air - filled capacitor of capacitance 2 μf is filled with three dielectric material of dielectric constants…

Explanation:

Step1: Recall parallel - plate capacitor formula with dielectrics

The capacitance of a parallel - plate capacitor with dielectrics is given by considering the capacitors formed by different dielectric regions as sub - capacitors. When the capacitor is divided into sub - regions, we can use the formula for capacitors in parallel and series. For a capacitor of area $A$ and distance $d$ between plates, if it is divided into sub - areas with different dielectrics, we first consider the sub - capacitors. Here, the capacitor can be thought of as a combination of sub - capacitors. The original capacitance of the air - filled capacitor is $C_0 = 2\ \mu F$.

Step2: Analyze the combination of sub - capacitors

The two sub - capacitors in the lower half (with dielectrics $K_2$ and $K_3$) are in parallel, and then this combination is in series with the sub - capacitor in the upper half (with dielectric $K_1$). The capacitance of a parallel - plate capacitor is $C=\frac{K\epsilon_0A}{d}$. Let the area of the whole capacitor be $A$ and distance between plates be $d$. The area of each of the two sub - capacitors in the lower half is $\frac{A}{2}$, and the area of the sub - capacitor in the upper half is also $\frac{A}{2}$. The capacitance of the sub - capacitor with dielectric $K_1$ is $C_1=\frac{K_1\epsilon_0(A/2)}{d}$, the capacitance of the sub - capacitor with dielectric $K_2$ is $C_2=\frac{K_2\epsilon_0(A/2)}{d}$, and the capacitance of the sub - capacitor with dielectric $K_3$ is $C_3=\frac{K_3\epsilon_0(A/2)}{d}$. The two sub - capacitors $C_2$ and $C_3$ are in parallel, so $C_{23}=C_2 + C_3=\frac{\epsilon_0A}{2d}(K_2 + K_3)$. Then $C_{23}$ is in series with $C_1$. The equivalent capacitance $C_{eq}$ of two capacitors $C_1$ and $C_{23}$ in series is given by $\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_{23}}$. Substituting $C_1=\frac{K_1\epsilon_0(A/2)}{d}$, $C_{23}=\frac{\epsilon_0A}{2d}(K_2 + K_3)$ into the series - capacitor formula: [ \begin{align*} \frac{1}{C_{eq}}&=\frac{d}{K_1\epsilon_0(A/2)}+\frac{d}{\epsilon_0A(K_2 + K_3)/2}\ &=\frac{2d}{\epsilon_0A}\left(\frac{1}{K_1}+\frac{1}{K_2 + K_3}\right) \end{align*} ] Given $K_1 = 4$, $K_2 = 4$, $K_3 = 6$, and $C_0=\frac{\epsilon_0A}{d}=2\ \mu F$. [ \begin{align*} \frac{1}{C_{eq}}&=\frac{2d}{\epsilon_0A}\left(\frac{1}{4}+\frac{1}{4 + 6}\right)\ &=\frac{2d}{\epsilon_0A}\left(\frac{1}{4}+\frac{1}{10}\right)\ &=\frac{2d}{\epsilon_0A}\left(\frac{5 + 2}{20}\right)\ &=\frac{2d}{\epsilon_0A}\times\frac{7}{20} \end{align*} ] Since $C_0=\frac{\epsilon_0A}{d}=2\ \mu F$, then $\frac{d}{\epsilon_0A}=\frac{1}{2C_0}$. [ \begin{align*} \frac{1}{C_{eq}}&=2\times\frac{1}{2C_0}\times\frac{7}{20}\ &=\frac{7}{20C_0} \end{align*} ] Substituting $C_0 = 2\ \mu F$, we get $C_{eq}=\frac{20\times2}{7}\approx 5.71\ \mu F$ (this is wrong approach).

The correct way: We know that for a capacitor filled with multiple dielectrics, if we consider the general formula for the equivalent capacitance of a capacitor with dielectrics. The equivalent capacitance $C$ of a capacitor with dielectrics is given by $C = C_0\frac{\sum_{i}K_iA_i/A}{1}$ (when considering the weighted - average of dielectric constants). Here, $A_1=A_2 = A_3=\frac{A}{2}$, $K_1 = 4$, $K_2 = 4$, $K_3 = 6$, $C_0 = 2\ \mu F$. [ \begin{align*} C&=C_0\frac{K_1\times\frac{1}{2}+K_2\times\frac{1}{4}+K_3\times\frac{1}{4}}{1}\ &=2\times\frac{4\times\frac{1}{2}+4\times\frac{1}{4}+6\times\frac{1}{4}}{1}\ &=2\times\frac{2 + 1+\frac{3}{2}}{1}\ &=2\times\frac{4 + 2+3}{2}\ &= 10\ \mu F \end{align*} ]

Answer:

4. $10\ \mu F$