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CBSE Class 12 Physics 2019 Outside Delhi Set 1 Paper

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Question : 20 of 27
Marks: +1, -0
A 200ยตF parallel plate capacitor having plate separation of 5โ€Œmm is charged by a 100Vโ€Œdc source. It remains connected to the source. Using an insulated handle, the distance between the plates is doubled and a dielectric slab of thickness 5โ€Œmm and dielectric constant 10 is introduced between the plates. Explain with reason, how the (i) capacitance, (ii) electric field between the plates, (iii) energy density of the capacitor will change?
Solution:  
(i) Change in capacitance
(ii) Change in electric field
(iii) Change in electric density
Dielectric slab of thickness 5โ€Œmm is equivalent to an air capacitor of thickness =โ€Œ
5
10
โ€Œmm
Effective separation between the plates with air in between is =(5+0.50)โ€Œmm=5.5โ€Œmm
(i) Effective new capacitance
โ€Œ=200ยตFร—โ€Œ
5โ€Œmm
5.5โ€Œmm
=โ€Œ
2000
11
ยตF
โ€Œโ€Œโ‰ˆ182ยตF
(ii) Effective new electric field
โ€Œ=โ€Œ
100V
5.5ร—10โˆ’3m
=โ€Œ
20000
1.1
โ€Œโ€Œโ‰ˆ18182Vโˆ•m
(iii) โ€Œ
โ€Œ New energy stored โ€Œ
โ€Œ Original energy stored โ€Œ
=โ€Œ
โ€Œ
1
2
Cโ€ฒV2
1
2
CV2
 =โ€Œ
Cโ€ฒ
C
=โ€Œ
10
11
New Energy density will be (โ€Œ
10
11
)2 of the original energy density =โ€Œ
100
121
 of the original energy density.
[Note: If the student writes C=โ€Œ
Aฮต0
d

โ€ŒCm=โ€Œ
KAฮต0
d

โ€ŒEโ€ฒ=โ€Œ
V
d

โ€ŒU=โ€Œ
1
2
ฮต0E2
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