Electrostatics problem related to polarization and a cylindrical dielectric

[AHSAN MUJTABA]

Homework Statement

A long cylindrical metal of radius R is filled with a linear dielectric with permittivity $\epsilon.$ A line charge with a constant $\lambda$ is placed inside the dielectric such that the line is parallel to the axis of symmetry of the cylinder (that is really the only way to place the line inside the cylinder). Compute the electric field and polarization everywhere. Also compute all relevant charge densities.

Now specialize to the case of the line charge being at the symmetry axis of the cylinder and show that your results are the same as those obtained using Gauss's law.

Relevant Equations

linear polarization equations
Gauss law of displacement field

I understand that the above eqs would be used but I clearly don't know how to use them. I am a bit confussed.

 

[Delta2]

You have to show us your attempt on a solution otherwise we can't help.

The difficulty of this problem (which makes it more interesting i have to say) is that the linear charge density is not necessarily placed at the symmetry axis of the cylinder, a fact which spoils up the symmetry of the problem and makes things a bit harder to calculate.

So let's start backwards and first view the simple case where the linear charge density is placed exactly on the symmetry axis of the cylinder. How would you then solve for the electric field E, inside the dielectric and on the surface of the cylinder, the Polarization field P in the dielectric, and the surface charge density on the surface of the cylinder?

 

[AHSAN MUJTABA]

Ok,
1) I assume that the cylinder radius is $R$ and the dielectric inside it has radius $a$.
2) Now, for inside I can simply write the field as $E=k int (lambda)(dl)/epsilon a^2$

 

[AHSAN MUJTABA]

I think the configuration is sort of a cylindrical capacitor.

 

[Delta2]

Ok,
1) I assume that the cylinder radius is $R$ and the dielectric inside it has radius $a$.
2) Now, for inside I can simply write the field as $E=k int (lambda)(dl)/epsilon a^2$

The problem statement says that the cylinder is filled with dielectric so the radius of the dielectric is also R. Do you mean that $a$ is the distance from the center of the cylinder at which we wish to evaluate the electric field $E$?
Anyway you got to fix your Latex cause I don't agree to what you write at 2).
Did you want to write $$E=k\int \frac{\lambda dl}{\epsilon a^2}$$?

 

[AHSAN MUJTABA]

No, I meant let the radius of dielectric inside as $a$ and here the confusing thing is that I don't have to use the Gauss's law directly. Also, I need to place some image charge density to calculate the $$E$$ and $$D$$.

 

[AHSAN MUJTABA]

If my linear charge density is at the centre then I can write the potential as
$$V=k\int \frac{\lambda dz} {\sqrt{z^2+r^2-2rz cos\phi}}$$
I hope I am correct in this
where r is $\rho$ by convention in the cylinder and $\phi$ is the azimuthal angle

 

[Delta2]

If my linear charge density is at the centre then I can write the potential as
$$V=k\int \frac{\lambda dz} {\sqrt{z^2+r^2-2rz cos\phi}}$$
I hope I am correct in this
where r is $\rho$ by convention in the cylinder and $\phi$ is the azimuthal angle

I see two problems with this formula:
1) A dependency on the azimuthal angle which is not justified given the symmetry of the problem when the charge density is placed at the center
2) I think the problem wants us to consider the linear charge density (and the cylinder) as infinite in the z-direction so i think this integral diverges to infinity.

No, I meant let the radius of dielectric inside as $a$ and here the confusing thing is that I don't have to use the Gauss's law directly. Also, I need to place some image charge density to calculate the $$E$$ and $$D$$.

Hold on, if you planning on solving this with the method of image charges i am afraid i can't be much of help (i had another approach in my mind). Maybe @vanhees71 can see this and offer us kindly his views on this problem.

 

[vanhees71]

Well, for the infinite line density you must use a more general version of the Helmholtz fundamental theorem. It's just the limiting case, where the integral is applicable. You must only subtract the naive expression at a constant point, i.e.,
$$V(\vec{x})=\frac{1}{4 \pi \epsilon} \int_{\mathbb{R}} \mathrm{d} z' \left [\frac{\lambda}{\sqrt{R^2+(z-z')^2}}-\frac{\lambda}{\sqrt{R_0^2+z^{\prime 2}}} \right].$$
This subtracts the logarithmic divergence. The potential is defined anyway only up to an additive constant.

Of course as almost always a direct solution of the Poisson equation is much easier than the ready-made integral formulas! Just write down the equation
$$-\Delta V=\frac{1}{\epsilon} \rho$$
in cylinder coordinates (using the symmetries of the charge distribution, i.e., non dependence on $\varphi$ nor on $z$). The line-charge can be written as a Dirac-$\delta$ distribution along the cylinder axis, but note that the cylinder coordinates have a coordinate singularity there. So just solve for $V$ away from the cylinder axis and choose the appropriate integration constants such as to get the right singularity along the cylinder axis.

 

[Fred Wright]

I offer some suggestions for solving this problem with respect to the "method of images". First, because the system cylindrical longitudinal symmetry, it can be reduced to two dimensions and we observe the the $\vec E$ field can only have a component normal to the cylindrical surface. That is, it can only have a component in the radial direction.
For the purposes of the method of images, in the diagram below, $q$ is the differential element of charge ($q=\lambda dz$) and $q'$ is the image charge. We take the cylinder to be grounded, that is $V=0$ on its surface. Let the potential due to $q$ be $V(r,\theta)$ and the potential of the image charge, $q'$, be $V'(r,\theta)$ where $r$ is the radial distance from the center of the circle. At any point on the interior the total potential $V_T=V+V'$. In order to find the unknown quantities $X'$ and $q'$ evaluate $V_T$ at point $P$ ($\theta =0$) and point $Q$ ($\theta =\pi$). You will get two equations for the two unknowns. Also, remember that $E_r = -\frac{\partial V_T}{\partial r}$

 

[AHSAN MUJTABA]

I think inside the integral can I take $r cos\phi = z$ because in my integral all the terms inside the root in the denominator have X and R and should our potential depend on z?

 

[AHSAN MUJTABA]

Well, for the infinite line density you must use a more general version of the Helmholtz fundamental theorem. It's just the limiting case, where the integral is applicable. You must only subtract the naive expression at a constant point, i.e.,
$$V(\vec{x})=\frac{1}{4 \pi \epsilon} \int_{\mathbb{R}} \mathrm{d} z' \left [\frac{\lambda}{\sqrt{R^2+(z-z')^2}}-\frac{\lambda}{\sqrt{R_0^2+z^{\prime 2}}} \right].$$
This subtracts the logarithmic divergence. The potential is defined anyway only up to an additive constant.

Of course as almost always a direct solution of the Poisson equation is much easier than the ready-made integral formulas! Just write down the equation
$$-\Delta V=\frac{1}{\epsilon} \rho$$
in cylinder coordinates (using the symmetries of the charge distribution, i.e., non dependence on $\varphi$ nor on $z$). The line-charge can be written as a Dirac-$\delta$ distribution along the cylinder axis, but note that the cylinder coordinates have a coordinate singularity there. So just solve for $V$ away from the cylinder axis and choose the appropriate integration constants such as to get the right singularity along the cylinder axis.

here my line is parallel to z axis. Does this not means that my Radial coordinate would change instead of z?

 

[vanhees71]

Sure, my expressions are for the most simple case, where the line charge is along the cylinder axis. With the charge along an axis parallel to the cylinder axis you need to fulfill the more complicated boundary conditions using the hints in #10.

 

[AHSAN MUJTABA]

In Jackson they mention that we have a famous potential due to line charge that is :
$\frac 1{4\pi\epsilon_o} ln\frac{R^2}{r^2}$
Can I use it directly?

 

[vanhees71]

Yes, sure (of course with an $\epsilon=\epsilon_0 \epsilon_r$ inside the dielectric cylinder and $\epsilon=\epsilon_0$ outside). You can derive this potential (which is for a unit-line-charge-density along the $z$ axis) most easily by writing down the Laplace equation in cylinder coordinates for a potential $\Phi=\Phi(r)$ (where $r=\sqrt{x_1^2+x_2^2}$, $x_1=r \cos \varphi$, $x_2=r \sin \varphi$, and $x_3=z$ is the definition of the cylinder coordinates $(r,\varphi,z)$) and the ansatz
$$\Phi(\vec{x})=\Phi(r).$$
Using the Laplace operator in cylinder coordinates gives
$$\Delta \Phi(r)=\frac{1}{r} \partial_r (r \partial_r \Phi)=0.$$
This gives
$$r \partial_r \Phi=A \; \Rightarrow \; \Phi=A \ln(r/R),$$
where $R>0$ is an arbitrary constant.

To get $A$ use Gauss's Law for a cylinder $Z$ of height $h$ and radius $a$ around the $z$ axis,
$$\int_{\partial Z} \mathrm{d}^2 \vec{f} \cdot \vec{\nabla} \Phi=-\frac{1}{\epsilon} \lambda h=\int_0^{2 \pi} \mathrm{d} \varphi \int_0^h \mathrm{d} z a \frac{A}{a} = 2 \pi A h \; \Rightarrow \; A=-\frac{\lambda}{2 \pi}.$$
So the potential for a line charge $\lambda$ along the $z$ axis is
$$\Phi(r)=-\frac{\lambda}{2 \pi \epsilon} \ln(r/R)=-\frac{\lambda}{4 \pi \epsilon} \ln(r/R).$$
For the original problem you need a line charge along a line parallel to the $z$ axis (take it at $x_1=a$ and $x_2=0$). I think it's easier to switch back to Cartesian coordinates and to use the method of image charges suggested in #10. Denote the potential within the cylinder with $\Phi_{<}$ and the potential outside with $\Phi_{>}$. Then the ansatz with the image-charge-line densities should be
$$\Phi_{<}(\vec{x})=-\frac{\lambda}{4 \pi \epsilon_0 \epsilon_r} \ln \left (\frac{(x_1-a)^2+x_2^2}{R} \right) - \frac{\lambda'}{4 \pi \epsilon_0 \epsilon_r} \ln \left (\frac{(x_1-a')^2+x_2^2}{R} \right)$$
and
$$\Phi_{>}(\vec{x}) = -\frac{\lambda''}{4 \pi \epsilon_0} \ln \left (\frac{(x_1-a)^2+x_2^2}{R} \right).$$
Now plug in the cylinder coordinates again (everything now depends only on $r$ and $\varphi$, not $z$ due to translation invariance) and solve for $\lambda'$, $\lambda''$, and $a'$ using the boundary conditions
$$\Phi_{<}(R,\varphi)=\Phi_{>}(R,\varphi), \quad \epsilon_0 \epsilon_r \partial_r \Phi_{<}(R,\varphi)=\epsilon_0 \partial_r \Phi_{>}(R,\varphi).$$

 

[Delta2]

@vanhees71 i don't understand how you introduce constant $R$ in post #15, the differential equation is $$\partial_r\Phi=\frac{A}{r}$$ which the way i see it has the general solution $$\Phi=A\ln r+C$$ where C a constant

 

[vanhees71]

This you NEVER ever should do. There is a dimensionful quantity as the argument of the logarithm in your solution. That's not even defined. That's why you should introduce the 2nd integration constant rather in the way with some arbitrary constant $R$ of the dimension of a length. This you can achieve by formaly setting $C=-A \ln R$ and then combining the logs to $A \ln(r/R)$, which is dimensionally correct. Note that $\partial_r \ln (r/R)=1/r$ for any $R>0$.

 

[Fred Wright]

$$\Delta \Phi(r)=\frac{1}{r} \partial_r (r \partial_r \Phi)=0.$$

This equation holds if you have longitudinal isotropic symmetry. However, if the symmetry is broken, as it is with the off axis line of charge, I maintain that the Laplacian must be
$$
\Delta \Phi(r,\theta)=\frac{1}{r} \frac{\partial}{\partial r}(r \frac{\partial \Phi}{\partial r})+\frac{1}{r^2}\frac{\partial^2 \Phi}{\partial \theta^2}=0
$$

 

[vanhees71]

Yes, but for the problem at hand the image-charge method is much simpler.

 

[Javed]

I offer some suggestions for solving this problem with respect to the "method of images". First, because the system cylindrical longitudinal symmetry, it can be reduced to two dimensions and we observe the the $\vec E$ field can only have a component normal to the cylindrical surface. That is, it can only have a component in the radial direction.
For the purposes of the method of images, in the diagram below, $q$ is the differential element of charge ($q=\lambda dz$) and $q'$ is the image charge. We take the cylinder to be grounded, that is $V=0$ on its surface. Let the potential due to $q$ be $V(r,\theta)$ and the potential of the image charge, $q'$, be $V'(r,\theta)$ where $r$ is the radial distance from the center of the circle. At any point on the interior the total potential $V_T=V+V'$. In order to find the unknown quantities $X'$ and $q'$ evaluate $V_T$ at point $P$ ($\theta =0$) and point $Q$ ($\theta =\pi$). You will get two equations for the two unknowns. Also, remember that $E_r = -\frac{\partial V_T}{\partial r}$
View attachment 271872

How would the potentials show the effect of the dielectric through the method of images?

 

[vanhees71]

Just consider #15. You'll see that outside the dielectric cylinder you have a potential looking as one where there's only a uniform line charge along the axis where the true line charge is present, but the effective line charge $\lambda''$ is different from the free line charge $\lambda$ due to the contribution from the polarization of the cylinder. Inside the cylinder you have the superposition of the true line charge and the image-line-charge located outside the cylinder.