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A parallel plate capacitor with plate area A and plate separation d is filled with a dielectric material of dielectric constant K = 4. The thickness of the dielectric material is x, where x < d.
Let C₁ and C₂ be the capacitance of the system for $x = \frac{1}{3} d$ and $x = \frac{2 d}{3}$ , respectively. If $C_{1} = 2 μF$ , then the value of C₂ is ____ $μF$ .

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answer is 3.

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Detailed Solution

$C = \frac{ε_{0} A}{d - (t - \frac{t}{K})}$

(t is thickness of dielectric

d is distance between the plates

K is dielectric constant

A is area of each plate)

$(i) given t = \frac{1}{3} d$

$C_{1} = \frac{ε_{0} A}{d - (\frac{1}{3} d - \frac{\frac{1}{3} d}{4})}$

$\begin{array}{l} = \frac{ε_{0} A}{\frac{2 d}{3} + \frac{d}{12}} \\ = \frac{ε_{0} A}{\frac{9 d}{12}} \\ C_{1} = \frac{4 ε_{0} A}{3 d} = 2 μ F \\ \frac{ε_{0} A}{d} = \frac{3}{2} μ F . . . . (1) \end{array}$

$(ii) given t = \frac{2 d}{3}$

$\begin{array}{l} C_{2} = \frac{ε_{0} A}{d - (\frac{2 d}{3} - \frac{\frac{2 d}{3}}{4})} \\ = \frac{ε_{0} A}{\frac{d}{3} + \frac{d}{6}} \end{array}$

$\begin{aligned} = \frac{ε_{0} A}{\frac{d}{2}} \\ C_{2} = \frac{2 ε_{0} A}{d} \end{aligned}$

From (1)

$\begin{array}{l} C_{2} = 2 \times \frac{3}{2} \\ C_{2} = 3 μ F \end{array}$

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