Question 518: Electrostats & Capacitance

Question

A parallel plate capacitor has plate area $40 \mathrm{~cm}^2$ and plate separation 2 mm . The space between the plates is filled with a dielectric medium of thickness 1 mm and dielectric constant 5 . The capacitance of the system is ( $\varepsilon_0=$ permittivity of vacuum)

Accepted Answer

For a parallel plate capacitor composed of two capacitors in series, the equivalent capacitance can be calculated using the formula:
$ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} $
The capacitance $ C_1 $ of the portion with the dielectric is given by:
$ C_1 = \frac{A \varepsilon_0 k}{t} $
And the capacitance $ C_2 $ of the portion without the dielectric is:
$ C_2 = \frac{A \varepsilon_0}{d-t} $
where:
$ A $ is the plate area,
$ \varepsilon_0 $ is the permittivity of vacuum,
$ k $ is the dielectric constant,
$ t $ is the thickness of the dielectric,
$ d $ is the total separation between the plates.
Substituting the given values:
$ \frac{1}{C_{eq}} = \frac{1}{\frac{40 	imes 10^{-4} \varepsilon_0 	imes 5}{1 	imes 10^{-3}}} + \frac{1}{\frac{40 	imes 10^{-4} \varepsilon_0}{2 	imes 10^{-3} - 1 	imes 10^{-3}}} $
Simplifying, we calculate:
$ \frac{1}{C_{eq}} = \frac{1}{20 \varepsilon_0} + \frac{1}{4 \varepsilon_0} $
Therefore, the equivalent capacitance $ C_{eq} $ is:
$ C_{eq} = \frac{10}{3} \varepsilon_0 \, 	ext{F} $

Answer

$24 \varepsilon_0 \mathrm{~F}$

Answer

$\frac{3}{10} \varepsilon_0 \mathrm{~F}$

Answer

$\frac{10}{3} \varepsilon_0 \mathrm{~F}$

Answer

$10 \varepsilon_0 \mathrm{~F}$

Electrostats & Capacitance

A parallel plate capacitor has plate area 40 \mathrm{~cm}^2 and plate separation 2 mm . The space between the plates is filled with a dielectric medium of thickness 1 mm and dielectric constant 5 . The capacitance of the system is ( \varepsilon_0= permittivity of vacuum)

Choose the correct answer: