Question

A slab of material of dielectric constant k has the same area A as the plates of a parallel plate capacitor and has a thickness (3/4d), where d is the separation of the plates. The change in capacitance when the slab is inserted between the plates is ______.

पर्याय

• `C = (Aε_0)/d ((k + 3)/(4k))`

• `C = (Aε_0)/d ((2k)/(k + 3))`

• `C = (Aε_0)/d ((k + 3)/(2k))`

• `C = (Aε_0)/d ((4k)/(k + 3))`

Answer 1

A slab of material of dielectric constant k has the same area A as the plates of a parallel plate capacitor and has a thickness (3/4d), where d is the separation of the plates. The change in capacitance when the slab is inserted between the plates is C = `(Aε_0)/d ((4k)/(k + 3))`.

Explanation:

The diagrammatic representation of the capacitor when the slab is inserted is shown below.

The above configuration can be considered as two capacitors C₁ and C₂ placed in series with each other, as shown in the diagram below.

Circuit configuration

The capacitance C₁ s the one with a dielectric with dielectric constant K and plate separation  3/4 d

C₁ = `(kε_0 A)/((3/4)d)`

C₁ = `(4kε_0 A)/(3d)`

The capacitance C₂ is the one with no dielectric and plate separation `(1/4)` d

C₂ = `(€_0 A)/((1/4) d)`

C₂ = `(4€_0 A)/d`

When two capacitors are placed in series with each other, the equivalent capacitance C is given as follows:

C = `(C_1C_2)/(C_1 + C_2)`

Substitute the values of capacitances of C₁ and C₂.

C = `(((kε_0 A)/((3/4)d)) * ((4ε_0 A)/d))/((kε_0 A)/((3/4) d) + (4ε_0 A)/d)`

C = `(((4k)/3) * (4) * ((ε_0 A)/d)^2)/(((4k)/3 + 4) * ((ε_0 A)/d))`

C = `(((4k)/3) * ((ε_0 A)/d))/((k + 3)/5)`

C = `(Aε_0)/d ((4k)/(k + 3))`