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Point charges and dielectric

  1. Aug 11, 2013 #1
    Hello. I am not sure if I must post here. The problem is not for homework, it derived from a classmate's question.

    1. The problem statement, all variables and given/known data
    We have a large quantity of liquid dielectric with relative permittivity [itex]\varepsilon [/itex] and a charge [itex]Q[/itex] immersed in the liquid in depth [itex]h[/itex]. We bring another charge [itex]q[/itex] in distance [itex]h[/itex] above the surface of the liquid, such as the line connecting [itex]Q[/itex] and [itex]q[/itex] is perpendicular at the surface. The charges are not free to move. We need to find the total force on the second charge [itex]\displaystyle{q}[/itex].

    2. Relevant equations
    • Linear dielectric polarization: [itex]\vec{P}=(\varepsilon -1)\varepsilon _{0}\vec{E_f}[/itex]
    • Bound charges: [itex]\rho _b=-\nabla \cdot \vec{P}[/itex] and [itex]\sigma _b=\vec{P} \cdot \hat{k}[/itex]
    • Coulomb's law

    3. The attempt at a solution
    You can find my attempt at the attachments. Sorry for not posting it in Latex but I already had written the PDF file.
    This year I finished high school and I will enroll to university. I have read alone some of the methods used in my solution so I am not so familiar with them (for example delta function and bound charges). That's the reason I post here and I want to ask if there is anything wrong in my attempt.

    Thanks in advance and sorry for the grammar mistakes here and in PDF (I am not good in English). :shy:
     

    Attached Files:

  2. jcsd
  3. Aug 11, 2013 #2

    TSny

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    Hello, Stealth95 and welcome to PF!

    Congratulations on your progress in studying physics. It is very impressive to see someone just out of high school working on a problem like this!!

    Overall, your handling of the mathematical manipulations looks very good. However, I think there is a mistake in how you wrote one of your starting equations. You wrote

    The electric field in this equation should be the total electric field ##\vec{E}## rather than the electric field due to the free charges alone, ##\vec{E_f}##. So, the equation should read
    $$\vec{P}=(\varepsilon -1)\varepsilon _{0}\vec{E}$$ where ##\vec{E}## is the total field produce by both the free and bound charges in the system. The bound charges will consist of the induced surface charge on the surface of the liquid and the induced charge in the immediate neighborhood of the charge Q immersed in the liquid.
     
    Last edited: Aug 11, 2013
  4. Aug 13, 2013 #3
    Thank you for your answer TSny.
    Unfortunately when I read the equation for linear dielectrics I didn't notice that [itex]\vec{E} [/itex] is the total field, so I made the problem much simpler. And also this mistake destroys the whole solution. :frown:

    Yesterday I tried again with the correct equation:

    Gauss law in region [itex]\displaystyle{L}[/itex] gives:
    [tex]\displaystyle{\varepsilon _0 \nabla \cdot \vec{E}=Q\delta ^3(\vec{r}+\vec{h})-\nabla \cdot \vec{P}\Rightarrow \varepsilon \varepsilon _0 \nabla \cdot \vec{E}=Q\delta ^3(\vec{r}+\vec{h})}[/tex]
    So:
    [tex]\displaystyle{\rho =Q\delta ^3(\vec{r}+\vec{h})-\nabla \cdot \vec{P}=\varepsilon _0\nabla \cdot \vec{E}=\frac{Q}{\varepsilon }\delta ^3(\vec{r}+\vec{h})}[/tex]
    The field of this is:
    [tex]\displaystyle{\vec{E_1}=\frac{Q}{4\pi \varepsilon \varepsilon _0}\int_{L}{\frac{\vec{h}-\vec{r}}{\left|\vec{h}-\vec{r} \right|^3}\delta ^3(\vec{r}+\vec{h})dV}=\frac{Q}{16\pi \varepsilon \varepsilon _0 h^2}\hat{k}}[/tex]
    This answer seems better than the previous one, since it's also the field in the case that the whole space is filled with the dielectric.
    But in our problem we have also [itex]\displaystyle{\vec{E_2}}[/itex] due to the surface charge [itex]\displaystyle{\sigma _b=\vec{P} \cdot \hat{k}}[/itex]. I didn't manage to find this field yet. I attempted to put [itex]\displaystyle{\sigma _b }[/itex] in a derivative using Divergence theorem:
    [tex]\displaystyle{4\pi \varepsilon _0 E_{2z}=\int_{S}{\sigma _b \frac{\vec{h}-\vec{r}}{\left|\vec{h}-\vec{r} \right|^3}\cdot d \vec{S}}=\oint_{\partial L}{\sigma _b \frac{\vec{h}-\vec{r}}{\left|\vec{h}-\vec{r} \right|^3}\cdot d \vec{S}}=...=\int_{L}{\frac{\vec{h}-\vec{r}}{\left|\vec{h}-\vec{r} \right|^3}\cdot \nabla \sigma _b \; dV}}[/tex]
    but this doesn't work because I cannot compute [itex]\displaystyle{\nabla \sigma _b }[/itex] and also I am not really sure if this method is correct.

    Maybe this problem is difficult for now... If I manage to find something better I'll write again.
     
  5. Aug 13, 2013 #4

    rude man

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    Bonehead reaction: F = (1/4
     
  6. Aug 13, 2013 #5

    TSny

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    Your result for ##\vec{E}_1## at the position of q looks good. You can see that the effect of the dielectric is to weaken the field due to Q by a factor of ##1/\varepsilon##.

    To find ##\sigma_b##, you know the relation ##\sigma_b = P_n = \varepsilon_0 (\varepsilon - 1)E_n## where ##E_n## is the normal component of the total electric field just inside the liquid at the surface. The total field is the superposition of the fields due to Q, q, and ##\sigma_b## itself. Thus, $$\sigma_b = P_n = \varepsilon_0 (\varepsilon - 1)(\vec{E}_Q/\varepsilon + \vec{E}_q +\vec{E}_{\sigma_b})\cdot\hat{k}$$
    Here, ##E_Q## is the electric field at the surface due to Q alone and the ##1/\varepsilon## factor takes into account that Q is immersed in the liquid (as you found when finding ##E_1##).

    ##\vec{E}_{\sigma_b}## is the electric field just inside the surface due to the surface charge density. If you can express ##\vec{E}_{\sigma_b}## in terms of ##\sigma_b##, then you see that you will have an equation that you can solve for ##\sigma_b##.

    [Side note: There is a completely different way to approach this problem using the "method of image charges". But your approach should also get the answer.]
     
  7. Aug 15, 2013 #6
    I think we can find [itex]\displaystyle{\vec{E}_b}[/itex] using the well-known way: Since it is the field infinitely close to the surface, we may assume that we have an infinite plane with homogenous surface charge [itex]\displaystyle{\sigma _b}[/itex] at this point. So using Gauss law:
    [tex]\displaystyle{\vec{E}_b=-\frac{\sigma _b}{2\varepsilon _0}\hat{k}}[/tex]
    Now we can find [itex]\displaystyle{\sigma _b}[/itex] as you said and then just integrate for [itex]\displaystyle{\vec{E}_2}[/itex].
    I think that the problem is solved.

    I know this method for the cases that we have conductors, for example infinite metallic plate or metallic sphere. Here I ended up in this equation:
    [tex]\displaystyle{E_2=\frac{\sigma _b (r^2+h^2)^{3/2}}{8\varepsilon _0 h^3}}[/tex]
    but I still have to calculate [itex]\displaystyle{\sigma _b}[/itex] to continue. But since I found an answer with the other method there's no problem.

    Thanks for helping me!
     
  8. Aug 15, 2013 #7

    rude man

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    I did not mean this to be posted.
     
  9. Aug 15, 2013 #8

    rude man

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    You don't have a uniformly charged surface. The only source of charge is Q, a single point charge.
    The surface charge far from Q is obviously less than that closest to Q. I don't see how you can apply Gauss' theorem in any way here.

    PS I suspect the answer is F = qQ/4πε0(2h)2, in other words, the dielectric has no effect on the force on q. The E field is less in the dielectric than in vacuo but q sits in vacuo.
     
    Last edited: Aug 15, 2013
  10. Aug 15, 2013 #9

    vanhees71

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    I think it's good to start solving the problem from scratch again. It's a nice example for applying the technique of "mirror charges" (as was mentioned before). Let's first set up the mathematical problem, simplifying it so that we can easily solve it.

    First we write down the macroscopic Maxwell equations for the electrostatic case (in Heaviside-Lorentz units):
    [tex]\vec{\nabla} \cdot \vec{D}=\rho=q \delta(\vec{x}-\vec{x}_0), \quad \vec{\nabla} \times \vec{E}=0, \quad \vec{D}=\epsilon \vec{E}.[/tex]
    Now we assume that the dielectric (with [itex]\epsilon=\text{const}[/itex] for [itex]z<0[/itex] and [itex]\epsilon=1[/itex] for [itex]z>0[/itex]) fills the entire half-space [itex]z<0[/itex] with [itex]\vec{x}=(x,y,z)[/itex] in Cartesian coordinates. According to the situation described in the OP, we set [itex]\vec{x}_0=(0,0,-h)[/itex].

    The mathematical prolem is not yet complete, because we have to find the boundary conditions for the fields involved. This is done by using the integral theorems appropriate for the two electrostatic laws above. The first equation, Gauß's Law, we integrate over an infinitesimal box parallel to the surface using Gauß Integral Theorem, [itex]z=0[/itex] (with the charge [itex]Q[/itex] outside this box) where one side is above the boundary and one inside. Since there are no free charges this means that the normal component of [itex]\vec{D}[/itex] (i.e., the [itex]z[/itex] component in our case) must be continuous across the surface:
    [tex]D_z(x,y,0^+)=D_z(x,y,0^-) \; \Rightarrow \; E_z(x,y,0^+)=\epsilon E_z(x,y,0^-).[/tex]

    The second equation we integrate over a little closed curve in an arbitrary plane perpendicular to the boundary with one side in the medium the other outside. It follows that the tangential components of [itex]\vec{E}[/itex] must be continuous, i.e.,
    [tex]E_j(x,y,0^+)=E_j(x,y,0^-) \quad \text{for} \quad j \in \{x,y \}.[/tex]

    Now the problem is completely formulated and we can start solving it. Since the curl of [itex]\vec{E}[/itex] vanishes, there exists (at least locally) a scalar potential [itex]\phi[/itex] such that
    [tex]\vec{E}=-\vec{\nabla} \phi.[/tex]
    Pluggin this in Gauß's Law, we find
    [tex]\Delta \phi=-\frac{1}{\epsilon} Q \delta^{(3)}(\vec{x}-\vec{x}_0).[/tex]
    Now, for symmetry reasons, we can try to solve the problem with the ansatz of mirror charges, i.e., for [itex]z<0[/itex] we assume that the electric field is given by the free charge and an image charge [itex]Q'[/itex] at [itex]z>0[/itex] which should be located at [itex]\vec{x}_M=(0,0,+h')[/itex].

    This means that for [itex]z<0[/itex] we make the ansatz
    [tex]\phi(\vec{x})=\frac{Q}{4 \pi \epsilon \sqrt{x^2+y^2+(z+h)^2}}+\frac{Q'}{4 \pi \sqrt{x^2+y^2+(z-h')^2}}, \quad z<0.[/tex]
    For [itex]z>0[/itex] cannot be any singularity so that there must be only the influence of the free charge inside the medium, but due to the dielectricum appearing as an effective charge [itex]Q''[/itex]:
    [tex]\phi(\vec{x})=\frac{Q''}{4 \pi \sqrt{x^2+y^2+(z+h)^2}}, \quad z>0.[/tex]
    Now we have to fulfill the boundary conditions. Because the tangential components of [itex]\vec{E}[/itex] at the surface are continuous and the jump of the normal component should be finite, the potential should be continuous across the surface. This means
    [tex]\frac{Q}{4 \pi \epsilon \sqrt{x^2+y^2+h^2}}+\frac{Q'}{4 \pi \sqrt{x^2+y^2+h'^2}}=\frac{Q''}{4 \pi \sqrt{x^2+y^2+h^2}}.[/tex]
    Since this should be correct for all [itex]x,y[/itex], we conclude that [itex]h'=h[/itex]. Setting [itex]x=y=0[/itex] then gives
    [tex]\frac{Q}{\epsilon}+Q'=Q''.[/tex]
    The continuity of [itex]D_z[/itex] across the boundary implies
    [tex]\epsilon \partial_z \phi(x,y,z=0^-)=\partial_z \phi(x,y,z=0^+).[/tex]
    After a little algebra this implies
    [tex]\epsilon \left (\frac{Q}{\epsilon}-Q' \right )=Q''.[/tex]
    Solving the two linear equations for [itex]Q'[/itex] and [itex]Q''[/itex] gives
    [tex]Q''=\frac{2Q}{1+\epsilon}, \quad Q'=\frac{Q(\epsilon-1)}{\epsilon(\epsilon+1)}.[/tex]
    This tells us that for [itex]z>0[/itex] there is a Coulomb field with an effective charge at the place of the real free charge inside the dielectric but screened by the factor [itex]2Q/(1+\epsilon)<1[/itex]. Inside the dieelectric the field is created by an also shielded free charge [itex]Q/\epsilon[/itex], and the mirror charge of the same sign outside the dielectric. In reality this additional field comes from the induced polarization of the boundary surface.
     
  11. Aug 15, 2013 #10
    Generally I am not really sure if it is correct. I know that the surface is not uniformly charged. But we need the field just under the surface (almost on it) so I think that we can make the assumption that in one specific point we have constant [itex]\displaystyle{\sigma _b}[/itex].

    Thank you for your solution. I have a question about the potential for [itex]\displaystyle{z>0}[/itex]. The function you write corresponds to a Coulomb field:
    [tex]\displaystyle{\vec{E}=\frac{Q''}{4\pi }\frac{\vec{x}-\vec{x}_0}{\left|\vec{x}-\vec{x}_0 \right|^3}}[/tex]
    So
    [tex]\displaystyle{\nabla ^2\varphi =-\nabla \cdot \vec{E}=-\frac{Q''}{4\pi }\nabla \cdot \frac{\vec{x}-\vec{x}_0}{\left|\vec{x}-\vec{x}_0 \right|^3}=0 \; \; \; \; \text{for} \; \; z>0}[/tex]
    But Poisson's equation is:
    [tex]\displaystyle{\nabla ^2\varphi =q\delta ^3(\vec{x}+\vec{x}_0) \; \; \; \; \text{for} \; \; z>0}[/tex]
    because there's another free charge above the dielectric. When calculating the force we should not add it's contribution at the total field for [itex]\displaystyle{z>0}[/itex] but I think that we should consider it now because it's field affects the bound charges. The potential is:
    [tex]\displaystyle{\phi(\vec{x})=\frac{q}{4 \pi \sqrt{x^2+y^2+(z-h)^2}}+\frac{Q''}{4 \pi \sqrt{x^2+y^2+(z+h)^2}}, \quad z>0}[/tex]
    That changes the equations for the charges. Continuity gives:
    [tex]\displaystyle{\frac{Q}{\epsilon }+Q'=q+Q''}[/tex]
    and also:
    [tex]\displaystyle{\epsilon \left.\frac{\partial \phi }{\partial z}\right|_{z=0^{-}}=\left.\frac{\partial \phi }{\partial z}\right|_{z=0^{+}}\Rightarrow \epsilon Q'-Q=q-Q''}[/tex]
    Only [itex]\displaystyle{Q''}[/itex] contributes at force so:
    [tex]\displaystyle{F=\frac{Q''q}{16\pi \varepsilon _0 h^2}=\frac{2Qq-(\varepsilon -1)q^2}{16\pi (\varepsilon +1) \varepsilon _0 h^2}}[/tex]
    I get the same answer using the other method (the one I discussed with TSny).
     
  12. Aug 15, 2013 #11

    TSny

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    I think that's right. Both q and Q'' contribute to ##\phi## for z > 0.

    I believe that's correct. As expected, both methods give the same answer.

    Very nice work!
     
    Last edited: Aug 15, 2013
  13. Aug 15, 2013 #12

    vanhees71

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    This is a very important point concerning the image-charge method. You cannot put any image charges at [itex]z>0[/itex] when you want to calculate the field (or its potential) there, because there are no charges there (by assumption). Thus for [itex]z>0[/itex] you indeed must have
    [tex]\vec{\nabla} \cdot \vec{D}=\epsilon \vec{\nabla} \cdot \vec{E}=-\Delta \phi=0, \quad z>0.[/tex]

    By symmetry we thus make the ansatz that the real charge at [itex](0,0,-h)[/itex] (note this is located at [itex]z=-h<0[/itex]) within the dielectric gives a field outside the dielectric as if it came from the same position but with a charge [itex]Q''[/itex] somehow screened by the polarization of the dielectric. That this is correct can only be seen from the fact that this ansatz satisfies the boundary conditions, I've explained in my previous posting. For sake of clarity, I'll give the complete electric field everywhere
    [tex]\vec{E}=-\vec{\nabla} \phi=\frac{Q}{4 \pi \epsilon [x^2+y^2+(z+h)^2]^{3/2}} \begin{pmatrix} x \\ y \\ z+h
    \end{pmatrix}+\frac{Q'}{4 \pi [x^2+y^2+(z-h)^2]^{3/2}} \begin{pmatrix}
    x \\ y \\ z-h
    \end{pmatrix}, \quad z<0,[/tex]
    [tex]\vec{E}=\frac{Q''}{4 \pi [x^2+y^2+(z+h)^2]^{3/2}} \begin{pmatrix} x \\ y \\ z+h
    \end{pmatrix}, \quad z>0.
    [/tex]
    The polarization is given by
    [tex]\vec{P}=\begin{cases}
    (\epsilon-1) \vec{E} & \text{for} \quad z<0, \\
    0 & \text{for} \quad z>0.
    \end{cases}
    [/tex]
     
  14. Aug 15, 2013 #13

    TSny

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    In the problem posted in the OP there are two real point charges Q and q. Q in z < 0 and q in z > 0.
     
  15. Aug 16, 2013 #14

    vanhees71

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    Ah, sorry. This I haven't seen. Of course, you can simply solve the same problem with [itex]q[/itex] at [itex]z>0[/itex] and then superpose the two solutions.
     
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