A parallel plate capacitor having area A and separation between plates d has capacitance C if the area of place and separation between plates is changed to 2A and d/2 respectively then the new capacitance is
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Answer:
Capacitance
d
ϵ
0
A
C
3
and C
4
are in series.
C
1
=
d
ϵ
0
S
,C
3
=
2
d
ϵ
1
2
S
, C
4
=
2
d
ϵ
2
2
S
,C
5
=
d
ϵ
1
2
S
,C
34
=
C
3
+C
4
C
3
C
4
C
2
=C
34
+C
5
=
3
7
d
ϵ
0
S
C
1
C
2
=
3
7
Capacitance
d
ϵ
0
A
C
3
and C
4
are in series.
C
1
=
d
ϵ
0
S
,C
3
=
2
d
ϵ
1
2
S
, C
4
=
2
d
ϵ
2
2
S
,C
5
=
d
ϵ
1
2
S
,C
34
=
C
3
+C
4
C
3
C
4
C
2
=C
34
+C
5
=
3
7
d
ϵ
0
S
C
1
C
2
=
3
7
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