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Electric Displacement (Gauss's Law)

  1. Apr 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Two long, conducting cylinders of thin metal with radii ##R## and ##5R##, respectively, are arranged concentrically so that their axes coincide. The smaller cylinder is placed inside the larger one and the space between them is filled with a dielectric whose relative permittivity ##\epsilon_r## increases with radius in a linear fashion:

    ##\epsilon_r (\rho) = \alpha \frac{\rho}{R}##​

    The potential difference between the cylinders is ##V_0## volts.

    Use the differential form of Gauss's law to show that the electric displacement D between the plates is:

    ##D_\rho (\rho) = \frac{C_0}{\rho_0}##​

    Where ##C_0## is some arbitrary constant.

    2. Relevant equations

    ##\nabla . D = \rho_{free}##​

    Here ##\rho## is the charge density (not radius as defined in the question).

    In cylindrical coordinates divergence is:

    ##\nabla . v = \frac{1}{\rho} \frac{\partial}{\partial \rho} (\rho v_\rho) + \frac{1}{\rho} \frac{\partial v_\phi}{\partial \phi} + \frac{\partial v_z}{\partial z}##​

    Polarization is: ##P=\epsilon_0 \chi_e E##

    Also ##\epsilon_r = \frac{\epsilon}{\epsilon_0}##.

    3. The attempt at a solution

    To avoid confusion, for radius I will use r instead of ##\rho##. In the cylindrical coordinates I took only the first term (since we only have a radial dependence):

    ##\nabla . D = \frac{1}{r} \frac{\partial}{\partial r} (r D_r) = \rho_{free}##​

    I can then solve the partial differential equation by integrating between R and 5R. But how can I find the charge density ##\rho## when I'm only given ##\epsilon_r##? :confused:

    (I've found that ##\chi_e = \epsilon_r - 1 = \frac{\alpha r}{R} - 1## but it doesn't seem to help.)

    So how can we relate the relative permittivity to charge density? Or is there another way for finding ##\rho##?

    Any help would be greatly appreciated.
     
  2. jcsd
  3. Apr 11, 2015 #2
    I've tried to use the relationship ##D=\epsilon E## by finding E first. Setting the free charge density to zero (not sure if this is right):

    ##\nabla . E = \frac{1}{r} \frac{\partial}{\partial r} (r E_r) = 0##

    ##E_r=\frac{C_0}{r}##​

    And ##\epsilon =\epsilon_r \epsilon_0 = \frac{\alpha r}{R \epsilon_0}##. So:

    ##\therefore D_r=\epsilon E = \frac{\alpha r}{R \epsilon_0} \frac{C_0}{r} = \frac{\alpha C_0}{R \epsilon_0}##​

    But the question says that the answer must be:

    ##D_r(r)=\frac{C_0}{r}##
    What is wrong here? How come they did not get the ##\alpha## and ##\epsilon_0## terms?
     
  4. Apr 11, 2015 #3
    I think you are on the right way, [itex]\bigtriangledown(\epsilon\vec{E})=0[/itex] because all of the charges between the cylinders are non-free. Then, I would use the formula [itex]\bigtriangledown(\epsilon\vec{E})=\bigtriangledown\epsilon\vec{E}+\epsilon div\vec{E}[/itex] and the relation between electric field and potential which will give you an ODE with respect to potential. The boundary conditions for this ODE are given in the task.
     
  5. Apr 12, 2015 #4
    Thank you very much for the response.

    But how does the potential ##\phi## factor into the calculation? And why should we being the potential into the calculation when the final answer does not include the potential (i.e. D(r)=C0/r)?

    I've used Poisson's equation as follows:

    ##\nabla^2 \phi = \frac{1}{r} \frac{\partial}{\partial r} (r \frac{\partial \phi}{\partial r}) = \frac{-\rho }{\epsilon_0}##​

    And using the relation ##E=-\nabla \phi##:

    ##\nabla \phi = \frac{\partial \phi}{\partial r} = \frac{C_0}{r} = - E##​

    Again, that's the same answer for E as before.

    And within the boundary condition R to 5R the potential is ##\therefore \phi = C_0 r + C_1 = V_0##.

    Since ##\epsilon = \epsilon_0 \epsilon_r = \frac{\alpha r \epsilon_0}{R}##, we have:

    ##D= E \epsilon = \frac{C_0}{r} \frac{\alpha r \epsilon_0}{R} = \frac{\alpha \epsilon_0 C_0}{R}##​

    But that's not the answer. Why is that? :confused:
     
  6. Apr 12, 2015 #5
    First,
    [itex]\bigtriangledown(\epsilon\vec{E})=\bigtriangledown\epsilon\vec{E}+\epsilon div\vec{E}=-(\bigtriangledown\epsilon \bigtriangledown\phi)-\epsilon \bigtriangleup\phi=0[/itex]
    Then,
    [itex]\bigtriangledown\epsilon=\frac{\alpha}{R}\vec{r_0}[/itex]
    Hence,
    [itex]r\frac{d^2\phi}{dr^2}+\frac{d\phi}{dr}=0[/itex]
     
  7. Apr 13, 2015 #6
    Thank you, but I have some questions.

    I'm not quite sure how you got the expression "##\nabla (\epsilon E) = \nabla \epsilon E + \epsilon (\nabla .E)##". I couldn't find it in my textbook, and according to Maxwell's equations it should simply be: ## \nabla (D) = \nabla (\epsilon E)=\rho_{free}##, right?

    Anyway, here is what I did (I've obtained a slightly different answer from yours):

    ##\nabla .D = \nabla \epsilon \nabla E + \epsilon (\nabla . E) = 0##

    ##\nabla .D= \frac{\alpha r}{R} . \nabla E + \epsilon (\frac{\rho}{\epsilon_0})= 0##​

    Now shouldn't the second term be zero (since ##\nabla . E = \rho/ \epsilon_0 = 0##)? :confused:

    Here's the rest of my working:

    ##\nabla .D = \frac{\alpha r}{R} . \frac{\partial E}{\partial r} = 0##​

    And in terms of potential it becomes:

    ##\nabla . D = - \frac{\alpha r}{R}. \frac{\partial^2 \phi}{\partial r} = 0##​

    What do we do from here? I am a little confused.
     
  8. Apr 13, 2015 #7
    The formula for [itex]\bigtriangledown(\epsilon \vec{E})[/itex] that you are asking about is general for any divergence,it is a pure mathematical thing(you can find it in any math books devoted to vector analysis or even in Wikipedia: http://en.wikipedia.org/wiki/Divergence). On the other hand, it is indeed equal to [itex]\rho_{free}[/itex] which in turn equals zero. That is exactly what I wrote above. Note, that in the first term [itex]\bigtriangledown\epsilon\vec(E)[/itex]the hamilton operator influences epsilon only(i.e. you need to calculate the gradient of epsilon), whereas in the second term [itex]\epsilon div\vec{E}[/itex] it influences the electrical field only(i.e. you need to calculate the divergence of E). Then we replace E with [itex]-\bigtriangledown\phi[/itex].By definition [itex]\bigtriangledown^2\phi[/itex] becomes [itex]\bigtriangleup\phi[/itex](i.e. laplasian of phi) . Divergence of E can not be zero because E is caused by both free and non free charges and there are non-free charges as there is a dielectric between the cylinders. Could you please read carefully about the difference between D and E? This is a subtle thing but if you grasp it, you will completely understand what is going on in this problem. You will also be able to guess which of them must depend on the distance and which can not.
     
    Last edited: Apr 13, 2015
  9. Apr 14, 2015 #8
    Thank you, that makes perfect sense now. I can see where it comes from.

    So basically I need to solve the homogeneous ODE for ##\phi##, and then find E from this potential, and finally use ##D=\epsilon E## to find D?

    I tried to solve the second order ODE:

    ##-\epsilon \nabla^2 \phi -\frac{\alpha r}{R} \nabla \phi =0##​

    Where the Laplacian was ##\nabla^2 \phi = \bigtriangleup \phi##, and ##\epsilon = \frac{\epsilon_0 \alpha r}{R}##. Therefore

    ##-\frac{\alpha r}{R} (\epsilon_0 \nabla^2 \phi + \phi) =0##​

    By solving the characteristic equation the solution would be: ##\phi = C_1 + C_2 e^{-1/\epsilon_0}##. Based on that E would be ##E= -\nabla \phi = - C_2 e^{-1/\epsilon_0}##. But multiplying this by ##\epsilon## does not yield the correct solution. Why is that?
     
  10. Apr 14, 2015 #9
    Yes, your general strategy is right. First,find potential,then E and D. But there is something strange with your equations...what does the gradient of epsilon equal? if [itex]\epsilon=\frac{\epsilon_0\alpha r}{R}[/itex] then [itex]\bigtriangledown\epsilon=\frac{\partial}{\partial r}(\frac{\epsilon_0\alpha r}{R})\vec{r_0}[/itex] where [itex]\vec{r_0}[/itex] is a unity vector in the direction of the radius(Look for the definition of gradient). The gradient of the potential must not transform into potential,it is equal to [itex]\frac{\partial\phi}{\partial r}\vec{r_0}[/itex].And you need to calculate the scalar product of these two gradients which is really easy.
     
  11. Apr 15, 2015 #10
    Oops, sorry I meant by differentiation we find ##\nabla \epsilon = \frac{\epsilon_0 \alpha}{R} \vec{r}##, so

    ##\therefore \nabla (\epsilon E) = - \frac{\epsilon_0 \alpha r}{R} \nabla^2 \phi - \frac{\epsilon_0 \alpha}{R} \nabla \phi##​

    Solving for ##\nabla \phi## we get -1 by the quadratic formula. Multiplying this E by ɛ again I get a wrong answer (D=ɛ0αr/R between the cylinders). Did I make some sort of mistake working out ##\nabla \phi##? :rolleyes:

    I think once we have an expression for E (or ##-\nabla \phi##), we can use V0 to solve for the constant C0. But we need to first show that D(r)=C0/r. Unfortunately I was unable to get that using this strategy so I'm not sure what is wrong here.
     
  12. Apr 15, 2015 #11
    Now your equation seems ok apart from the fact that you have already substituted the equation for the gradient of the potential, so you should replace [itex]\bigtriangledown\phi[/itex] with [itex]\frac{d\phi}{dr}[/itex] otherwise it seems that you add a vector(gradient) to a scalar(laplasian) which makes no sense. On the left side you have divergence which is a scalar. Note, we calculate the gradient of a scalar and get a vector, we calculate the divergence of a vector and get a scalar. The problem is that you can not simply use the quadratic formula to solve the second order differential equation. Laplasian is the second order derivative and you have the first order derivative in the other term. You should use methods appropriate for solving differential equations(Hint:you can introduce a new variable which will be equal to the first derivative of the potential,this will give you the first order differential equation). The correct potential is proportional to ln(r). As concerns the boundary conditions, voltage is by definition is the difference between potentials. The potential in turn is defined up to an additive constant. So, you can assume that the inner cylinder has a zero potential whereas the outer one has the potential which is equal to [itex]V_0[/itex]. Now, I am sure you can derive the correct equation for the potential by yourself.
     
    Last edited: Apr 15, 2015
  13. Apr 17, 2015 #12
    Yes, I introduced a new variable 'x', so for the DE above the auxiliary equation is:

    ##\frac{-\epsilon_0 \alpha r}{R} x^2 - \frac{\epsilon_0 \alpha}{R} x = 0##​

    So to find the roots we must use the quadratic formula:

    ##x= \frac{{-\epsilon_0 \alpha}{R} \sqrt{(\frac{-\epsilon_0 \alpha}{R})^2}}{-2(\frac{\epsilon_0 \alpha r}{R})}##​

    Clearly the two solutions are x= 0 and -1/r. Then the solution to the DE can be written as the function:

    ##\phi = C_0 + C_1 e^{(-1/r)r}##​

    To get ##E=-\nabla \phi## we can differentiate this. However, the solution above can't be correct because the derivative would be zero.

    But since we know ɛ and what D has to be, from the relationship D=ɛE then E must be equal to RC00αr2. Did you also get this solution by solving the DE above? Could you explain how you arrived at this answer?
     
  14. Apr 17, 2015 #13
    As I wrote above, you can not use the quadratic formula to solve a second order differential equation. This is wrong! You need to study math books in order to learn how to solve this equation, there is no physics here.
     
  15. Apr 18, 2015 #14
    My calculus textbook says:

    For a second order differential equation ##ay'' + by'+cy=0##, if r1 and r2 are the roots of the characteristic equation ##ar^2+br+c=0##, then the general solution is ##y=C_1 e^{r_1 x}+ C_2 e^{r_2 x}##.
    Since I had a homogeneous 2nd order DE, I followed exactly the same steps. Aside from the quadratic formula, there was no other way of finding the roots.

    I also tried your method of reducing this to a first order DE by introducing a new variable ##x=\nabla \phi##:

    ##\frac{-\epsilon \alpha r}{R} \nabla x - \frac{\epsilon \alpha}{R} x = 0##
    By separation of variables we have:

    ##\int \frac{1}{x} dx =\int^{5R}_R \frac{-R}{\epsilon \alpha r} \frac{\epsilon \alpha}{R} . dr##
    ##ln (x) = ln (\frac{5R}{R})##

    Unfortunately as you can see it didn't work (please correct me if I am wrong). Are you able to to get the correct ##E## using this method?
     
    Last edited: Apr 18, 2015
  16. Apr 18, 2015 #15
    You should take an indefinite integral(without limits). Thus,[itex]x=\frac{C_1}{r}[/itex],where C1 is a constant(to be determined through the boundary conditions). Then you remember that [itex]x=\bigtriangledown\phi[/itex], so you have a new differential equation:[itex]\frac{d\phi}{dr}=\frac{C_1}{r}[/itex] whose solution is [itex]\phi=C_1ln(r)+C_2[/itex].Now, use the fact that [itex]\phi(R)=0[/itex] and [itex]\phi(5R)=V_0[/itex] to determine the two constants. You can always check that a certain solution is correct by substituting it into the differential equation.
    If you use the characteristic equation, you must remember that its roots are constants(i.e. independent of r).
     
    Last edited: Apr 18, 2015
  17. Apr 20, 2015 #16
    Thank you very much for the guidance. But could you please explain how you obtained ##x=C_1/r## in the beginning?
    Because after separating the variables and rewriting in differential form, once I integrate there are actually two constants of integration:

    ##\frac{1}{x} dx = - \frac{1}{r} dr##, and integrating:

    ##ln (x) + C_0 = - ln (r) + C_1##

    ##ln (x) = - ln (r) + C_1 - C_0 \implies x = \nabla \phi = -r - e^{C_0} + e^{C_1}##
     
  18. Apr 20, 2015 #17
    Yes, you're right, I must have admitted a mistake. Though you can replace [itex]e^C_0+e^C_1[/itex] with just D, where D is an arbitrary constant(because both C0 and C1 are arbitrary constants). Thus, x=-r+D is the solution.
     
  19. Apr 22, 2015 #18
    Thanks. For the constants I'm going to use C instead of D to avoid confusion with electric displacement:

    ##x= \nabla \phi = -r + C##

    ##\phi = \frac{-r^2}{2} + C_0 r + C_1##
    Applying the boundary conditions:

    ##\phi (R) = \frac{-R^2}{2} + C_0 R + C_1 = 0##
    ##\phi (5R) = \frac{-(5R)^2}{2} + C_0 (5R) + C_1 = \frac{-R^2}{2} + C_0 R + C_1 = _0##

    ##\therefore C_0 = \frac{V_0}{4R} + 3R, \ C_1 = -(\frac{5R^2}{2} + \frac{V_0}{4})##​

    Now substituting back:

    ##\phi = \frac{-r^2}{2} + (\frac{V_0}{4R} + 3R)r - (\frac{5R^2}{2}) - \frac{V_0}{4}##

    ##\implies \nabla \phi = -E = -r + \frac{V_0}{4R} + 3R##

    But ##D= \epsilon E = \frac{\epsilon_0 \alpha r}{R} (r - \frac{V_0}{4R} - 3R)##​

    This is wrong since the question says we must have D(r)=C/r. Am I missing something here? :confused:

    Are you sure this is the correct method? Isn't it possible to somehow just use the general solution to Laplace’s equation in cylindrical coordinates, instead of the ##\nabla \epsilon E + \epsilon (\nabla . E)## formula? I'm not sure if this method will work...
     
  20. Apr 26, 2015 #19
    So I've tried a different method:

    ##\nabla .(\epsilon E) = 0 \implies \frac{1}{r} \frac{\partial}{\partial r} (r E_r) = 0 \implies E_r = \frac{C_0}{r}##

    The constant C0 can be found from the voltage condition:

    ##\int^{5R}_R E_r dr = C_0 (ln r)^{5R}_R = V_0##

    ##\therefore \ C_0 = \frac{V_0}{ln (5R/R)} = \frac{V_0}{ln(5)}##

    Electric displacement should be:

    ##D(r)=\epsilon E = \frac{\alpha \epsilon_0 r}{R} \frac{C_0}{r} = \frac{\alpha \epsilon_0 C_0}{R}##

    This is wrong as the answer must be D=C/r. So I tried to find polarization and use the relationship ##D=\epsilon_0 E + P##:

    ##\chi_e = \epsilon_0-1 \implies P = \frac{\epsilon_0 C_0}{R}- \frac{C_0 \epsilon_0}{r}##

    ##\therefore D= \frac{\epsilon_0 C_0}{r} + \frac{\epsilon_0 C_0}{R}- \frac{C_0 \epsilon_0}{r} = \frac{\alpha \epsilon_0 C_0}{R} = \frac{C}{R}##

    I was told for this problem I must start from the general solution to Laplace’s equation in cylindrical form. So, why am I not getting the right answer? Why am I unable to get an ##r## dependence in the equation for ##D(r)##?
     
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