|May5-12, 05:06 AM||#1|
dielectric slab in between plates of capacitor
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?
|May5-12, 05:15 AM||#2|
making demands on people for answers is not the wisest way to engender help
definately doesnt encourage me to help you :(
|May5-12, 06:26 AM||#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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