quinbrium ?
12. The capacitance of parallel plate capacitor becomes 7/6 times its original value it a
dielectric slab of thickness =.=d introduced in between the plates (d = separation
between plates). Write the dielectric constant of slab.
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Answer:
The capacitance is given by C
1
=
d
ϵ
0
A
...(i)
If electric slab of constant k of thickness t=
3
2
d is introduced, then capacitance becomes
C
2
=
d−t(1−
k
1
)
ϵ
0
A
=
d−
3
2
d(1−
k
1
)
ϵ
0
A
...(ii)
Eq. (i) becomes
C
2
=
d[
(1−
3
2
)+
3k
2
]
ϵ
0
A
⇒C
2
=
(
3
1
+
3k
2
)
C
1
⇒
6
7
C
1
=
3
1
+
3k
2
C
1
⇒
3
1
+
3k
2
=
7
6
⇒
3k
2
=
7
6
−
3
1
=
21
11
or 33k=21×2
k=
33
42
=
11
14
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