if I am wrong.
I'm not quite sure what you did there. Walk us through your work, step by step. (I'm pretty sure the surface area of your Gaussian surface should fit in there somewhere, by the way).

So, I understand that the Gaussian surface does not contain the outer pipe. However, is the Electric field I obtain from that the answer this problem is looking for? I feel as though the outer field would affect the inner field, as it makes the inner field stronger.
No, [the field] from the outer pipe doesn't. That's because if a particular charge is outside the Gaussian surface then every field line originating from that charge that enters the surface must also exit the surface. All field lines originating outside the surface have an equal number of exit and entry points, which ultimately cancel. Only field lines that originate [or terminate] from inside the surface don't [necessarily] cancel. :smile:

[Edited slightly for grammar and clarity.]
 

ktw

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Thank you for the help. I finally got the problem! Well.. at least part A. I know from personal experience that it would be best for me to post my work after I got the answer, so here I have it attached:
uIhExar5qRCjbpHhrTybbcIMNa4mv3YcKhIQ=w1250-h938-no.jpg
 

collinsmark

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'Looks good to me. Good job! :smile:
 
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