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Effective distance in vacuum

  1. Dec 14, 2015 #1
    http://postimg.org/image/p7sz481cv/

    I now know what is effective distance
    Suppose you have two charges embedded in a dielectric medium with constant k (so instead of empty space there's a "sea" of dielectric media around them). They are separated by a distance d in the medium, so that Coulomb's law would be:
    ##F##=##\frac{1}{k 4 \pi ε_o}####\frac{q_1 q_2}{d^2}##

    Now we want to know what the effective distance would be in empty space to produce the same force. So:

    ##\frac{1}{k~4 \pi ε_o}## ##\frac{q_1 q_2}{d^2}## = ##\frac{1}{4 \pi ε_o} \frac{q_1 q_2}{d_{eff}^2}##

    which reduces to

    ##k~d^2## = ##d_{eff}^2##

    If we look at the image I have given link of how the net distance between the charges has been calculated there?
     
  2. jcsd
  3. Dec 14, 2015 #2

    cnh1995

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    Let's assume the charges q1 and q2 are in free space and the slab inserted has a dielectric constant k. If the slab weren't there, the net distance between the charges would be Dnet=r. If ,instead of free space, the charges were in this material with dielectric constant k and thickness t, the effective distance between them in vacuum would be t√k. But here, both free space and that material are present. So, net distance between the charges(in vacuum)=
    actual distance in vacuum+equivalent of the thickness of the slab in vacuum.
    Hence,Dnet=(r-t)+t√k
    where r-t is the actual distance in vacuum and t√k is the equivalent of the slab's thickness in vacuum.
    This is as per my understanding.
     
  4. Dec 14, 2015 #3
    You mean
    this.png

    rather than
    edit.png
     
  5. Dec 14, 2015 #4

    cnh1995

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    Last edited: Dec 14, 2015
  6. Dec 14, 2015 #5
    I don't think he means the second situation.
    It is the same as in your problem. Some of the space between the two charges is empty, some is filled with dielectric.
    The distance through the empty space is (r-t), the distance through dielectric is t. These are geometric distances.
    In order to find effective distances you multiply each by the square of the dielectric constants: 1 for the empty part and √k for the dielectric part.
    Then you just add them to get total effective distance.
     
  7. Dec 15, 2015 #6
    Why?Is this a technique to solve such questions?
     
  8. Dec 15, 2015 #7

    cnh1995

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    this-png.93341.png
    Here, let the force between the charges be F. To have the same force F in vacuum, the distance between the charges must be t√k in vacuum. Hence, t√k is the equivalent of slab's thickness in vacuum. So, in general, if there are some more media in between the charges, equivalent of their thickness in vacuum should be added to the 'actual'(or you can say 'effective', since k=1) distance in vacuum.
    I think so. You can try a number of problems and verify.
     
  9. Dec 15, 2015 #8
    Did you mean square root of the dielectric constants
     
  10. Dec 16, 2015 #9
    Yes, square root. Sorry. :smile:
     
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