An electric field of 200 v/m exists un the region b/w the plates of a parallel plate capacitor of plate separation 5 cm. The potential difference when a slab of dielectric constant 4 and thickness 1 cm is inserted b/w the plates is
Answers
Answer:
Initial voltage=200V
q=C.V
q=
d
ϵ
0
.A
V (1)
When dielectric is inserted q remains constant and V decreases by k.
q=
d
′
+t−
k
t
ϵ
0
A
V (2)
From equation 1 and 2 we have,
d=d
′
+t−
k
t
d−d
′
=1.6mm=2−
k
2
solving,
k=5
Answer:8.5volt
Given: Electric field=200v/m
Seperation = Distance between parallel plates= 5cm
Dielectric constant (K) = 4
Thickness between the plates = 1cm
Explanation: Potential Difference before dielectric is inserted is:
Vo=Ed=200×5×10-²
Vo=10v
C=€oA/d
C=EoA/0.05. ( value of d is 0.05)
=> €oA=0.05C ------->(1)
When Dielectric is introduced for partially filled capacitance, we know that
C'= €oA/d-t(1-1/k)
=> 0.05C/0.05-0.01(1-1/4)
By simplyfying this,we get
C'= 5C/5-1(3/4)
C'= 20C/17
Then, Q=CoVo/C'
Q=Co×10/(20/17)×Co
V' = 17×10/20
V' = 8.5V