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Homework Help: Force on dielectric medium

  1. May 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Two point charges of magnitude q and 3q are located at a distance of 'a' from each other within an infinite dielectric medium of dielectric constant k. The force by the larger charge 3q, on the dielectric equals
    B)##\frac{1}{4\pi \epsilon_o} \left(1-\frac{1}{k}\right)\frac{3q^2}{a^2}##
    C)##\frac{1}{4\pi \epsilon_o}(k-1)\frac{3q^2}{a^2}##

    2. Relevant equations

    3. The attempt at a solution
    Honestly, I don't quite understand the question. How someone finds force on the medium in which it is placed? :confused:

    Any help is appreciated. Thanks!
    Last edited: May 9, 2013
  2. jcsd
  3. May 9, 2013 #2
    I am not sure, but i think they are talking about the contact force between the dielectric and the point charge. Like in normal reactions in mechanics, which will be equal to the force acting between the two charges... I guess it must be this... But not sure...
  4. May 9, 2013 #3
    One of my friends said that u can consider the dielectric medium to be made of many dipoles, find the force on the dipoles and hence on the medium. And i am not sue how to do that either. Vector integration is involved, which i have no idea how to apply! :-(
  5. May 9, 2013 #4
    Well, that would be ugly. :yuck:

    But I am sure that this doesn't involve those methods as this question is from my test paper (I don't know how to describe it) which doesn't include too much difficult questions.
  6. May 9, 2013 #5
    Yes that's true. I felt that this would have been worked out already using vector integration, and a formula would have been derived. I tried searching "force on a dielectric medium" on google. But i did not find anything given in simple terms. Everywhere they have used vector integration!

    And this was the only thing i actually understood. But seriously, applying vector integration for this would be such a drag!
  7. May 9, 2013 #6


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    Homework Helper

    What is the force between two point charges in a medium, different from vacuum?

  8. May 9, 2013 #7
    Cool question! Use the expression for total force on 3q-

    ##\frac{1}{4\pi \epsilon_ok} \frac{3q^2}{a^2}##

    where k is the dielectric constant. Now the force on 3q due to charge q is-

    ##\frac{1}{4\pi \epsilon_o} \frac{3q^2}{a^2}##

    Now as k>1 the net force is less than the coulombic force. The "extra" force towards q is nothing but the force exerted by the medium ! With this information I believe you can find the answer.

    Btw I think this kind of question can come in Jee Advanced. Oh while i typed echild already replied sorry!
    Last edited: May 9, 2013
  9. May 9, 2013 #8
    Wow! That's nice. :smile: I understood something more than i used to know! Thank you consciousness and ehild! Pranav hope you get the answer! and all the best for your Jee advanced!
    Last edited: May 9, 2013
  10. May 9, 2013 #9
    The force between the two charges in a different medium is ##\frac{1}{4\pi \epsilon_ok}\frac{3q^2}{a^2}##.

    You mean that the net force on 3q is sum of force from dielectric and the charge q, i.e

    [tex]F_{dielectric}+\frac{1}{4\pi \epsilon_o} \frac{3q^2}{a^2}=\frac{1}{4\pi \epsilon_ok} \frac{3q^2}{a^2}[/tex]
    [tex]\Rightarrow F_{dielectric}=\frac{1}{4\pi \epsilon_o} \left(\frac{1}{k}-1\right)\frac{3q^2}{a^2}[/tex]
    This is the force exerted by dielectric on 3q, hence the force on the dielectric is the ##-F_{dielectric}##. Does this look good?
  11. May 9, 2013 #10
    Yeah It looks good. I am learning all this nowadays in electrostatics.
  12. May 9, 2013 #11
    Thanks! :smile:
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