A parallel plate capacitor is of area 6 cm² and separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constant K₁ = 10, K₂ = 12 and K₃ = 14. The dielectric constant of a material which when fully inserted in above capacitor gives same capacitance would be?

Let the dielectric constant of material used be K.

The capacitance of a parallel plate capacitor with a dielectric material is given by:

C = (ε₀KA)/d

where:

• C is the capacitance
• ε₀ is the permittivity of free space
• K is the dielectric constant
• A is the area of the plates
• d is the separation between the plates

In this case, the gap is filled with three dielectric materials of equal thickness. Let the thickness of each dielectric be d/3. The equivalent capacitance of the combination is given by:

1/Ceq = (d/3)/(ε₀K₁A) + (d/3)/(ε₀K₂A) + (d/3)/(ε₀K₃A)

1/Ceq = d/(3ε₀A) * (1/K₁ + 1/K₂ + 1/K₃)

Ceq = (3ε₀A)/[d * (1/K₁ + 1/K₂ + 1/K₃)]

Substituting the given values:

Ceq = (3ε₀ * 6 * 10⁻⁴ m²)/[3 * 10⁻³ m * (1/10 + 1/12 + 1/14)]

Ceq = (6ε₀)/(1/10 + 1/12 + 1/14) = (6ε₀)/(0.1 + 0.0833 + 0.0714) = (6ε₀)/0.2547 ≈ 23.56ε₀

Now, if a single dielectric material with dielectric constant K is used to fill the entire gap, the capacitance would be:

C = (ε₀KA)/d

For this capacitance to be equal to Ceq, we have:

(ε₀KA)/d = 23.56ε₀

K = 23.56 * (d/A) = 23.56 * (3 * 10⁻³ m)/(6 * 10⁻⁴ m²) = 23.56 * 5 = 117.8

This value is not among the choices given. However, there might be a simpler way to solve this problem. We can calculate the equivalent dielectric constant as follows:

1/K_eq = (1/3)(1/K₁ + 1/K₂ + 1/K₃) = (1/3)(1/10 + 1/12 + 1/14) = (1/3)(0.25476) ≈ 0.0849

K_eq = 1/0.0849 ≈ 11.78

This value is also not in the given options, there may be an error in the question or options provided.