Question 516: Electrostats & Capacitance

Question

A parallel plate air capacitor, with plate separation ' d ' has a capacitance of 9 pF . The space between the plates is now filled with two dielectrics, the first having $\mathrm{K}_1=3$ and thickness $\mathrm{d}_1=\mathrm{d} / 3$, while the $2^{	ext {nd }}$ has $\mathrm{K}_2=6$ and thickness $d_2=2 \mathrm{~d} / 3$. The capacitance of the new capacitor is :

Accepted Answer

For a parallel plate capacitor with a dielectric material, the capacitance is described by the formula:
$ C = \frac{A \varepsilon_0 k}{d} $
where,
$ C $ is the capacitance,
$ A $ is the area of one of the plates,
$ \varepsilon_0 $ is the permittivity of free space,
$ k $ is the dielectric constant of the material,
$ d $ is the separation between the plates.
Calculating Individual Capacitances
For each section with a different dielectric:
First Dielectric:
Dielectric constant $ k_1 = 3 $
Thickness $ d_1 = \frac{d}{3} $
The capacitance $ C_1 $ is:
$    C_1 = \frac{A \varepsilon_0 k_1}{d_1} = \frac{A \varepsilon_0 	imes 3}{d/3} = 9 \frac{A \varepsilon_0}{d} = 9C    $
Second Dielectric:
Dielectric constant $ k_2 = 6 $
Thickness $ d_2 = \frac{2d}{3} $
The capacitance $ C_2 $ is:
$    C_2 = \frac{A \varepsilon_0 k_2}{d_2} = \frac{A \varepsilon_0 	imes 6}{2d/3} = 9 \frac{A \varepsilon_0}{d} = 9C    $
Calculating Total Capacitance
Since these two sections are in series, the total capacitance $ C_{	ext{total}} $ is given by:
$ C_{	ext{total}} = \frac{C_1 	imes C_2}{C_1 + C_2} = \frac{9C 	imes 9C}{9C + 9C} = \frac{81C^2}{18C} = \frac{9}{2}C $
Given that the original capacitor had a capacitance of $ 9 \, 	ext{pF} $, substitute this into the total capacitance equation:
$ C_{	ext{total}} = \frac{9}{2} 	imes 9 	imes 10^{-12} 	ext{F} $
Thus,
$ C_{	ext{total}} = 40.5 \, 	ext{pF} $
Therefore, the capacitance of the new capacitor is $ 40.5 \, 	ext{pF} $.

Answer

3.8 pF

Answer

20.25 pF

Answer

40.5 pF

Answer

45 pF

Electrostats & Capacitance

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