Question

A parallel plate capacitor has plate area $40\, cm ^2$ and plates separation $2\, mm$ The space between the plates is filled with a dielectric medium of a thickness $1 \,mm$ and dielectric constant $5$ The capacitance of the system is :

• A. $\frac{10}{3} \varepsilon_0 F$
• B. $\frac{3}{10} \varepsilon_0 F$
• C. $24 \varepsilon_0 F$
• D. $10 \varepsilon_0 F$

Answer 1

The Correct Option is A

This can be seen as two capacitors in series combination so
$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$
$=\frac{1}{\frac{K{\in }_{0}A}{t}}+\frac{1}{\frac{{\in }_{0}A}{d-t}}$
$=\frac{t}{K{\in }_{0}A}+\frac{d-t}{{ϵ}_{0}A}$
$=\frac{1\times 10^{-3}}{5{ϵ}_0\times 40\times 10^{-4}}+\frac{1\times 10^{-3}}{{ϵ}_{0}40\times 10^{-4}}$
$\frac{1}{C_{eq}}=\frac{1}{20{ϵ}_0}+\frac{1}{4{ϵ}_0}$
$C_{eq}=\frac{20\times 4{ϵ}_0}{24}=\frac{10{ϵ}_0}{3}F$

Answer 2

1. The system is equivalent to two capacitors in series:
- Capacitor with dielectric (\(C_1\)): \(d_1 = 1 \, \text{mm}, \, \kappa = 5\).
- Capacitor without dielectric (\(C_2\)): \(d_2 = 1 \, \text{mm}\).
2. Individual capacitances: \[ C_1 = \frac{\varepsilon_0 A \kappa}{d_1}, \quad C_2 = \frac{\varepsilon_0 A}{d_2}. \]
3. Substitute \(A = 40 \, \text{cm}^2 = 40 \times 10^{-4} \, \text{m}^2\): \[ C_1 = \frac{\varepsilon_0 \times 40 \times 10^{-4} \times 5}{1 \times 10^{-3}} = 200 \varepsilon_0, \quad C_2 = \frac{\varepsilon_0 \times 40 \times 10^{-4}}{1 \times 10^{-3}} = 40 \varepsilon_0. \]
4. Combine in series: \[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}. \]
\[ C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2} = \frac{200 \varepsilon_0 \times 40 \varepsilon_0}{200 \varepsilon_0 + 40 \varepsilon_0} = \frac{8000}{240} = \frac{10}{3} \varepsilon_0. \]
Thus, the capacitance is \(\frac{10}{3} \varepsilon_0 \, \text{F}\). For capacitors with multiple dielectrics, calculate individual capacitances and combine them using series or parallel formulas.