Question 508: Electrostats & Capacitance

Question

Air capacitor has capacitance of $1 \mu \mathrm{~F}$. Now the space between two plates of capacitor is filled with two dielectrics as shown in figure. The capacitance of the capacitor is [ $\mathrm{d}=$ distance between two plates of capacitor, $\mathrm{K}_1$ and $\mathrm{K}_2$ are dielectric constants of first dielectric and second dielectric respectively]

Accepted Answer

To determine the capacitance of the capacitor after inserting dielectrics, we start with the original capacitance of the air-filled parallel plate capacitor, given by:
$ C = \frac{\varepsilon_0 A}{d} = 1 \, \mu 	ext{F} $
Once the space between the plates is filled with two dielectrics, the scenario changes. The dielectrics divide the space between the plates into two sections, each with its own capacitance.
For the first dielectric with dielectric constant $ K_1 = 4 $, the capacitance becomes:
$ C_1 = K_1 \cdot \frac{\varepsilon_0 A}{2d} = 4 \cdot \frac{1}{2} = 2 \, \mu 	ext{F} $
Similarly, for the second dielectric with dielectric constant $ K_2 = 2 $, the capacitance is:
$ C_2 = K_2 \cdot \frac{\varepsilon_0 A}{2d} = 2 \cdot \frac{1}{2} = 1 \, \mu 	ext{F} $
Given that both sections are in parallel, the effective capacitance $ C_{	ext{eff}} $ of the capacitor is the sum of the two individual capacitances:
$ C_{	ext{eff}} = C_1 + C_2 = 2 \, \mu 	ext{F} + 1 \, \mu 	ext{F} = 3 \, \mu 	ext{F} $

Answer

$3 \mu \mathrm{~F}$

Answer

$6 \mu \mathrm{~F}$

Answer

$8 \mu \mathrm{~F}$

Answer

$12 \mu \mathrm{~F}$

Electrostats & Capacitance

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