CBSE Class 12-science Answered
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![question image](http://images.topperlearning.com/topper/new-ate/top_mob16559943602097211373037af338-df51-412d-9705-674baa0f238e.jpg)
Asked by ekanathtanpure77 | 23 Jun, 2022, 19:56: PM
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/67f4508e9866503e19e4f8829462db4862b4a903ea1de9.55531028f2.png)
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](https://images.topperlearning.com/topper/tinymce/cache/ff510c6ceb3bfd8f05db51f93633a07d.png)
( 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](https://images.topperlearning.com/topper/tinymce/cache/875724d2de62916d823861e5a3ab681f.png)
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](https://images.topperlearning.com/topper/tinymce/cache/534711c9e68193e7a07987ac41cba3f4.png)
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](https://images.topperlearning.com/topper/tinymce/cache/16125f36f13a503b355b931219ce9a02.png)
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](https://images.topperlearning.com/topper/tinymce/cache/0c218a1a365049c5fa563ed0905d8993.png)
Answered by Thiyagarajan K | 23 Jun, 2022, 23:51: PM
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