Answer in Brief

Sum

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?

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#### 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`

Concept: The Parallel Plate Capacitor

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