The plates of a parallel plate capacitor are separated by d . Two slabs of different dielectric constant K_1 and K_2 with thickness 3/8d 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 K_1 =

Initial capacitance C_0 = _0 A/d .New capacitance C = _0 A/_iK_i = _0 A/3d/8K_1 + d/2/K_2 + d/8/1 .Given C = 2C_0 : _0 A/d ≤ft( 38K_1 + 1/2K_2 + 1/8 ) = 2_0 A/d 3/8K_1 + 1/2K_2 + 1/8 = 1/2 3/8K_1 + 1/2K_2 = 3/8 .Divide by 3/8 : 1/K_1 + 4/3K_2 = 1 .Given K_1 = 1.25 K_2 = 5/4 K_2 K_2 = 4/5 K_1 . 1/K_1 + 4/3(4/5 K_1) = 1 1/K_1 + 5/3K_1 = 1 . 8/3K_1 = 1 K_1 = 8/3 ≈ 2.66 .

The plates of a parallel plate capacitor are separated by d . Two slabs of different dielectric constant K_1 and K_2 with thickness 3/8d and d/2 , res - Physics MCQ Physics | ExamTest.live

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The plates of a parallel plate capacitor are separated by d . Two slabs of different dielectric constant K_{1} and K_{2} with thickness \frac{3}{8} d and \frac{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 K_{1} = 1.25 K_{2}, the value of K_{1} is:

A $2.66$
B $2.33$
C $1.60$
D $1.33$

Solution & Step-by-step Explanation

Initial capacitance C_{0} = \frac{ϵ _{0} A}{d} .New capacitance C = \frac{ϵ _{0} A}{\sum \frac{t _{i}}{K _{i}}} = \frac{ϵ _{0} A}{\frac{3 d /8}{K _{1}} + \frac{d /2}{K _{2}} + \frac{d /8}{1}} .Given C = 2 C_{0} : \frac{ϵ _{0} A}{d ( \frac{3}{8 K _{1}} + \frac{1}{2 K _{2}} + \frac{1}{8} )} = \frac{2 ϵ _{0} A}{d} \frac{3}{8 K _{1}} + \frac{1}{2 K _{2}} + \frac{1}{8} = \frac{1}{2} ⟹ \frac{3}{8 K _{1}} + \frac{1}{2 K _{2}} = \frac{3}{8} .Divide by 3/8 : \frac{1}{K _{1}} + \frac{4}{3 K _{2}} = 1 .Given K_{1} = 1.25 K_{2} = \frac{5}{4} K_{2} ⟹ K_{2} = \frac{4}{5} K_{1} . \frac{1}{K _{1}} + \frac{4}{3 ( 4/5 K _{1} )} = 1 ⟹ \frac{1}{K _{1}} + \frac{5}{3 K _{1}} = 1 . \frac{8}{3 K _{1}} = 1 ⟹ K_{1} = \frac{8}{3} \approx 2.66 .

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The plates of a parallel plate capacitor are separated by d . Two slabs of different dielectric constant K_{1} and K_{2} with thickness \frac{3}{8} d and \frac{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 K_{1} = 1.25 K_{2}, the value of K_{1} is:

2.66

2.33

1.60

1.33

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