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Question
A parallel plate air condenser has a capacity of 20µF. What will be a new capacity if:
1) the distance between the two plates is doubled?
2) a marble slab of dielectric constant 8 is introduced between the two plates?
Solution
Capacitance between the parallel plates of the capacitor, C = 20 pF
Initially, distance between the parallel plates was d and it was filled with air. Dielectric constant of air, k = 1
Capacitance C, is given by the formula,
C = `(k∈_0A)/d`
= `(∈_0A)/d`
C = `(Ain_0)/d = 20muF`
if `d' -> 2d :. C' -> C/2 = 10 muF`
If k = 8 introduce
C" ` = (Ain_0)/d xxk = 20 xx 8 = 160 muF`
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