dielectric slab in between plates of capacitor

by karanbir
Tags: capacitor, dielectric, plates, slab
karanbir is offline
May5-12, 05:06 AM
P: 7
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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davenn is offline
May5-12, 05:15 AM
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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 :(

Hassan2 is offline
May5-12, 06:26 AM
P: 404
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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