# hello

### Asked by ekanathtanpure77 | 23rd Jun, 2022, 07:56: PM

Expert Answer:

###
Figure shows parallel plate capacitor that has dielectric in between plates .
Charges on plates of capacitor are free charges q_{f} and
charges on the surface of dielectric are bound charges q_{b} .
If we calculate Electric field using gaussian surface as shown in figure, we get
( Electric field present only in the space between plates and dielectric slab )

Hence electric field E is given as
..............................(1)

Dielectric material reduces the electric field by a factor (1 / k ) , where k is dielectric constant .
If E_{o} is electric field between capacitor plates without dielectric then we have
............................(2)
If we equate (1) and (2), then we have
From above expression, we get after simplification

_{f}and

_{b}.

_{o}is electric field between capacitor plates without dielectric then we have

### Answered by Thiyagarajan K | 23rd Jun, 2022, 11:51: PM

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