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Effect on electrostatic force due to dielectric medium.

  1. May 2, 2013 #1
    Hi! I have a high school physics question. I know the concept of induction, and I also know what happens when a dielectric slab is placed between 2 plates of a capacitor.

    2 charged particles are kept at a distance r. The mutual force of attraction/repulsion is F.

    What happens to the force when -
    1) both particles are now kept in dielectric medium (K)
    2) only one charge is kept in dielectric medium (k) other is in vacuum
    3) when a dielectric slab (K) of thickness r/2 is kept between the 2 particles
    4) when a random shaped dielectric solid body is placed between the 2 particles.

    Also, I found this question is test:

    3 concentric spherical shells are placed in vacuum. The region between the inner 2 shells is filled with dielectric material (K). What is the potential on the surface of middle shell?

    The answer is independent of K. There is a possibility that it is incorrect. Can someone please explain?

    Thank you!
  2. jcsd
  3. May 27, 2013 #2
    Well this is disappointing
    Should i repost this?
  4. May 28, 2013 #3

    Andrew Mason

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    Have you studied Gauss' law? What is the field inside any charged conducting shell?

    The dielectric will only have an effect if there is an electric field. But Gauss' law shows that there is no electric field inside these concentric conducting spherical shells.

    [Note: in order to get a response you should follow the posting rules: You have to state the problem, the applicable principles/equations, and you have to show us what you have done to solve the problem].

  5. May 29, 2013 #4
    Yes I know Gauss' Law.
    The answer to your question would be zero.

    Your post answers my last question.
    But the first four are the ones that really stumped me.

    Applicable Principles: Induction, Capacitance, Gauss Law.

    The questions I asked are theoretical. I'm not sure how to prove that I am not asking homework (if that's what you are worried about). I just thought of all possible cases.
    Here are the answers according to me:
    1) F/K on both
    2) F on the particle outside dielectric medium. F/K on the other one.
    3) F on both
    4) F on both
    The logic I have used is that the net force changes only when the particle in inside the medium. If the dielectric medium is inbetween the particles, it will act as a capacitor (inducing equal-opposite charges on two surfaces, and a capacitor doesn't have electric field outside it.
  6. May 29, 2013 #5

    Andrew Mason

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    I assume that K the dielectric constant of the medium.

    For a charged particle placed in a dielectric medium, apply Gauss' law using a spherical Gaussian surface that encloses the charge and the medium. Is the enclosed charge any different due to the presence of the dielectric medium? So how does the field outside the dielectric medium compare to the field at the same distance from the charge without the dielectric medium?
    Just apply Gauss' law. With the dielectric medium between the charges, take a gaussian surface (sphere) around each charge that just touches but does not include the dielectric medium. That will give you the field on either side of the dielectric medium. Can you relate those forces on the medium to the forces between the charges?

  7. May 29, 2013 #6

    Meir Achuz

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    1. F'=F/K
    2, 3, and 4 are complicated problems, because the dielectric gets polarized.
    2 and 3 can be solved with 'images'.
  8. May 29, 2013 #7
    Didn't read forum guidelines regarding homework questions?
  9. May 29, 2013 #8
    Dielectric mediums get polarized by electric fields and create counter-fields. Thus the forces between the particles should decrease (since the electric fields decrease by a factor of k, F'=F/k) whenever a dielectric encompasses both.

    If a dielectric is placed between them without touching any particles, the electric field only changes (by a factor of 1/k) between them, and so the force remains the same.

    If the dielectric encompasses one of them but leaves the other in vacuum, then the force will only change on one of the particles.

    per gauss law, none of the dielectrics will produce an electric flux, so they are not very complex.
    Last edited: May 29, 2013
  10. May 29, 2013 #9

    Andrew Mason

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    I am not sure how that can happen. Newton's third law still applies. The force on each charge has to be equal and opposite to the force on the other.

    Gauss' law always gives the correct answer. So the field of charge A (surrounded in dielectric) at the surface of B (in vacuum) is the same as if the dielectric was not present (since the net charge on the dielectric is 0). So the force of A + dielectric on B is the same as the force of just A on B with no dielectric. So the force of B on A + dielectric must also be the same.

  11. May 29, 2013 #10
    You're probably right, but assuming both particles have equal charge, shouldn't the dielectric repel the outside particle as well? So in sum the repelling force on it is larger than on the particle inside the dielectric? This will be without Newton's 3rd law being broken, as both the particles act on each other with the same resultant forces, but the presence of the dielectric changes things.

    /// Note: I'm only in my 1st year of uni, so don't take what I say for good salt
    Last edited: May 29, 2013
  12. May 30, 2013 #11
    How do i know that the dielectric hasn't disturbed the uniform distribution of flux in the gaussial sphere?
    Your method can only hold true if it can be proven that the electric field will be uniform in all dS areas of the gaussian sphere.

    That sounds reasonable only if it can be proven that electric field will be equal at all the elemental dS areas of a guassian surface.
  13. Jun 2, 2013 #12

    Andrew Mason

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    If there is an asymmetry you would have to use Coulomb's law. The effect of the dielectric will be to superimpose an electric dipole field on the field of the charge.

    The field of the charge will be uniform (Coulomb's law). But but the induced electric field of the dielectric would have to be superimposed to find the total field.

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