MCQ
A parallel plate capacitor with air between the plates has a capacitance of $9\,pF$. The separation between its plates is $'d'$. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $K_1=3$ and thickness $\frac {d}{3}$ while the other one has dielectric constant $K_2 = 6$ and thickness $\frac {2d}{3}$ . Capacitance of the capacitor is now........$pF$
  • A
    $1.8$
  • B
    $45$
  • $40.5$
  • D
    $20.25$

Answer

Correct option: C.
$40.5$
c
$\frac{\varepsilon_{\mathrm{o}} \mathrm{A}}{\mathrm{d}}=9 \mathrm{\,PF}$

$\mathrm{C}^{\prime}=\frac{\frac{\varepsilon_{\mathrm{o}} \times 3 \mathrm{A}}{\mathrm{d} / 3} \times \frac{\varepsilon_{\mathrm{o}} \times 6 \mathrm{A}}{2 \mathrm{d} / 3}}{\frac{\varepsilon_{\mathrm{o}} \times 3 \mathrm{A}}{\mathrm{d} / 3}+\frac{\varepsilon_{\mathrm{o}} \times 6 \mathrm{A}}{2 \mathrm{d} / 3}}=\frac{\frac{9 \varepsilon_{\mathrm{o}} \mathrm{A}}{\mathrm{d}} \times \frac{9 \varepsilon_{\mathrm{o}} \mathrm{A}}{2 \mathrm{d}}}{\frac{18 \varepsilon_{\mathrm{o}} \mathrm{A}}{2 \mathrm{d}}}=\frac{9}{2} \frac{\varepsilon_{\mathrm{o}} \mathrm{A}}{\mathrm{d}}$

$C^{\prime}=\left(\frac{9}{2} \times 9\right) \,p F=40.5 \,p F$

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