Air capacitor has capacitance of $1 \mu \mathrm{~F}$. Now the space between two plates of capacitor is…

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 $\cdot \mathrm{K}_2$ are dielectric constants of first dielectric and second dielectric respectively]
  1. $3 \mu \mathrm{~F}$
  2. $6 \mu \mathrm{~F}$
  3. $8 \mu \mathrm{~F}$
  4. $12 \mu \mathrm{~F}$

Solution

Capacitance of a parallel plate capacitor, $\mathrm{C}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=1 \mu \mathrm{~F}$ ... (given) After inserting the dielectrics, $\mathrm{C}_1=\mathrm{K}_1 \cdot \frac{\varepsilon_0 \mathrm{~A}}{2 \mathrm{~d}}=4 \times \frac{1}{2}=2 \mu \mathrm{~F}$ ... (given, $\mathrm{K}_1=4$ ) $\begin{array}{ll} & \mathrm{C}_2=\mathrm{K}_2 \cdot \frac{\varepsilon_0 \mathrm{~A}}{2 \mathrm{~d}}=2 \times \frac{1}{2}=1 \mu \mathrm{~F} \\ \therefore \quad & \mathrm{C}_{\text {eff }}=\mathrm{C}_1+\mathrm{C}_2=3 \mu \mathrm{~F} \end{array}$ ...(given, $\mathrm{K}_2=4$ )

Asked in: MHT CET 2024 (09 May Shift 2)

Practice more Electrostatics questions on Aicharya