Question

A certain capacitor is rated at 20 nF. Currently, it uses paper as a dielectric, which has κ = 3.7. If the capacitor is made with the same dimensions but using glass as dielectric, which has κ = 5.6, what will the new capacitance be?

C =  nF

What if we double the thickness of the glass, what will the new capacitance be?

C =  nF

Answer 1

We know that capacitance is given by:

C = (Kε₀A) / (d)

Where, K is dielectric constant,

A is area of plates.

d is distance of separation of plates.

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PART 1:

A, and d are constant.

So the new capacitance is given by:

C₁ = (C/3.7) x (5.6)

C₁ = (20 nF/3.7) x (5.6)

C₁ = 30.27 nF --------------- (Answer)

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PART 2:

d₁ = 2d

C = (Kε₀A) / (d)

C₁ is inversely proportional to d.

C₂ = C₁/2

C₂ = (30.27 nF) / (2)

C₂ = 15.14 nF --------------- (Answer)