Question

In a capacitor of capacitance 20 µF the distance between the plates is 5 mm. If a dielectric slab of width 2 mm and dielectric constant 3 is inserted between the plates, then the new capacitor will be ______.

पर्याय

• 22.3 µF

• 25 µF

• 19 µF

• 27.2 µF

Answer 1

In a capacitor of capacitance 20 µF the distance between the plates is 5 mm. If a dielectric slab of width 2 mm and dielectric constant 3 is inserted between the plates, then the new capacitor will be 27.2 µF.

Explanation:

The new capacitance will be when a dielectric slab is added between the plates.

`"C"_1 = (epsilon_"o""A")/("d - t"(1 - 1/"K"))`

`therefore "C"_1/"C" = (epsilon_"o""A")/("d - t"(1 - 1/"K"))`

`"C"_1/"C" = "d"/("d - t"(1 - 1/"K"))`

`"C"_1/"C" = (5 xx 10^-3)/(5 xx 10^-3 - 2 xx 10^-3 (1 - 1/3))`

`"C"_1/"C" = (5 xx 10^-3)/(5 xx 10^-3 - 2 xx 10^-3 (2/3))`

`"C"_1/"C" = (5 xx 10^-3)/(5 xx 10^-3 - 1.33 xx 10^-3)`

`"C"_1 = (5 xx 20)/3.67 = 27.2 mu"F"`