Q.2
The capacitance of a parallel plate capacitor having air
between its plates is 10 uf if the distance between the plates is
reduced to half and the medium between plates is replaced by
a medium of dielectric constant k=5, then find the value how
much increase or decrease in the capacitance of the capacitor.
Answers
Answer:
100μf
Explanation:
C= KA€₀/d
where K is dielectric constant
A is the area between plates
d is the distance between plates
€₀ is the permittivity
for air, K = 1
⇒ C₀ = A€₀/d = 10μf
Given distance is reduced to half and medium dielectric constant K = 5 is inserted
d¹ = d/2
⇒ C = KA€₀/d¹
= 5A€₀/(d/2)
= 5A€₀*2/d
= 10A€₀/d
= 10C₀ = 10*10μf = 100μf
∴The new capacitance increases 10 times the initial capacitance
Given : The capacitance of a parallel plate capacitor having air between its plates is 10 µf . the distance between the plates is reduced to half and the medium between plates is replaced by a medium of dielectric constant k=5,
To Find : the value how much increase or decrease in the capacitance of the capacitor.
Solution:
C = Kε₀ A / d
The capacitance of a parallel plate capacitor having air
between its plates is 10 µf
C = 10 µf
K = 1 for dielectric constant of air
=> 1. ε₀ A / d = 10 µf
=> ε₀ A / d = 10 µf
distance between the plates is reduced to half
=> New d = d/2
medium between plates is replaced by a medium of dielectric constant k=5
=> k = 5
new C = 5 ε₀ A / (d/2)
=> new C = 10 ε₀ A / d
=> new C = 10 * 10 µf
=> new C = 100 µf
increase in capacitance = 100 - 10 = 90 µf
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