Question 626: Electrostats & Capacitance

Question

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:

Accepted Answer

Using $C_{e q}=\frac{\varepsilon_0 A}{\frac{t_1}{K_1}+\frac{t_2}{K_2}+\frac{t_3}{K_3}}$
here $C_0=\frac{\varepsilon_0 A}{d}, t_1=\frac{3 d}{8}, t_2=\frac{d}{2}, t_3=\frac{d}{8}$
$$K_1=K_1, \quad K_2=\frac{K_1}{1.25} 	ext { and } K_3=1$$
Given $C_{	ext {eq }}=2 C_0$
$$\begin{aligned}
& \Rightarrow 2 C_0=\frac{\varepsilon_0 A}{\frac{3 d}{8 k_1}+\frac{d 	imes 1.25}{2 k_1}+\frac{d}{8}} \
& \Rightarrow \frac{2 \varepsilon_0 A}{d}=\frac{\varepsilon_0 A}{\frac{3 d}{8 k_1}+\frac{d}{2 k_1} 	imes \frac{5}{4}+\frac{d}{8}} \
& \Rightarrow 2=\frac{1}{\frac{3}{8 k_1}+\frac{5}{8 k_1}+\frac{1}{8}} \Rightarrow k_1=\frac{8}{3}=2.66
\end{aligned}$$

Answer

1.60

Answer

1.33

Answer

2.66

Answer

2.33

Electrostats & Capacitance

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