Question 114: Electrostats & Capacitance

Question

Three parallel plate capacitors each with area $A$ and separation $d$ are filled with two dielectric ( $k_1$ and $k_2$ ) in the following fashion. Which of the following is true?
$$ \left(k_1>k_2\right) $$

Accepted Answer

The capacitance of a parallel plate capacitor is $C=\frac{k \epsilon_0 A}{d}$, where $k$ is the dielectric constant of the material, $A$ is the area of the plate and d is the separation between the plates.
When capacitors are in parallel then the equivalent capacitance is given as
$$ C_{e q}=C_1+C_2+\cdots+C_n $$
When capacitors are in series then the equivalent capacitance is given as
$$ \frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\cdots+\frac{1}{\mathrm{C}_n} $$
Let $\mathrm{C}_0=\frac{\epsilon_0 \mathrm{~A}}{\mathrm{~d}}$.
For Capacitor (A)
The capacitance, $C_{A 1}=\frac{\mathrm{k}_1 \epsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=\frac{2 \mathrm{k}_1 \epsilon_0 \mathrm{~A}}{\mathrm{~d}}=2 \mathrm{k}_1 \mathrm{C}_0$
The capacitance, $C_{A 2}=\frac{k_1 \epsilon_0\left(\frac{A}{2}ight)}{d / 2}=\frac{k_1 \epsilon_0 A}{d}=k_1 C_0$
The capacitance, $\mathrm{C}_{\mathrm{A} 3}=\frac{\mathrm{k}_2 \epsilon_0 \mathrm{~A} / 2}{\mathrm{~d} / 2}=\frac{\mathrm{k}_2 \epsilon_0 \mathrm{~A}}{\mathrm{~d}}=\mathrm{k}_2 \mathrm{C}_0$
$\mathrm{C}_{\mathrm{A} 2}$ and $\mathrm{C}_{\mathrm{A} 3}$ are in parallel,
$$ \mathrm{C}_{\mathrm{A}(2,3)}=\mathrm{k}_1 \mathrm{C}_0+\mathrm{k}_2 \mathrm{C}_0=\left(\mathrm{k}_1+\mathrm{k}_2ight) \mathrm{C}_0 $$
Now, $C_{A 1}$ and $C_{A(2,3)}$ are in series, so the equivalent capacitance of capacitor $A$ is
$$ \frac{1}{C_A}=\frac{1}{C_{A 1}}+\frac{1}{C_{A(2,3)}}=\frac{1}{2 k_1 C_0}+\frac{1}{\left(k_1+k_2ight) C_0} $$
$$\Rightarrow $$ $$ \frac{1}{\mathrm{C}_{\mathrm{A}}}=\frac{\mathrm{k}_1+\mathrm{k}_2+2 \mathrm{k}_1}{2 \mathrm{k}_1\left(\mathrm{k}_1+\mathrm{k}_2ight) \mathrm{C}_0}=\frac{3 \mathrm{k}_1+\mathrm{k}_2}{2 \mathrm{k}_1\left(\mathrm{k}_1+\mathrm{k}_2ight) \mathrm{C}_0} $$
So, $C_A=\frac{2 \mathrm{k}_1\left(\mathrm{k}_1+\mathrm{k}_2ight) \mathrm{C}_0}{3 \mathrm{k}_1+\mathrm{k}_2}$.
For Capacitor (B)
The capacitance, $C_{B 1}=\frac{\mathrm{k}_2 \epsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=\frac{2 \mathrm{k}_2 \epsilon_0 \mathrm{~A}}{\mathrm{~d}}=2 \mathrm{k}_2 \mathrm{C}_0$
The capacitance, $C_{B 2}=\frac{\mathrm{k}_1 \epsilon_0\left(\frac{\mathrm{~A}}{2}ight)}{\mathrm{d} / 2}=\frac{\mathrm{k}_1 \epsilon_0 \mathrm{~A}}{\mathrm{~d}}=\mathrm{k}_1 \mathrm{C}_0$
The capacitance, $C_{B 3}=\frac{\mathrm{k}_2 \epsilon_0 \mathrm{~A} / 2}{\mathrm{~d} / 2}=\frac{\mathrm{k}_2 \epsilon_0 \mathrm{~A}}{\mathrm{~d}}=\mathrm{k}_2 \mathrm{C}_0$
$\mathrm{C}_{\mathrm{B} 2}$ and $\mathrm{C}_{\mathrm{B} 3}$ are in parallel,
$$ \mathrm{C}_{\mathrm{B}(2,3)}=\mathrm{k}_1 \mathrm{C}_0+\mathrm{k}_2 \mathrm{C}_0=\left(\mathrm{k}_1+\mathrm{k}_2ight) \mathrm{C}_0 $$
Now, $C_{B 1}$ and $C_{B(2,3)}$ are in series, so the equivalent capacitance of capacitor $B$ is
$$ \frac{1}{C_B}=\frac{1}{C_{B 1}}+\frac{1}{C_{B(2,3)}}=\frac{1}{2 \mathrm{k}_2 \mathrm{C}_0}+\frac{1}{\left(\mathrm{k}_1+\mathrm{k}_2ight) \mathrm{C}_0} $$
$$\Rightarrow $$ $$ \frac{1}{\mathrm{C}_{\mathrm{A}}}=\frac{\mathrm{k}_1+\mathrm{k}_2+2 \mathrm{k}_2}{2 \mathrm{k}_2\left(\mathrm{k}_1+\mathrm{k}_2ight) \mathrm{C}_0}=\frac{\mathrm{k}_1+3 \mathrm{k}_2}{2 \mathrm{k}_2\left(\mathrm{k}_1+\mathrm{k}_2ight) \mathrm{C}_0} $$
So, $C_B=\frac{2 \mathrm{k}_2\left(\mathrm{k}_1+\mathrm{k}_2ight) \mathrm{C}_0}{\mathrm{k}_1+3 \mathrm{k}_2}$.
For Capacitor (C)
The capacitance, $C_{C 1}=\frac{\mathrm{k}_1 \epsilon_0 \mathrm{~A} / 2}{\mathrm{~d} / 2}=\frac{\mathrm{k}_1 \epsilon_0 \mathrm{~A}}{\mathrm{~d}}=\mathrm{k}_1 \mathrm{C}_0$
The capacitance, $C_{C 2}=\frac{\mathrm{k}_2 \epsilon_0 \mathrm{~A} / 2}{\mathrm{~d} / 2}=\frac{\mathrm{k}_2 \epsilon_0 \mathrm{~A}}{\mathrm{~d}}=\mathrm{k}_2 \mathrm{C}_0$
The capacitance, $C_{C 3}=\frac{\mathrm{k}_2 \epsilon_0 \mathrm{~A} / 2}{\mathrm{~d} / 2}=\frac{\mathrm{k}_2 \epsilon_0 \mathrm{~A}}{\mathrm{~d}}=\mathrm{k}_2 \mathrm{C}_0$
The capacitance, $\mathrm{C}_{\mathrm{C} 4}=\frac{\mathrm{k}_1 \epsilon_0 \mathrm{~A} / 2}{\mathrm{~d} / 2}=\frac{\mathrm{k}_1 \epsilon_0 \mathrm{~A}}{\mathrm{~d}}=\mathrm{k}_1 \mathrm{C}_0$
$\mathrm{C}_{\mathrm{C} 1}$ and $\mathrm{C}_{\mathrm{C} 2}$ are in parallel,
$$ \mathrm{C}_{\mathrm{C}(1,2)}=\mathrm{k}_1 \mathrm{C}_0+\mathrm{k}_2 \mathrm{C}_0=\left(\mathrm{k}_1+\mathrm{k}_2ight) \mathrm{C}_0 $$
Now, $\mathrm{C}_{\mathrm{C} 3}$ and $\mathrm{C}_{\mathrm{C} 4}$ are in parallel,
$$ \mathrm{C}_{\mathrm{C}(3,4)}=\mathrm{k}_2 \mathrm{C}_0+\mathrm{k}_1 \mathrm{C}_0=\left(\mathrm{k}_1+\mathrm{k}_2ight) \mathrm{C}_0 $$
Now, $\mathrm{C}_{\mathrm{C}(1,2)}$ and $\mathrm{C}_{\mathrm{C}(3,4)}$ are in series, so the equivalent capacitance of capacitor C is
$$ \frac{1}{\mathrm{C}_{\mathrm{C}}}=\frac{1}{\mathrm{C}_{\mathrm{C}(1,2)}}+\frac{1}{\mathrm{C}_{\mathrm{C}(3,4)}}=\frac{1}{\left(\mathrm{k}_1+\mathrm{k}_2ight) \mathrm{C}_0}+\frac{1}{\left(\mathrm{k}_1+\mathrm{k}_2ight) \mathrm{C}_0} $$
$$\Rightarrow $$ $\frac{1}{\mathrm{C}_{\mathrm{C}}}=\frac{2}{\left(\mathrm{k}_1+\mathrm{k}_2ight) \mathrm{C}_0}$
So, $\mathrm{C}_{\mathrm{C}}=\frac{\left(\mathrm{k}_1+\mathrm{k}_2ight) \mathrm{C}_0}{2}$.
It is given that $\mathrm{k}_1>\mathrm{k}_2$, let $\mathrm{k}_1=2$ and $\mathrm{k}_2=1$
$$ C_A=\frac{2 	imes 2 	imes 3}{7} C_0=\frac{12}{7} C_0 \approx 1.71 C_0 $$
$C_B=\frac{2 	imes 1 	imes 3}{5} C_0=\frac{6}{5} C_0=1.2 C_0$
$\mathrm{C}_{\mathrm{C}}=2 	imes 2 	imes \frac{1}{3} \mathrm{C}_0=\frac{4}{3} \mathrm{C}_0 \approx 1.33 \mathrm{C}_0$
$$ \Rightarrow \mathrm{C}_{\mathrm{A}}>\mathrm{C}_{\mathrm{C}}>\mathrm{C}_{\mathrm{B}} $$
Hence, the correct option is (C).

Answer

$C_C>C_B>C_A$

Answer

$C_B>C_C>C_A$

Answer

$C_A>C_C>C_B$

Answer

$C_C>C_A>C_B$

Electrostats & Capacitance

Three parallel plate capacitors each with area A and separation d are filled with two dielectric ( k_1 and k_2 ) in the following fashion. Which of the following is true? $ \left(k_1>k_2\right)

Choose the correct answer: