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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.........$pF$
$1.8$
$45$
$40.5$
$20.25$
Solution
${\frac{\varepsilon_{\mathrm{o}} \mathrm{A}}{\mathrm{d}}=9 \mathrm{\,PF}} $
${{\rm{C}}^1} = \frac{{\frac{{{\varepsilon _{\rm{o}}} \times 3{\rm{A}}}}{{{\rm{d}}/3}} \times \frac{{{\varepsilon _{\rm{o}}} \times 6{\rm{A}}}}{{2{\rm{d}}/3}}}}{{\frac{{{\varepsilon _{\rm{o}}} \times 3{\rm{A}}}}{{{\rm{d}}/3}} + \frac{{{\varepsilon _{\rm{o}}} \times 6{\rm{A}}}}{{2{\rm{d}}/3}}}} = \frac{{\frac{{9{\varepsilon _{\rm{o}}}{\rm{A}}}}{{\rm{d}}} \times \frac{{9{\varepsilon _{\rm{o}}}{\rm{A}}}}{{2{\rm{d}}}}}}{{\frac{{18{\varepsilon _{\rm{o}}}{\rm{A}}}}{{2{\rm{d}}}}}} = \frac{9}{2}\frac{{{\varepsilon _{\rm{o}}}{\rm{A}}}}{{\rm{d}}}$
$ {\mathrm{C}^{1}=\left(\frac{9}{2} \times 9\right) \mathrm{\,pF}=40.5 \mathrm{\,pF}}$
