a slab of material of dielectric constant k has the same area as that of the plate of parallel plate capacitor but has the thickness the d bye to where d is a separation between the plates find out the expression for it capacitance where the slab is inserted between the plates of capacitor
Answers
Answer:
Initially when there is vacuum between the two plates, the capacitance of the capictor is
C
0
=
d
ε
0
A
Where, A is the area of parallel plates
Suppose that the capacitor is connected to a battery, an electric field E
0
is produced
now if we insert the dielectric slab of thickness t=d/2 the electric field reduce to E
Eis producedow, if we insert the dielectric slab of thickness t=
t=d2the electric field reduces to ENow, the gap between plates is divided in two parts, for distance there is electric field E and for the remaining distance
(d-t) the electric field is E
0
0
ε0 If V be the potential difference between the plates of the capacitor, then
V=E
t
+E
0
(d−t)
V=
2
Ed
+
2
E
0
d
=
2
d
(E+E
0
)(t=
2
d
)
V=
2
d
(
K
E
0
+E
0
)=
2K
dE
0
(K+1)(as
E
E
0
=K)
nowE
0
=
ε
0
σ
=
ε
0
A
q
⇒V=
2K
d
ε
0
A
q
(K+1)
weknowC=
V
q
=
(K+1)d
2Kε
0
A
Explanation:
Hope it helps you~
Answer:
Initially when there is vacuum between the two plates, the capacitance of the capictor is
C
0
=
d
ε
0
A
Where, A is the area of parallel plates
Suppose that the capacitor is connected to a battery, an electric field E
0
is produced
now if we insert the dielectric slab of thickness t=d/2 the electric field reduce to E
Eis producedow, if we insert the dielectric slab of thickness t=
t=d2the electric field reduces to ENow, the gap between plates is divided in two parts, for distance there is electric field E and for the remaining distance
(d-t) the electric field is E
0
0
ε0 If V be the potential difference between the plates of the capacitor, then
V=E
t
+E
0
(d−t)
V=
2
Ed
+
2
E
0
d
=
2
d
(E+E
0
)(t=
2
d
)
V=
2
d
(
K
E
0
+E
0
)=
2K
dE
0
(K+1)(as
E
E
0
=K)
nowE
0
=
ε
0
σ
=
ε
0
A
q
⇒V=
2K
d
ε
0
A
q
(K+1)
weknowC=
V
q
=
(K+1)d
2Kε
0
A
Explanation: