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  • A parallel plate capacitor with plate area A and plate separation d is filed with a dielectric material of dielectric constant K = 4. The thickness of the dielectric material is x, where x &

A parallel plate capacitor with plate area A and plate separation d is filed with a dielectric material of dielectric
constant K = 4. The thickness of the dielectric material is x, where x < d.

Let C_1and C_2 be the capacitance of the system forx=\frac{1}{3} d and x=\frac{2 d}{3}, respectively. If C_1=2 \mu Fthe value of C_2 is \mu \mathrm{F}

Option: 1

3


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & C_1=\frac{\frac{\epsilon_0 A}{\frac{2 d}{3}} \times \frac{4 \epsilon_0 A}{\frac{d}{3}}}{\frac{\epsilon_0 A}{2 d / 3}+\frac{4 \epsilon_0 A}{d / 3}}=\frac{18}{\frac{3}{2}+12} \frac{\epsilon_0 A}{d}=18 \times \frac{2}{27} \frac{\epsilon_0 A}{d}=\frac{4}{3} \frac{\epsilon_0 A}{d} \\ & \text { According to } \mathrm{qn}, \frac{4}{3} \frac{\epsilon_0 A}{d}=2 \Rightarrow \frac{\epsilon_0 A}{d}=\frac{3}{2}----(i) \\ & \text { Now, } C_2=\frac{\frac{\epsilon_0 A}{\frac{d}{3}} \times \frac{4 \epsilon_0 A}{\frac{2 d}{3}}}{\frac{\varepsilon_0 A}{d / 3}+\frac{4 \epsilon_0 A}{2 d / 3}}=\frac{18}{3+6} \frac{\epsilon_0 A}{d}=2 \times \frac{\epsilon_0 A}{d}=2 \times \frac{3}{2}=3 \end{aligned}

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Kuldeep Maurya

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