Dielectric problem -- two line charges inside a dielectric cylinder

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Homework Statement


Consider a cylindrical hole of radius a and infinite length cut into a dielectric medium with relative electric permittivity ε (the interior can be treated as a vacuum). Inside the hole there are two line charges of infinite length with line charge densities λ and −λ, respectively. These line charges are arranged parallel to the z−axis, but displaced from it by an amount ±d/2 along the direction φ = 0 (where d ≪ a).

Show that the electrostatic potential due to the line charges in the range d ≪ r ≪ a (i.e. ignoring the effects of the dielectric medium outside) is given to lowest order by

V(r,φ)≃ ##\frac{λ dcosφ}{2πε_0 r}##

Determine the electrostatic potential everywhere for r ≫ d.


Homework Equations




The Attempt at a Solution


for the first part
##\int{E\dot dS}=\frac{Q}{\epsilon_0}##
##E=\frac{\lambda}{2\pi \epsilon_0 r'}##
##V=\int{E\dot dr'}##
##V=\frac{\lambda}{2\pi \epsilon_0}\ln{r'}##

##V_{total}=\frac{\lambda}{2\pi\epsilon_0}[\ln{|\vec{r}-\frac{d}{2}\vec{x}}-\ln{|\vec{r}+\frac{d}{2}\vec{x}}]##
##|\vec{r}+\frac{d}{2}\vec{x}| ~r+\frac{d}{2}cos\phi##

##V_{total}=\frac{\lambda}{2\pi \epsilon_0}[\ln{r-\frac{d}{2} cos\phi}-\ln{r+\frac{d}{2} cos\phi}] ##
##V_{total}=-\frac{\lambda}{2\pi \epsilon_0}\frac{cos\phi}{r} ##

I don't know why I get a minus here- i assume I've done something wrong.

Im really struggling with the next part of the question. I know the boundary conditions are that the parallel component of E must be continuous and also the perpendicular component of D must be continuous since we have no free charges at the boundary.

Many thanks
 
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Answers and Replies

  • #2
kuruman
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A minus sign is missing from the electric field equation. It should be ##V = -\int{\vec{E} \cdot d\vec{r}}##. Also can you please use the "PREVIEW" button and proofread your equations before posting them? They are hard to read.
 
  • #3
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A minus sign is missing from the electric field equation. It should be ##V = -\int{\vec{E} \cdot d\vec{r}}##. Also can you please use the "PREVIEW" button and proofread your equations before posting them? They are hard to read.
Thank you. I wondered where that minus sign came from. Do you happen to know how to do the next part?
 
  • #4
kuruman
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Thank you for fixing the equations. Your ##V_{total}## is incorrect. Other than the negative sign, there is a ##d## missing in the numerator. I think it's because you did not treat ##r'## correctly. The electric field is

$$ \vec{E}= \frac{\lambda}{2 \pi \epsilon_0} \left[ \frac{(x-d/2)\hat{x}+y\hat{y}}{ [ (x-d/2)^2+y^2]^{1/2} } -\frac{(x+d/2)\hat{x}+y\hat{y}}{ [ (x+d/2)^2+y^2]^{1/2} } \right]$$
If ##r_{\pm} = [ (x \pm d/2)^2+y^2]^{1/2}## and ##r=(x^2+y^2)^{1/2}##, what does ##r_{\pm}## become in the limit ##d << r << a##?

On edit: Define ##x=r \cos \phi,~~y=r \sin \phi## and do a series expansion for ##d/r << 1##.
 
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  • #5
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Thank you for fixing the equations. Your ##V_{total}## is incorrect. Other than the negative sign, there is a ##d## missing in the numerator. I think it's because you did not treat ##r'## correctly. The electric field is

$$ \vec{E}= \frac{\lambda}{2 \pi \epsilon_0} \left[ \frac{(x-d/2)\hat{x}+y\hat{y}}{ [ (x-d/2)^2+y^2]^{1/2} } -\frac{(x+d/2)\hat{x}+y\hat{y}}{ [ (x+d/2)^2+y^2]^{1/2} } \right]$$
If ##r_{\pm} = [ (x \pm d/2)^2+y^2]^{1/2}## and ##r=(x^2+y^2)^{1/2}##, what does ##r_{\pm}## become in the limit ##d << r << a##?

On edit: Define ##x=r \cos \phi,~~y=r \sin \phi## and do a series expansion for ##d/r << 1##.
Thank you you spotting that. That was just a typo. I do get that factor of d with my method
 
  • #6
kuruman
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Can you finish the problem now?
 
  • #7
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Can you finish the problem now?
No, it was the next bit that I struggled with
 
  • #8
kuruman
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Have you studied the Uniqueness Theorem and solutions to Laplace's Equation in cylindrical symmetry? The potential you found is only approximate. One of the boundary conditions is continuity of the potential across the dielectric boundary. You cannot satisfy that without adding another term to the potential. Look for one that becomes very small at r ≈ 0.
 
  • #9
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Have you studied the Uniqueness Theorem and solutions to Laplace's Equation in cylindrical symmetry? The potential you found is only approximate. One of the boundary conditions is continuity of the potential across the dielectric boundary. You cannot satisfy that without adding another term to the potential. Look for one that becomes very small at r ≈ 0.
We know that V-> 0 as r-> ##\infinity## inside the dielectric so the potential here must take the form
##V=\sum{r^{-n}(c_nsin(n\phi)+d_ncos(n\phi)}##

inside the cavity we have
##V=a_0+b_0ln(r)+\sum{(a_nr^n+b_nr^{-n})(c_nsin(n\phi)+d_ncos(n\phi))}##

Am i right in assuming that we can say that V must be finite at r=0 so ##a_0=0##, ##b_0=0## and ##b_n=0## ? If i do this then my solution for V does not match the lowest order approximation
 
  • #10
kuruman
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Backtrack a bit. You still have not derived that ##V(r,\phi)=\frac{λ dcosφ}{2πε_0 r}##. You need to find an approximate expression for the electric field as I indicated in post #4 and do the integral correctly. This ##V_{total}=\frac{\lambda}{2\pi\epsilon_0}[\ln{|\vec{r}-\frac{d}{2}\vec{x}}-\ln{|\vec{r}+\frac{d}{2}\vec{x}}]## is not correct because you got sloppy with the arguments of the logarithms. You need to redo this.

Also note that the expression ##V(r,\phi)=\frac{λ dcosφ}{2πε_0 r}## is an approximate expression for r >>d. It looks to me like the b1 term in the expansion with c1=0. Another point to note is that because the +λ line is to the right of the -λ line, the potential must change sign right to left of the perpendicular bisector between the charge lines. What does this say about the coefficients cn and dn and about the possible values of n?
 

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