CBSE Class 12-science Answered
hello
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
( 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 Eo 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
Answered by Thiyagarajan K | 23 Jun, 2022, 11:51: PM
Concept Videos
CBSE 12-science - Physics
Asked by khankaifi178 | 08 Jan, 2024, 10:12: PM
ANSWERED BY EXPERT
CBSE 12-science - Physics
Asked by bmahalik21 | 05 Mar, 2023, 08:23: PM
ANSWERED BY EXPERT
CBSE 12-science - Physics
Asked by s3043632 | 22 Jan, 2023, 06:45: PM
ANSWERED BY EXPERT
CBSE 12-science - Physics
Asked by ekanathtanpure77 | 23 Jun, 2022, 07:56: PM
ANSWERED BY EXPERT
CBSE 12-science - Physics
Asked by manandhiman023 | 13 Jun, 2022, 01:52: PM
ANSWERED BY EXPERT
CBSE 12-science - Physics
Asked by yashrajgharte24.12dgatl | 06 Aug, 2021, 09:02: PM
ANSWERED BY EXPERT
CBSE 12-science - Physics
Asked by yashrajgharte24.12dgatl | 03 Aug, 2021, 03:31: PM
ANSWERED BY EXPERT
CBSE 12-science - Physics
Asked by hitanshu04 | 21 May, 2021, 07:05: PM
ANSWERED BY EXPERT
CBSE 12-science - Physics
Asked by hitanshu04 | 18 May, 2021, 07:05: PM
ANSWERED BY EXPERT
CBSE 12-science - Physics
Asked by xantyyyyxd | 17 Mar, 2021, 04:26: AM
ANSWERED BY EXPERT