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

To find the capacitance of the parallel plate capacitor with a dielectric medium, we should first understand the structure of the system. The arrangement consists of three serially connected regions: a region with air, the dielectric region, and another region with air. Let us evaluate the capacitance considering the system as a series of capacitors.

1. The given parameters are:
   • Plate area, \(A = 40 \, \text{cm}^2 = 40 \times 10^{-4} \, \text{m}^2\)
   • Plate separation, \(d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m}\)
   • Dielectric thickness, \(t = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m}\)
   • Dielectric constant, \(\kappa = 5\)
2. There are effectively two regions of capacitors:
   • The region with dielectric thickness \(t\)
   • The region filled with air with thickness \((d-t)\)
3. Calculate the capacitance of the dielectric-filled region:

The capacitance with dielectric, \(C_1 = \frac{\kappa \varepsilon_0 A}{t}\)

Plugging the values, \(C_1 = \frac{5 \varepsilon_0 (40 \times 10^{-4})}{1 \times 10^{-3}} = 200 \varepsilon_0 \, \text{F}\)

1. Calculate the capacitance of the air-filled region:

The capacitance without dielectric, \(C_2 = \frac{\varepsilon_0 A}{d - t}\)

Substitute the values, \(C_2 = \frac{\varepsilon_0 (40 \times 10^{-4})}{1 \times 10^{-3}} = 40 \varepsilon_0 \, \text{F}\)

1. Equivalent capacitance of the system:

Capacitances are in series, thus \(\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}\)

Substitute the capacitances to find \(C_{\text{eq}}\): \(\frac{1}{C_{\text{eq}}} = \frac{1}{200 \varepsilon_0} + \frac{1}{40 \varepsilon_0} = \frac{1 + 5}{200 \varepsilon_0} = \frac{6}{200 \varepsilon_0} = \frac{3}{100 \varepsilon_0}\)

Solve for capacitance, \(C_{\text{eq}} = \frac{100}{3} \varepsilon_0 = \frac{10}{3} \varepsilon_0 \, \text{F}\)

1. Conclusion: The capacitance of the system with the dielectric is \(\frac{10}{3} \varepsilon_0 \, \text{F}\), which matches one of the provided options.

Therefore, the correct answer is

$\frac{10}{3} \varepsilon_0 \, \text{F}$