Question

The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K₁ and K₂ with thickness `3/8  d and d/2`, respectively, are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K₁ = 1.25 K, the value of K₁ is:

Options

• 2.66

• 2.33

• 1.60

• 1.33

Answer 1

2.66

Explanation:

`C_1 = K_1(ε_0A)/(3d//8)`

= `K_1(ε_0A)/d ((8  K_1)/3)`

`C_2 = K_2(ε_0A)/(d//2)`

= `K_2(ε_0A)/d (2  K_2)`

`C_3 = (ε_0A)/(d//8)`

= `(ε_0A)/d (8)`

C₁, C₂ and C₃ are in series.

`1/C_(PQ) = 1/(C_1) + 1/(C_2) + 1/(C_2)`

= `d/(ε_0A) (3/(8  K_1) + 1/(2  K_2) + 1/8)`

Given K₁ = 1.25 K₂

= `d/(ε_0A) (3/(8 xx 1.25  K_2) + 1/(2  K_2) + 1/8)`

= `d/(ε_0A) (3/(10  K_2) + 1/(2  K_2) + 1/8)`

= `d/(ε_0A) ((3 + 5)/(10  K_2) + 1/8)`

= `d/(ε_0A) (8/(10  K_2) + 1/8)`

= `d/(ε_0A) ((64 + 10  K_2)/(80  K_2))`

`C_(PQ) = (ε_0A)/d ((80  K_2)/(64 + 10  K_2)) = 2(ε_0A)/d   ...(given)`

40 K₂ = 64 + 10 K₂

40 K₂ − 10 K₂ = 64

 30 K₂ = 64

`K_2 = 64/30`

`K_1 = 1.02  K_2`

= `1.25 xx 64/30`

= `80/30`

= 2.66