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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
$\frac{\varepsilon_{\mathrm{o}} \mathrm{A}}{\mathrm{d}}=9 \mathrm{\,PF}$
${{\rm{C}}^\prime } = \frac{{\frac{{{\varepsilon _{\rm{o}}} \times 3{\rm{A}}}}{{{\rm{d}}/3}} \times \frac{{{\varepsilon _{\rm{o}}} \times 6{\rm{A}}}}{{2{\rm{d}}/3}}}}{{\frac{{{\varepsilon _{\rm{o}}} \times 3{\rm{A}}}}{{{\rm{d}}/3}} + \frac{{{\varepsilon _{\rm{o}}} \times 6{\rm{A}}}}{{2{\rm{d}}/3}}}}$
$ = \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}}}$
$\mathrm{C}^{\prime}=\left(\frac{9}{2} \times 9\right) \mathrm{pF}=40.5 \mathrm{\,pF}$
Similar Questions
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