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CBSE Class 12-science Answered

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Asked by ekanathtanpure77 | 23 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 qf and
charges on the surface of dielectric are bound charges qb .
 
If we calculate Electric field using gaussian surface as shown in figure, we get
 
begin mathsize 14px style contour integral space E times d A space equals space E cross times A space equals space fraction numerator q subscript f space minus space q subscript b over denominator space epsilon subscript o end fraction end style
( Electric field present only in the space between plates and dielectric slab )

Hence electric field E is given as
 
begin mathsize 14px style E space equals space fraction numerator q subscript f space minus space q subscript b over denominator A space space epsilon subscript o end fraction end style  ..............................(1)
Dielectric material reduces the electric field by a factor (1 / k )  , where k is dielectric constant .
 
If Eo is electric field between capacitor plates without dielectric then we have
 
begin mathsize 14px style E space equals space E subscript o over k space equals space fraction numerator q subscript f over denominator k space epsilon subscript o A end fraction end style  ............................(2)
If we equate (1) and (2), then we have
 
begin mathsize 14px style fraction numerator q subscript f over denominator k space epsilon subscript o A end fraction space equals space fraction numerator q subscript f space minus space q subscript b over denominator space epsilon subscript o A end fraction end style
From above expression, we get after simplification
 
begin mathsize 14px style q subscript b space equals space q subscript f open parentheses 1 space minus space 1 over k close parentheses end style

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

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