Question

In a capacitor of capacitance 20 µF, the distance between the plates is 2 mm. If a dielectric slab of width 1 mm and dielectric constant 2 is inserted between the plates, what is the new capacitance?

Answer 1

Data: C = 20 µF = 20 x 10^-6 F, k = 2

d = 2 mm = 2 x 10^-3 m,

t = 1 mm = 1 x 10^-3 m

To find: Capacitance when the slab is inserted between the plates =?

Formulae: (i) `C = (Aε_0)/d`

and (ii) C' = `(Aε_0)/(d - t + t/k)`

Dividing (ii) by (i) , we get

`"C'"/C = d/ (d - t + t/k)`

C' =`[(2xx10^-3)/((2 - 1 + 1/2) xx 10 ^-3)] xx 20 xx10^-6` 

∴ C' = 26.67 x 10^-6 F = 26.67 µF

∴ The new capacitance is 26.67 µF