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AIIMS 25 May 2019 Morning Question Paper with solutions for online practice
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© examsnet.com
Question : 25
Total: 166
A capacitor of capacitance 9nF having dielectric slab of
ε
r
=
2.4
dielectric strength 20 MV/m and P.D=20V calculate area of plates.
2.1
×
10
−
4
m
2
4.2
×
10
−
4
m
2
1.4
×
10
−
4
m
2
2.4
×
10
−
4
m
2
Validate
Solution:
C
=
9
n
F
,
ε
r
=
2.4
,
V
=
20
volt
Dielectric strength = 20MV/m
Let separation between plants = d
E
=
v
d
20
×
10
6
=
20
d
d
=
10
−
6
m
Now,
C
=
∊
0
A
∊
r
d
9
×
10
−
9
=
8.85
×
10
−
12
×
A
×
2.4
10
−
6
A
=
9
×
10
−
15
8.85
×
2.4
×
10
−
12
A
=
0.42
×
10
−
3
A
=
4.2
×
10
−
4
m
2
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