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Electrostatics Problem

  1. Jun 16, 2014 #1
    1. The problem statement, all variables and given/known data

    Two similar point charges q1 and q2 are placed at a distance r apart in the air. A dielectric slab of thickness t(<<r) having dielectric constant K is placed between the charges. Calculate the coulomb force of repulsion between the charges.

    Now assume that a slab of thickness half the separation between the charges and the Coulomb's repulsive force is reduced in the ratio 9:4. Calculate K for such a slab.


    2. Relevant equations

    Coulomb's law :

    F=q1q2/4πεr2

    where ε=εoK



    3. The attempt at a solution

    I am not sure. For such a system how should I apply Coulomb's law ? We have a dielectric slab between the system. So how to make that law apply here ?

    I just thought that I have to transform the given system to its equivalent vacuum one, but how should I do that ?

    Please help !!

    Thanks in advance...:smile:
     
  2. jcsd
  3. Jun 17, 2014 #2
    Can someone help me please ? I am really clueless. :(
     
  4. Jun 17, 2014 #3
    Hi Sankalp

    You have the right idea .

    First convert the slab width 't' into effective air separation . Can you do that ?
     
  5. Jun 17, 2014 #4
    Thanks..

    If we hypothetically bring the two charges with the separation "t", then

    q1q2/4πεoKt2 = q1q2/4πεox2

    Solving we get x=√(K)t

    Then in the actual system, effective separation is d-t+x ? Right ?

    BTW, are you an IIT aspirant ?
     
  6. Jun 17, 2014 #5
    Correct.

    r-t+x
     
  7. Jun 17, 2014 #6
    Ya ya... I mistyped... I always do... d for distance...

    Now with this information I solved both the parts completely....

    Thanks a lot !!! The answer matches with the key. :)
     
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