Question 488: Electrostats & Capacitance

Question

Air capacitor has capacitance ' $\mathrm{C}_1$ '. The space between two plates of capacitor is filled with two dielectrics as shown in figure. The new capacitance of the capacitor is ' $\mathrm{C}_2$ '. The ratio $\frac{C_1}{C_2}$ is $(d=$ distance between two plates of capacitor, $\mathrm{K}_1$ and $\mathrm{K}_2$ are dielectric constants of two dielectrics respectively)

Accepted Answer

When an air capacitor has a capacitance of $ C_1 $, it is represented as:
$ C_1 = \frac{A \varepsilon_0}{d} $
where $ A $ is the area of the plates, $ \varepsilon_0 $ is the permittivity of free space, and $ d $ is the distance between the plates.
When the space between the plates of this capacitor is filled with two different dielectric materials, it affects the overall capacitance. The two dielectrics are inserted as shown, and this setup functions as capacitors in series. To find the new capacitance $ C_2 $ with dielectrics, we calculate as follows:
First, the capacitance with each dielectric in place is calculated:
For the first dielectric with constant $ K_1 $:
$ C_{K1} = \frac{A \varepsilon_0 K_1}{\frac{d}{2}} $
For the second dielectric with constant $ K_2 $:
$ C_{K2} = \frac{A \varepsilon_0 K_2}{\frac{d}{2}} $
The capacitance of the configuration can then be represented as capacitors in series:
$ \frac{1}{C_2} = \frac{1}{C_{K1}} + \frac{1}{C_{K2}} $
Substitute the expressions for $ C_{K1} $ and $ C_{K2} $:
$ \frac{1}{C_2} = \frac{1}{\frac{A \varepsilon_0 K_1}{d/2}} + \frac{1}{\frac{A \varepsilon_0 K_2}{d/2}} $
$ \frac{1}{C_2} = \frac{\frac{d}{2}}{A \varepsilon_0 K_1} + \frac{\frac{d}{2}}{A \varepsilon_0 K_2} $
Simplify to:
$ \frac{1}{C_2} = \frac{d}{2 \varepsilon_0 A} \left( \frac{1}{K_1} + \frac{1}{K_2} \right) $
$ C_2 = \frac{2 \varepsilon_0 A}{d} \left( \frac{K_1 K_2}{K_1 + K_2} \right) $
Therefore, the ratio of the original capacitance $ C_1 $ to the new capacitance $ C_2 $ is:
$ \frac{C_1}{C_2} = \frac{K_1 + K_2}{2 K_1 K_2} $

Answer

$\mathrm{K}_1+\mathrm{K}_2$

Answer

$\frac{\mathrm{K}_1+\mathrm{K}_2}{\mathrm{~K}_1-\mathrm{K}_2}$

Answer

$\frac{2 \mathrm{~K}_1 \mathrm{~K}_2}{\mathrm{~K}_1+\mathrm{K}_2}$

Answer

$\frac{\mathrm{K}_1+\mathrm{K}_2}{2 \mathrm{~K}_1 \mathrm{~K}_2}$

Electrostats & Capacitance

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