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  • The plates of alparallel plate capacitor are separated by $d$.

Q.30) The plates of alparallel 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 arerinserted in the capacitor. Due to this, thercapacitance becomes two times larger that when there is nothing between the plates.
If $K_1=1.25 K_2$, the value of $K_1$ is :
 

A)
1.33

B)  2.66

C) 2.33

D) 1.60

 

Answers (1)

best_answer

Given $C=2 C_0$, so:

$$
\frac{3}{8 K_1}+\frac{1}{2 K_2}+\frac{1}{8}=\frac{1}{2}
$$


Substitute $K_1=1.25 K_2 \Rightarrow K_2=\frac{K_1}{1.25}$, solve:

$$
\frac{3}{K_1}+\frac{4 \cdot 1.25}{K_1}=3 \Rightarrow \frac{8}{K_1}=3 \Rightarrow K_1=\frac{8}{3} \approx 2.66
$$


Answer: (2) 2.66

Posted by

Saumya Singh

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