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Electrostatic force in a medium of non uniform permittivity

  1. Apr 15, 2014 #1

    PhysicoRaj

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    1. The problem statement, all variables and given/known data
    Two point charges, [itex]+4 μC[/itex] and [itex]-10 μC[/itex] are placed 10 cm apart in air. A dielectric slab of large area and thickness 5 cm is placed between the charges. Find the force of attraction between the charges, if the dielectric has a dielectric constant of 9.

    2. Relevant equations
    [itex]F=\frac{1}{4πε}\frac{q_1q_2}{r^2}[/itex]

    [itex]ε=ε_0{ε_r}[/itex]

    3. The attempt at a solution
    The electric field exists in a region of non uniform permittivity. Half of the region is free space while the other half is εr=9. Now should I consider the effective permittivity of the space b/n the charges? Or should I calculate the electric fields in the two different media or something like that?
    Thanks.
     
    Last edited: Apr 16, 2014
  2. jcsd
  3. Apr 16, 2014 #2

    PhysicoRaj

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    Well, I was searching for an equation for effective permittivity and I got it as
    [itex]ε_{eff}=\frac{ε_r}{1+ε_r\frac{L}{L_{ε_r}}}[/itex]
    Plugging in 9 for [itex]ε_r,[/itex] 5cm each for [itex]L[/itex] and [itex]L_{ε_r},[/itex] I get [itex]ε_{eff}=0.9[/itex]
    Now
    [itex]\frac{1}{4πε}=\frac{9\times10^9}{ε_{eff}}[/itex][itex]=10^{10}[/itex]

    So [itex]F=\frac{10^{10}\times40\times10^{-12}}{(0.1)^2}[/itex]
    [itex]F=40N[/itex] and its [itex]wrong[/itex]:grumpy:
     
    Last edited: Apr 16, 2014
  4. Apr 17, 2014 #3

    PhysicoRaj

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    Anybody please?
     
  5. Jun 17, 2014 #4
    What is the correct answer as per answer key ?
     
  6. Jun 17, 2014 #5

    PhysicoRaj

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    Ok, I have to search it, its been a long time. Wait..
     
  7. Jun 17, 2014 #6

    PhysicoRaj

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    They have not given the answer.. there are 4 options: 9N, 11N, 5N, 12N.
     
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