Question 478: Electrostats & Capacitance

Question

A parallel plate air capacitor has capacitance $C_P$. It is equally filled with parallel layers of materials of dielectric constants $\mathrm{K}_1$ and $\mathrm{K}_2$. Now its capacity becomes $C_K$. The ratio $C_P$ to $C_K$ is

Accepted Answer

When dielectrics are placed in series, the equivalent dielectric constant is:
$$ \begin{aligned} & \frac{1}{\mathrm{~K}_{\mathrm{eq}}}=\frac{1}{2 \mathrm{~K}_1}+\frac{1}{2 \mathrm{~K}_2} \ & \frac{1}{\mathrm{~K}_{\mathrm{eq}}}=\left(\frac{\mathrm{K}_1+\mathrm{K}_2}{2 \mathrm{~K}_1 \mathrm{~K}_2}ight) \ & \mathrm{K}_{\mathrm{eq}}=\left(\frac{2 \mathrm{~K}_1 \mathrm{~K}_2}{\mathrm{~K}_1+\mathrm{K}_2}ight) \end{aligned} $$
So, new capacitance:
$$ \begin{aligned} & C_K=K_{e q} \cdot C_P=\left(\frac{2 K_1 K_2}{K_1+K_2}ight) C_P \ & \frac{C_P}{C_K}=\frac{K_1+K_2}{2 K_1 K_2} \end{aligned} $$

Answer

$\mathrm{K}_1+\mathrm{K}_2$

Answer

$\frac{K_1+K_2}{K_1 K_2}$

Answer

$\frac{K_1+K_2}{2 K_1 K_2}$

Answer

$\frac{2 K_1 K_2}{K_1+K_2}$

Electrostats & Capacitance

A parallel plate air capacitor has capacitance C_P. It is equally filled with parallel layers of materials of dielectric constants \mathrm{K}_1 and \mathrm{K}_2. Now its capacity becomes C_K. The ratio C_P to C_K is

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