Question 342: Electrostats & Capacitance

Question

The space between the plates of a parallel plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants $K_1=2, K_2=3$ and $K_3=5$ (as shown in figure). If new capacitance is $\frac{n}{3} C$ then the value of $n$ is $\_\_\_\_$ .

Accepted Answer

The capacitance of a parallel plate capacitor without any dielectric is given by:
$$ C=\frac{\epsilon_0 A}{d} $$
where A is the area of the plates and d is the separation between them.
Capacitor $\mathrm{C}_1$
Dielectric constant $\mathrm{K}_1=2$.
Thickness $\mathrm{d}_1=\frac{\mathrm{d}}{2}$.
Area $\mathrm{A}_1=\mathrm{A}$
So, the capacitance is;
$$ C_1=\frac{K_1 \epsilon_0 A_1}{d_1}=\frac{2 \epsilon_0 A}{d / 2}=4\left(\frac{\epsilon_0 A}{d}\right)=4 C $$
Capacitor $\mathrm{C}_2$
Dielectric constant $\mathrm{K}_2=3$.
Thickness $\mathrm{d}_2=\frac{\mathrm{d}}{2}$.
Area $\mathrm{A}_1=\mathrm{A} / 2$
So, the capacitance is;
$\mathrm{C}_2=\frac{\mathrm{K}_2 \epsilon_0 \mathrm{~A}_2}{\mathrm{~d}_2}=\frac{3 \epsilon_0(\mathrm{~A} / 2)}{\mathrm{d} / 2}=3\left(\frac{\epsilon_0 \mathrm{~A}}{\mathrm{~d}}\right)=3 \mathrm{C}$
Capacitor $\mathrm{C}_3$
Dielectric constant $\mathrm{K}_3=5$.
Thickness $\mathrm{d}_3=\frac{\mathrm{d}}{2}$.
Area $\mathrm{A}_2=\mathrm{A} / 2$
So, the capacitance is;
$$ \mathrm{C}_3=\frac{\mathrm{K}_3 \epsilon_0 \mathrm{~A}_3}{\mathrm{~d}_3}=\frac{5 \epsilon_0(\mathrm{~A} / 2)}{\mathrm{d} / 2}=5\left(\frac{\epsilon_0 \mathrm{~A}}{\mathrm{~d}}\right)=5 \mathrm{C} $$
Capacitors $C_2$ and $C_3$ are in parallel.
$$ C_{23}=C_2+C_3 $$
$$\Rightarrow $$ $$ C_{23}=3 C+5 C=8 C $$
This combined section $\left(\mathrm{C}_{23}\right)$ is in series with the first capacitor $\left(\mathrm{C}_1\right)$.
$\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_{23}}$
$$\Rightarrow $$ $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{4 \mathrm{C}}+\frac{1}{8 \mathrm{C}}$
$$\Rightarrow $$ $\frac{1}{C_{\mathrm{eq}}}=\frac{2+1}{8 C}=\frac{3}{8 C}$
$$\Rightarrow $$ $$ C_{\mathrm{eq}}=\frac{8 C}{3}=\frac{n}{3} C \Rightarrow n=8 $$
Therefore, the value of n is 8 . Hence, the correct answer is 8 .

Electrostats & Capacitance

The space between the plates of a parallel plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants K_1=2, K_2=3 and K_3=5 (as shown in figure). If new capacitance is \frac{n}{3} C then the value of n is \_\_\_\_ .

Choose the correct answer: