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Force on dielectric in parallel plate capicitor

  1. Jul 21, 2012 #1
    Force act on dielectric slab on inserting between the parallel plate = [8.85 * 10^-12]b(k-1)(V^2)/2d

    where b = width of the plate , d = distance b/w the plates , V is the constant potential difference across the plates and k = dielectric constant

    Which force is acting on dielectric slab in this case and who is acting this force on the slab and how this value comes ?
     
  2. jcsd
  3. Jul 21, 2012 #2
    Hi nik jain!
    What do you think which type of force should act?
    To obtain this value of force, start by making a diagram of the capacitor when the dielectric is being inserted in the capacitor, suppose that x length of dielectric is inside the capacitor. The dimensions of capacitor are l and b. What is the equivalent capacity when the x length of dielectric is inside it.
     
  4. Jul 22, 2012 #3
    I got it .

    It is the electrostatic force of attraction b/w the charges .

    One more question : Why its value remains constant as the distance(r) b/w the charges goes on decreasing and magnitude of force of attraction is inversely proportional to r^2 ?
     
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