Help with a Gauss' law problem -- A long cylinder with a uniform charge density throughout

  • #1
hey guys and gals, I'm normally pretty good at my physics homework, but this problem has my STUMPED, I think im starting it right, but I have no idea how to proceed or if I am on a good path.

Homework Statement


A Very Long Cylinder with a radius of 7.2cm (.072m) has a uniform density of charges throughout it. If the charge density of the cylinder is 1,264 nC.m^3, inside the cylinder for what distances x (less the the radius) from the center of the cylinder, the magnitude of the electric field equal to Ax^n where the electric field is in nC and x is measured in meters. What is the value of A, and what is the value of N


Homework Equations


intE*dA=Qin/Epsilon naught


The Attempt at a Solution


Ok, so I know I am working in three detentions, so my charge density is not going to be n=q/4piR^2, but rho (shown as P here on out) I was able to assume that the intE*dA=E*2pi*R*L
and that lead me to

Qin (charge)=int P*Dv and Dv=2piL


but im still confused and have no idea how I am supposed to get a numerical value for these things.


please help
 

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  • #2
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Given: ... a very long uniformly charged cylinder of radius ##R = 7.2~\rm cm = 0.072~\rm m~##... the volume charge density is ##ρ = 1264~\rm{nC} ⋅ \rm m^{-3} = 1.264~×~10^{-6}~\rm C ⋅ \rm m^{-3}~##...
Unknown: ... the values of##~A~##and##~n~##for the electric field with magnitude##~E = Ar^n~##for##~r~\lt~R~## inside the cylinder ...
Solution: ... easier to obtain by applying Gauss's law of electricity than by direct integration using ##\vec E = \int d\vec E## ...
##\dots## Gauss's law##~\Rightarrow~Φ_E = ~\begin{cases}\begin{align} &\oint_{\rm S} \vec E \cdot d \vec s = \oint_{\rm S} \vec E \cdot \hat {\mathbf n}ds = \oint_{\rm S} E(ds)(\cos θ)\dots\nonumber \\ & q/ε_0~\Leftarrow~q =~\text {total charge enclosed by surface S}\dots\nonumber \end{align}\end{cases}##
where ##θ## is the angle between ##\vec E## and the outward unit normal vector ##\hat {\mathbf n}## perpendicular to the area element ##ds## of the imaginary Gaussian surface ##\rm S##. The field point where the electric field ##\vec E## given above is being evaluated must lie on the Gaussian surface itself.
At this point I would like to point out the advantage of using Gauss's law in the form shown above involving ##\vec E##. The integral on the right hand side is in general complicated. But by choosing properly the imaginary Gaussian surface ##\rm S##, one in which the magnitude of ##\vec E## and ##θ## become constant in ##\rm S##, then both ##E## and ##\cos θ## can be taken out of the integral sign, leaving only ##Φ_E = E\cos θ \oint_{\rm S} ds## which reduces the complication in the evaluation of the integral.
In the case of a circular cylindrical Gaussian surface
... ##Φ_E## = ##Φ_{E\rm t}## + ##Φ_{E\rm b}## + ##Φ_{E\rm s}~\Leftarrow
\begin{cases}\begin{align} & Φ_{E\rm t} = \text {flux through the top circular end}\dots \nonumber \\ & Φ_{E\rm b} = \text {flux through the bottom circular end}\dots \nonumber \\ & Φ_{E\rm s} = \text {flux through the lateral side} \dots\nonumber\end{align} \end{cases}##
With##~\vec E~##radially outward for a positive cylindrical charge distribution, ##\hat {\mathbf n}## at the two circular ends are perpendicular to ##\vec E## , meaning ##\vec E## is parallel to the circular faces so that no ##\vec E## vectors pass through those surfaces, making ##Φ_{E\rm t} = Φ_{E\rm b} =## 0. Only ##Φ_{E\rm s}## has nonzero contribution to ##Φ_E## since there ##\hat {\mathbf n}## is parallel to ##\vec E## giving ##θ = 0°## with ##\cos ( 0°) = 1## that causes the ##\vec E## vectors pass through the lateral side of the cylinder.
It follows that according to Gauss's law, we get after carrying out the intergration $$Φ_E = Φ_{E\rm s} =~\begin{cases}\begin{align} & \int E(ds)\cos 0° =\iint (A{r_0}^n)(r_0dθdz) ~\dots\nonumber \\ & A{r_0}^{n + 1}\iint dθdz = 2πAL{r_0}^{n + 1}~\dots \nonumber\end{align}\end{cases}$$since ##0\leq θ\leq 2π## , ##-L/2 \leq z\leq +L/2~## , and where ##r_0\lt R## is one particular radius of the imaginary Gaussian surface ##\rm S## inside the charge distribution.
Evaluation of total charge enclosed by Gaussian surface S with ##ρ ## = constant follows:
$$q = \int ρdV = ρ\iiint (rdθdrdz) = ρ\int_0^{r_0} rdr \int_0^{2π} dθ \int_{-L/2}^{+L/2} dz$$##\Rightarrow~q = (1/2)ρ{r_0}^2(2π)L = πρL{r_0}^2~\Rightarrow~Φ_E = q/ε_0 = πρL{r_0}^2/ε_0~\dots##
Combining the two expressions obtained for ##Φ_E##, we finally get
##\dots~2πAL{r_0}^{n + 1} = πρL{r_0}^2/ε_0~\Rightarrow~2A{r_0}^{n - 1} = ρ/ε_0~\dots##
By looking at the units only, without the numerical values of the physical quantities involved in the preceding equation, we get the unknown value of ##n##. Use the angle brackets to denote the units only##~\dots~\langle⋅\rangle =## unit(s)##~\dots##
##\Rightarrow~\langle{ 2A{r_0}^{n - 1} }\rangle = \langle{ ρ/ε_0 }\rangle~\Rightarrow~\langle A \rangle \langle{{r_0}^{n - 1}}\rangle = {\langleρ\rangle}/{\langle{ε_0}\rangle} = {\langleρ\rangle} {\langle{ε_0} \rangle}^{-1}~\dots##
##\dots~\langle A \rangle \rm {m^{n - 1}} = (\rm {C ⋅ m^{-3}})/\langle{ε_0} \rangle = \rm {C ⋅ m^{-3}} {\langle{ε_0} \rangle}^{-1}~\dots##
##\dots~##Coulomb force##~F = q_1q_2/(4πε_0r^2)~\Rightarrow~\langle F \rangle = \langle q_1q_2 \rangle/[\langle ε_0 \rangle\langle r^2 \rangle]##
##\Rightarrow~\rm N = \rm {C^2}/[\langle{ε_0} \rangle \rm {m^2}] = \rm {C^2}{\langle{ε_0} \rangle}^{-1} \rm {m^{-2}}~\Rightarrow~{\langle{ε_0} \rangle}^{-1} = \rm N ⋅ \rm {C^{-2}}⋅\rm {m^{+2}}##
It follows that
##\dots~\langle A \rangle \rm {m^{n - 1}} = \rm {C ⋅ m^{-3}} {\langle{ε_0} \rangle}^{-1} = \rm {C ⋅ m^{-3}}⋅\rm N ⋅ \rm {C^{-2}} ⋅\rm {m^{+2}} = \rm m^{-1}⋅\rm N ⋅ \rm {C^{-1}}~\dots##
##\Rightarrow~\rm {m^{n - 1}} = \rm m^{-1}~\Rightarrow~n - 1 = -1~\Rightarrow~n = 0~\Rightarrow~\langle A \rangle \rm {m^{-1}} = \rm m^{-1}⋅\rm N ⋅ \rm {C^{-1}}~\dots##
##\Rightarrow~\langle A \rangle = \rm N ⋅ \rm {C^{-1}}~\Rightarrow~##a unit of the electric field
Going back to the result we saw by combining the two expressions obtained earlier for ##Φ_E~##, we find that
##\dots~n = 0~\Rightarrow~2A{r_0}^{-1} = ρ/ε_0~\dots##
##\Rightarrow~A{r_0}^{-1} = \begin{cases}\begin{align} & (0.5)(1.264×10^{-6}~\rm C ⋅ \rm m^{-3})/(8.85 ×10^{-12}~\rm C^2 ⋅ N^{-1} ⋅ \rm m^{-2})~\dots \nonumber \\ & 7.141243 ×10^4~\rm C^{-1} ⋅ N ⋅ \rm m^{-1}~\dots\nonumber \end{align}\end{cases}##
##\Rightarrow~A = 7.141243 ×10^4~\rm N ⋅ \rm C^{-1}~\Leftarrow~\text{value of }## ##A## ##\text{ in }## ##E = Ar_0^n~\dots##
so that ##n = 0~\Rightarrow~E = A## ##~\text {at the center of the cylindrical charge distribution}##.
The given value ##R = 7.2~\rm cm = 0.072~\rm m~## has no use, unless ##r_0 = 7.2~\rm cm = 0.072~\rm m~##and it turns out that##~n \neq 0~##.
 
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  • #3
kuruman
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This thread is almost 9 years old. It cannot be of much help to the OP.
 
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berkeman
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This thread is almost 9 years old. It cannot be of much help to the OP.
Yeah, but it might help someone else who finds the thread via a Google search. As long as a schoolwork thread has been dormant for a while, it's okay to post a full solution for future reference. :smile:
 
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  • #5
kuruman
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Yeah, but it might help someone else who finds the thread via a Google search. As long as a schoolwork thread has been dormant for a while, it's okay to post a full solution for future reference. :smile:
In that case, a better solution is required for the record because the solution in #2 meanders about and comes up with the nonsensical conclusion
so that ##n = 0~\Rightarrow~E = A## ##~\text {at the center of the cylindrical charge distribution}##.
At the center of the distribution, the field is zero, not equal to the constant ##A##. Furthermore, the radial dependence is linear inside the distribution. Here is a compact solution.

We construct a cylindrical Gaussian surface of radius ##r## that is coaxial with the cylinder. For a length ##L## of the cylinder, Gauss's law says
$$E 2\pi rL=\frac{\rho\pi r^2L}{\epsilon_0} \implies E=\frac{\rho r}{2\epsilon_0 }.$$Thus, if we are to write ##E=Ar^n##, we must identify ##A =\dfrac{\rho}{2\epsilon_0}## and ##n=1##.
 
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  • #6
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Okay, I was wrong. I'm sorry. I accept my mistake. But even without the mathematics, it is clear that ##E = 0## at the center of the uniformly charged cylinder because ##\vec E## run radially away from the axis so that pairs of them moving away opposite one another cancel out. I failed to see that. I wonder where I made a mistake in my computation.
Another thing though. I don't think that giving a solution that meanders about is not at all that bad, so long as one comes to sensible results of course. Poor students who want to learn physics will not be able to grasp everything in compact solutions that kuruman seems to be so fond of.
 
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  • #7
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Could somebody familiar with the way mathematicians argue please take a look at this and make a comment? Thank you. I'm not a mathematician. What follows is about the electric field at the axis of a long uniformly charged right circular cylinder.
... a better solution is required for the record because the solution in #2 meanders about and comes up with the nonsensical conclusion ...
... At the center of the distribution, the field is zero, not equal to the constant ##A##. Furthermore, the radial dependence is linear inside the distribution. Here is a compact solution.
We construct a cylindrical Gaussian surface of radius ##r## that is coaxial with the cylinder. For a length ##L## of the cylinder, Gauss's law says
$$E 2\pi rL=\frac{\rho\pi r^2L}{\epsilon_0} \implies E=\frac{\rho r}{2\epsilon_0 }.$$Thus, if we are to write ##E=Ar^n##, we must identify ##A =\dfrac{\rho}{2\epsilon_0}## and ##n=1##.
Following his arguments in post #5, he obtained ##~E2πrL = ρπr^2L/ε_0~\dots##
##\Rightarrow~E = ρr/(2ε_0)~\dots## which he got, presumably, by just dividing by ##2πrL##. Equivalently, one could write his result as ##~2πrL[ E - ρr/(2ε_0) ]## = 0 ##~\Rightarrow~r[ E - ρr/(2ε_0) ]## = 0##~\dots~## which one obtains by dividing by ##2πL## involving constants only, leaving the common variable factor ##r## untouched to avoid division by 0, which everyone should know is not allowed in mathematics. Now, along the cylindrical axis where ##r = 0##, is it correct to say that ##[ E - ρr/(2ε_0) ]## = 0 is valid too, so that one gets ##E = ρr/(2ε_0)##? Or is it more mathematically sound to conclude that, assuming ##r = 0##, then ##[ E - ρr/(2ε_0) ]~\neq~0## so that ##E~\neq~ρr/(2ε_0)~##, and that therefore, one cannot say that ##E = 0~##when ##r = 0##, something that one gets only when ##[ E - ρr/(2ε_0) ] = 0?## I'm sorry if my questions are somewhat confusing. I don't know how to ask them in a better way. I'm not a native English language speaker.
 
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Could somebody familiar with the way mathematicians argue please take a look at this and make a comment? Thank you. I'm not a mathematician. What follows is about the electric field at the axis of a long uniformly charged right circular cylinder.

Following his arguments in post #5, he obtained ##~E2πrL = ρπr^2L/ε_0~\dots##
##\Rightarrow~E = ρr/(2ε_0)~\dots## which he got, presumably, by just dividing by ##2πrL##. Equivalently, one could write his result as ##~2πrL[ E - ρr/(2ε_0) ]## = 0 ##~\Rightarrow~r[ E - ρr/(2ε_0) ]## = 0##~\dots~## which one obtains by dividing by ##2πL## involving constants only, leaving the common variable factor ##r## untouched to avoid division by 0, which everyone should know is not allowed in mathematics. Now, along the cylindrical axis where ##r = 0##, is it correct to say that ##[ E - ρr/(2ε_0) ]## = 0 is valid too, so that one gets ##E = ρr/(2ε_0)##? Or is it more mathematically sound to conclude that, assuming ##r = 0##, then ##[ E - ρr/(2ε_0) ]~\neq~0## so that ##E~\neq~ρr/(2ε_0)~##, and that therefore, one cannot say that ##E = 0~##when ##r = 0##, something that one gets only when ##[ E - ρr/(2ε_0) ] = 0?## I'm sorry if my questions are somewhat confusing. I don't know how to ask them in a better way. I'm not a native English language speaker.
I am a mathematician, and I think you are partially correct on this comment, that is for ##r\neq 0## it is ##E=\rho r/2\epsilon_0## but for ##r=0## we cant tell what is happening, all we can say is that $$\lim_{r\to 0}E=0$$
 
  • #9
kuruman
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Could somebody familiar with the way mathematicians argue please take a look at this and make a comment? Thank you. I'm not a mathematician. What follows is about the electric field at the axis of a long uniformly charged right circular cylinder.

Following his arguments in post #5, he obtained ##~E2πrL = ρπr^2L/ε_0~\dots##
##\Rightarrow~E = ρr/(2ε_0)~\dots## which he got, presumably, by just dividing by ##2πrL##. Equivalently, one could write his result as ##~2πrL[ E - ρr/(2ε_0) ]## = 0 ##~\Rightarrow~r[ E - ρr/(2ε_0) ]## = 0##~\dots~## which one obtains by dividing by ##2πL## involving constants only, leaving the common variable factor ##r## untouched to avoid division by 0, which everyone should know is not allowed in mathematics. Now, along the cylindrical axis where ##r = 0##, is it correct to say that ##[ E - ρr/(2ε_0) ]## = 0 is valid too, so that one gets ##E = ρr/(2ε_0)##? Or is it more mathematically sound to conclude that, assuming ##r = 0##, then ##[ E - ρr/(2ε_0) ]~\neq~0## so that ##E~\neq~ρr/(2ε_0)~##, and that therefore, one cannot say that ##E = 0~##when ##r = 0##, something that one gets only when ##[ E - ρr/(2ε_0) ] = 0?## I'm sorry if my questions are somewhat confusing. I don't know how to ask them in a better way. I'm not a native English language speaker.
Perhaps you missed the relevance of the opening statement in my compact proof, so I will point it out, "We construct a cylindrical Gaussian surface of radius ##r## that is coaxial with the cylinder." If the radius of this surface were exactly equal to zero, it would not be a surface but an infinitely thin straight line and would not qualify as a closed Gaussian surface enclosing volume ##V##. Thus, ##r\neq 0## and it is safe to divide both sides of the equation by ##r##.
 
  • #10
PeroK
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... the electric field at the centre of the cylinder must be ##0## by symmetry. If it were non-zero, in which direction would it point?

And if that observation saves a page of meandering and messy algebra, then all the better. For the instructor and the student.
 
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  • #11
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@Delta2 ... Thanks for your comments.
I am a mathematician, and I think you are partially correct on this comment, that is for ##r\neq 0## it is ##E=\rho r/2\epsilon_0## but for ##r=0## we cant tell what is happening, all we can say is that $$\lim_{r\to 0}E=0$$
But if ##~E=\rho r/2\epsilon_0~## for ##~r\neq 0~##, how did you get ##~\lim_{r\to 0}E=0~## for ##~r = 0~?~## Which expression for ##E~## did you use in evaluating the limit?
 
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@Delta2 ... Thanks for your comments.

But if ##~E=\rho r/2\epsilon_0~## for ##~r\neq 0~##, how did you get ##~\lim_{r\to 0}E=0~## for ##~r = 0~?~## Which expression for ##E~## did you use in evaluating the limit?
This expression:##~E=\rho r/2\epsilon_0~##
 
  • #13
Delta2
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Yes as @PeroK said, when evaluating a limit we don't care what happens at exactly the point of the limit (here we don't care what happens for r=0, in order to evaluate the ##\lim_{r\to 0}E##. Only case we care of course is if we want to prove that the function is continuous.
 
  • #14
PeroK
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Yes as @PeroK said, when evaluating a limit we don't care what happens at exactly the point of the limit (here we don't care what happens for r=0, in order to evaluate the ##\lim_{r\to 0}E##. Only case we care of course is if we want to prove that the function is continuous.
In this case, an assumption of the continuity of ##E## at ##r = 0## would also lead to ##E(r = 0) = 0##.
 
  • #15
Delta2
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In this case, an assumption of the continuity of ##E## at ##r = 0## would also lead to ##E(r = 0) = 0##.
Yes I was thinking along the same lines but cant fully justify that E has to be continuous at r=0. I know usually discontinuity of the E-field is where we have discontinuity of the charge density but I don't know if this is a sound argument.
 
  • #16
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... when evaluating a limit we don't care what happens at exactly the point of the limit (here we don't care what happens for r=0, in order to evaluate the ##\lim_{r\to 0}E##. Only case we care of course is if we want to prove that the function is continuous.
So, if one knows what happens in the neighborhood of the point in question, then one can make conclusions about the limit at exactly that point? Then, if one wants to talk next about the limit of the magnitude of the electric field ##~E~## at a point, one also has to examine the continuity of ##~E~## at that point by making some assumptions. So how does one make these assumptions? Where does one get them? Can one base these assumptions on what the neighboring points around the point in question do? If yes, then knowing that ##~\lim_{r\to 0}E~## in the neighborhood of ##~r = 0~## can be used to justify the continuity of ##~E~## there and come to the conclusion that ##~E = 0~## at ##~r = 0~##, just like what PeroK did in #14.
 
  • #17
Delta2
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So, if one knows what happens in the neighborhood of the point in question, then one can make conclusions about the limit at exactly that point? Then, if one wants to talk next about the limit of the magnitude of the electric field ##~E~## at a point, one also has to examine the continuity of ##~E~## at that point by making some assumptions. So how does one make these assumptions? Where does one get them? Can one base these assumptions on what the neighboring points around the point in question do? If yes, then knowing that ##~\lim_{r\to 0}E~## in the neighborhood of ##~r = 0~## can be used to justify the continuity of ##~E~## there and come to the conclusion that ##~E = 0~## at ##~r = 0~##, just like what PeroK did in #14.
I don't know what assumption we should made to ensure that the E-field is continuous at a point. All I know is that if the charge density has some discontinuity then the E-field also has discontinuity at that point, but I am not sure about the inverse.
 
  • #18
kuruman
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@Delta2 ... Thanks for your comments.

But if ##~E=\rho r/2\epsilon_0~## for ##~r\neq 0~##, how did you get ##~\lim_{r\to 0}E=0~## for ##~r = 0~?~## Which expression for ##E~## did you use in evaluating the limit?
If someone gave you the expression $$y=a~x$$ and asked you "what is the value of ##y## at ##x=0## would you answer "zero" or would you start talking about the "continuity of ##y## in the neighborhood of ##x=0##" and the "##~\lim_{x \rightarrow 0} {y}~##"? The relation ##E=\frac{\rho}{2\epsilon_0}r## is a linear relation just like ##y=ax## or ##V=IR##. In this equation, ##E## is a positive scalar quantity such that ##E \geq 0## and should be treated as such.

The vector field ##\vec E## is continuous. The plot below shows the dependence of the magnitude ##E## as a function of the radial distance from the axis. There is no discontinuity anywhere.

EPlot.png


Alternatively, you can plot the Cartesian x-component (see below). Likewise, there is no discontinuity anywhere.
E_x.png
 
  • #19
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@kuruman ... I know fully well what open surfaces and closed surfaces are and how they differ from lines. I also happen to know what coaxial cylinders are. I wasn't talking about reducing the ##~r~## of the imaginary Gaussian surface ##~\rm S~## to zero in post #7. I was referring to the range of values ##~0\leq r\lt R~##. Again, the equations obtained from Gauss's law
##~~E2πrL = ρπr^2/(2ε_0)~\Rightarrow~2πrL[ E - ρr/(2ε_0) ] = 0~\Rightarrow~r[ E - ρr/(2ε_0) ] = 0##
are valid for any point on ##~\rm S~## where ##~0\leq r\lt R~## so that, strictly speaking, you cannot divide the last equation by ##~r~## to get ##~[ E - ρr/(2ε_0) ] = 0.~##I think it should remain $$ r[ E - ρr/(2ε_0) ] = 0~\Rightarrow\begin{cases}\begin{align} & r = 0 ~\rm {but}~[ E - ρr/(2ε_0) ] \neq 0
~,\rm {or}\dots\nonumber \\ & r\neq 0~\rm {but}~[ E - ρr/(2ε_0) ] = 0\dots\nonumber \end{align}\end{cases} $$ According to Delta2 in post #8:
I am a mathematician, and I think you are partially correct ... that is for ##r\neq 0## it is ##E=\rho r/2\epsilon_0## but for ##r=0## we cant tell what is happening, all we can say is that ##\lim_{r\to 0}E=0~.##
Therefore, the only valid conclusion that one can make from Gauss's law concerning what we're talking about is to say that ##~E = ρr/(2ε_0)~## when ##~r\neq 0~.## The conclusion regarding ##~E~## when ##~r = 0~## is left hanging. The graphs that you're showing me now makes sense only to those who accept the result of Gauss's law obtained by dividing by zero an expression that does not include zero in its domain. I'm not saying that physics is wrong. I know that what you're arguing for is already an old established fact in physics. What I'm looking for is a justification of what you want me to accept that is not obtained by an illegal procedure in mathematics like dividing by zero. What I want is something like the idea proposed by PeroK in post #14.
In this case, an assumption of the continuity of ##E## at ##r = 0## would also lead to ##E(r = 0) = 0##.
I thought of something in post #16:
So, if one knows what happens in the neighborhood of the point in question, then one can make conclusions about the limit at exactly that point? Then, if one wants to talk next about the limit of the magnitude of the electric field ##~E~## at a point, one also has to examine the continuity of ##~E~## at that point by making some assumptions. So how does one make these assumptions? Where does one get them? Can one base these assumptions on what the neighboring points around the point in question do? If yes, then knowing that ##~\lim_{r\to 0}E~## in the neighborhood of ##~r = 0~## can be used to justify the continuity of ##~E~## there and come to the conclusion that ##~E = 0~## at ##~r = 0~##, just like what PeroK did in #14.
and asked Delta2 about it in post #17 but his reply was
I don't know what assumption we should made to ensure that the E-field is continuous at a point. All I know is that if the charge density has some discontinuity then the E-field also has discontinuity at that point, but I am not sure about the inverse.
so that nothing more came out of it and my dissatisfaction regarding the present state of affairs continues.
 
  • #20
kuruman
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Again, the equations obtained from Gauss's law
##~~E2πrL = ρπr^2/(2ε_0)~\Rightarrow~2πrL[ E - ρr/(2ε_0) ] = 0~\Rightarrow~r[ E - ρr/(2ε_0) ] = 0##
are valid for any point on ##~\rm S~## where ##~0\leq r\lt R~## so that, strictly speaking, you cannot divide the last equation by ##~r~## to get ##~[ E - ρr/(2ε_0) ] = 0.~##I think it should remain $$ r[ E - ρr/(2ε_0) ] = 0~\Rightarrow\begin{cases}\begin{align} & r = 0 ~\rm {but}~[ E - ρr/(2ε_0) ] \neq 0
~,\rm {or}\dots\nonumber \\ & r\neq 0~\rm {but}~[ E - ρr/(2ε_0) ] = 0\dots\nonumber \end{align}\end{cases} $$
Look at the first case. You say that if ##r=0##, ##[ E - ρr/(2ε_0) ] \neq 0##. Fair enough. What could ##[ E - ρr/(2ε_0) ]## conceivably be? First of all, we have already accepted that ##r=0## therefore ##[ E - ρr/(2ε_0) ]## reduces to ##E##. So we have ##rE=0## with ##r=0## which mathematically says that ##E## could be one of two choices: (a) is non-zero and has the value ##E_0## at ##r=0##; (b) ##E## is also zero at ##r=0##. This is as far as the math will take you.

At this point, we set the math aside and consider the physical situation. Construct a Gaussian surface whose radial boundaries are ##a \leq r \leq b## (##a,~b~<R##) that has thickness ##\Delta z## and subtends angle ##\Delta \phi## with the ##z-##axis. The volume enclosed by this surface is $$V=\frac{1}{2}(b^2-a^2)\Delta \phi \Delta z~~ \text{and the enclosed charge is }q_{\text{enc}}=\frac{1}{2}\rho(b^2-a^2)\Delta \phi \Delta z.$$The faces perpendicular to the electric field have areas ##E_a=a\Delta \phi \Delta z## and ##E_b=b\Delta \phi \Delta z##.

Now let's make the assumption that the radial field at ##r=0## is non-zero and has the value ##E_r=E_0##. By superposition, the field inside is ##E_r=E_0+\dfrac{\rho~ r}{2\epsilon_0}##. The corresponding fluxes are
##\Phi_a=-(E_0+\dfrac{\rho~ a}{2\epsilon_0})(a\Delta \phi \Delta z)##
##\Phi_b=(E_0+\dfrac{\rho~ b}{2\epsilon_0})(b\Delta \phi \Delta z)##
and
##\Phi_{net}=\Phi_a+\Phi_b=\left[E_0(b-a)+\dfrac{1}{2\epsilon_0}\rho(b^2-a^2)\right]\Delta \phi \Delta z.## Gauss's law says ##\Phi_{net}=\dfrac{q_{enc}}{\epsilon_0}.## Substituting from above the following must be true $$\left[E_0(b-a)+\frac{1}{2\epsilon_0}\rho(b^2-a^2)\right]\Delta \phi \Delta z=\frac{1}{\epsilon_0}\times \frac{1}{2}\rho(b^2-a^2)\Delta \phi \Delta z.$$It will be true only if ##E_0(b-a)=0##. By construction of the Gaussian surface, ##b>a## which means that (##b-a##) cannot be zero. Therefore ##E_0## must be zero. In other words, if we assume that the field does not vanish on the axis, Gauss's law is violated. Conversely, if you find an expression for the field using Gauss's law, you cannot superpose another field to what you found without violating the law.
 
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  • #21
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I'm sorry but I already disagree with you right at the beginning of your arguments.
$$\dots~{\rm {assumption~1:}} ~r = 0~ {\rm with} ~[E - ρr/(2ε_0)] \neq0~\Rightarrow~E\neq0~\dots$$ $$\dots~{\rm {assumption~2:}}~rE = 0~\Rightarrow~\begin{cases}\begin{align} & ~(i)~r = 0~{\rm {but}}~E\neq0~,~{\rm {or}} ~\dots\nonumber \\ & ~(ii)~r\neq0~{\rm {but}}~E = 0~\dots \nonumber\end{align}\end{cases}$$If your assumption 2 follows from assumption 1, then I don't think that ##~(ii)~## is a valid conclusion.
 
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  • #22
kuruman
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I'm sorry but I already disagree with you right at the beginning of your arguments.
$$\dots~{\rm {assumption~1:}} ~r = 0~ {\rm with} ~[E - ρr/(2ε_0)] \neq0~\Rightarrow~E\neq0~\dots$$ $$\dots~{\rm {assumption~2:}}~rE = 0~\Rightarrow~\begin{cases}\begin{align} & ~(i)~r = 0~{\rm {but}}~E\neq0~,~{\rm {or}} ~\dots\nonumber \\ & ~(ii)~r\neq0~{\rm {but}}~E = 0~\dots \nonumber\end{align}\end{cases}\nonumber$$If your assumption 2 follows from assumption 1, then I don't think that ##~(ii)~## is a valid conclusion.
Perhaps you did not understand the significance of what I posted in #20. Starting with ##rE=0##, I considered case (i) ##r=0## and ##E=E_0\neq 0## at ##r=0##. I then showed that if that is true, Gauss's law is violated. At this point one is faced with two options:$$\text{Choose}~\begin{cases}\begin{align} & ~(i)~r = 0~{\rm {and}}~E\neq0~\text{and violate Gauss's law.}~ \nonumber \\ & ~(ii)~r\neq0~{\rm {and}}~E = 0~\text{in accordance with Gauss's law.} \nonumber\end{align}\end{cases}$$ I chose to uphold the law, you chose otherwise. Watch out for the Gauss's law police.
 
  • #23
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In the second part of your arguments, the one with (math+physics), I don't get this:
... has thickness ##\Delta z## ... subtends angle ##\Delta \phi## with ##z-##axis ... volume enclosed by this surface is ##V= (1/2)(b^2-a^2)\Delta \phi \Delta z~~ \text{and the enclosed charge is }q_{\text{enc}}=(1/2)\rho(b^2-a^2)\Delta \phi \Delta z## ...
Isn't ##~\Delta \phi~## on the plane ⊥ to the ##~z~## axis subtending an ∠ with the ##~x~## and ##~y~## axes? And why do you have the factor 1/2 in the volume enclosed? For an infinitesimal volume, its simply ##~V##= length × width × thickness without an obvious need for a constant factor. Then you assumed a nonzero field ##E_0## at the center of the cylinder so that the total field ##E_{\rm T}## at any points away from the cylinder axis becomes ##E_{\rm T}## = ## E_0 + E_r~##=##~E_0 + ρr/(2ε_0)##. Then you took the flux through the vertical sections of the lateral side of the volume enclosed ##~V_{\rm {enc}}~##. Regarding that volume, with inward flux ##~\Phi_a\gt0~## and outward flux ##~\Phi_b\lt0~## having opposite signs to what you did, one should get the total flux$$\begin{align}\Phi_{\rm T} = \Phi_a + \Phi_b & = [E_0 +ρa/(2ε_0)]a\Delta\phi\Delta z - [E_0 +ρb/(2ε_0)]b\Delta\phi\Delta z~\dots\nonumber \\ & = [E_0(a - b) - (b^2 - a^2)/(2ε_0)]\Delta\phi\Delta z~\dots \nonumber\ \\ & = [ρ(b^2 - a^2)/ε_0]\Delta\phi\Delta z~\dots \nonumber\end{align}$$ $$\Rightarrow~E_0 - ρ(b + a)/(2ε_0) = ρ(b + a)/ε_0~\Rightarrow~E_0 = 3ρ(b + a)/(2ε_0)\neq0~\dots$$ At any rate, that did not allow me to successfully apply the reductio ad absordum that you used to show that ##~\vec E = 0~## at the center of the cylinder.
So, after finally understanding the second part of your argument, I have these two misgivings: ##(i)~##the use of the sign convention for the flux of ##V_{\rm {enc}}## that is opposite to what I would normally use, and ##~(ii)~##the factor (1/2) in your expression for ##~V_{\rm {enc}}~##. I wouldn't have used the area of a triangle for infinitesimal area. But then, why not? You were able to remove the annoying term by using that area formula. Anyway, I'm somewhat satisfied now. Thanks for your patience and mathematical-physical insights. But up to now, I still don't know how pure mathematics could help in removing that obvious contradiction.
Oh, that reminds me. I hope the Gauss's Law Polizei won't look for me anymore.
gau_cyl5.jpg
 
  • #24
kuruman
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Thanks to your nice drawing I can answer your questions. First, I use the convention ##\Delta \phi## to convey the idea of a finite difference, e.g. (60o-43o) as opposed to an infinitesimal difference that I would write as ##d\phi##. The area of the sector on top that has radius ##a## is a fraction of a circle of radius ##a##. That fraction is proportional to the subtended angle: ##f=\dfrac{\Delta \phi}{2\pi}## which becomes 1 when ##\Delta \phi=2\pi##. Makes sense, no? Thus, the area of that sector is $$A_{\text{top,a}}=f \pi a^2=\frac{\Delta \phi}{2\pi}\pi~a^2=\frac{1}{2}(\Delta \phi) a^2.$$Simlarly, the area of the larger sector is ##A_{\text{top,b}}=\frac{1}{2}(\Delta \phi) b^2.## Thus, the area of the top face of the Gaussian surface is the difference, ##A_{\text{top}}=A_{\text{top,b}}-A_{\text{top,a}}=\frac{1}{2}(\Delta \phi) (b^2-a^2).##

Now for the net flux. The flux through the 4 flat faces is zero because the electric field has no component perpendicular to them. The flux through the front face is negative because the outward normal is in the negative radial direction, toward to axis whilst the field is away from the axis. The curved area at ##a## is ##A_a=a~\Delta\phi \Delta z.## Thus, the total flux through the front curved area is $$\Phi_a=-\left(E_0+\dfrac{\rho~ a}{2\epsilon_0}\right)(a\Delta \phi \Delta z)=\left(-E_0a-\dfrac{\rho~ a^2}{2\epsilon_0}\right)(\Delta \phi \Delta z)$$ The flux through the front curved piece is positive, $$\Phi_b=\left(E_0+\dfrac{\rho~ b}{2\epsilon_0}\right)(b\Delta \phi \Delta z)=\left(E_0 b+\dfrac{\rho~ b^2}{2\epsilon_0}\right)(\Delta \phi \Delta z)$$When you add the two, you get,$$\Phi_{\text{net}}=\left[E_0 b+\dfrac{\rho~ b^2}{2\epsilon_0}+(-E_0a-\dfrac{\rho~ a^2}{2\epsilon_0})\right](\Delta \phi \Delta z)=\left[E_0(b-a)+\frac{1}{2\epsilon_0}\rho(b^2-a^2)\right](\Delta \phi \Delta z).$$I hope that, with this, you will be completely satisfied. We aim to please.
 
  • #25
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I don't understand this part of your last explanations:
... the area of the sector on top that has radius ##a## is a fraction of a circle of radius ##a##. That fraction is proportional to the subtended angle: ##f=\dfrac{\Delta \phi}{2\pi}## which becomes 1 when ##\Delta \phi=2\pi##. Makes sense, no? Thus, the area of that sector is $$A_{\text{top,a}}=f \pi a^2=\frac{\Delta \phi}{2\pi}\pi~a^2=\frac{1}{2}(\Delta \phi) a^2.$$Simlarly, the area of the larger sector is ##A_{\text{top,b}}=\frac{1}{2}(\Delta \phi) b^2.## Thus, the area of the top face of the Gaussian surface is the difference, ##A_{\text{top}}=A_{\text{top,b}}-A_{\text{top,a}}=\frac{1}{2}(\Delta \phi) (b^2-a^2)## ...
Why is the fraction ##~f~## not ##~f\neq a^2\Delta \phi/(πa^2)?## You used one that is half as small that I think it should be. But if one uses the formula for the area of a triangle ½×Base×Height, which I already mentioned in post #23, then one gets also $$A_{\text{top,a}}=½(a\Delta \phi)a = ½(a^2\Delta \phi)~$$like you did. I think its OK to do that for an infinitesimal area element. All the rest will come out similar to what you got.
I was so focused looking at the flux direction, thinking which is coming in and going out of ##V_{\rm {enc}}## that I totally forgot the more important relation between ##\vec E## and the outward normal ##\hat {\rm n}~.##
Yes, I think I can say I'm fully satisfied now except for the purely math part. I'm really thankful for your patience and efforts, and the kindness of Physics Forums staff for enabling me to continue learning even after the normal years of basic education. I'm glad that you liked my drawing. Here is another one similar to the first. I enjoy making geometric drawings using the computer.
gau_cyl_sec2a.jpg
 

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