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Dielectric slab in between plates of capacitor

  1. May 5, 2012 #1
    i'm reposting this question as nobody answered it previously. please answer it this time.

    in given type of arrangement we normally take it equivalent to two capacitors(plate 1 and surface 2) and (surface 2 and plate 3) in series. My question is that a capacitor is formed with two plates of equal and opposite charge but the induced charge on dielectric slab(surface 2) is less than the charge on plates 1 and 3 which is clear by the formula q(ind)=q(1-1/k), then how can we assume plate 1 and surface 2 as one capacitor?
     

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  3. May 5, 2012 #2

    davenn

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    making demands on people for answers is not the wisest way to engender help

    definately doesnt encourage me to help you :(

    Dave
     
  4. May 5, 2012 #3
    Here is one way to justify this method:
    Place a hypothetical conducting plate (with infinitesimal thinness) on the other face of the dielectric and you have two capacitors in series now! Since the plate doesn't change the electric field E , the voltage difference between the two REAL plates remains the same. So does charge Q on each plane which depends on the normal component of E. Thus the ratio Q/V=C is also remains the same.
     
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