A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be
400%
66.6%
33.3%
200%
A
200%
B
66.6%
C
33.3%
D
400%
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Solution
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Initial capacitance=ϵ∘Ad
When it is half filled by a dielectric of dielectric constant k then
C1=kϵ∘Ad2=2kϵ∘Ad
and C2=ϵ∘Ad2=2ϵ∘Ad
1C=1C1+1C2
=d2ϵ∘A(1k+1)
=d2ϵ∘A(15+1)
=d2ϵ∘A(65)
=3d5ϵ∘A
∴C=5ϵ∘A3d
Hence, % increase in capacitance is
=⎛⎜
⎜
⎜⎝5ϵ∘Ad−ϵ∘Adϵ∘Ad⎞⎟
⎟
⎟⎠×100=23×100=66.6%
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