Question

A parallel plate capacitor with air between the plates has a capacitance of $9pF$. 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 $d/3$ while the other one has dielectric constant $K_2=6$ and thickness $2d/3.$ Capacitance of the capacitor is now

• A. 20.25 $pF$
• B. 1.8 $pF$
• C. 45 $pF$
• D. 40.5 $pF$

Answer 1

Answer: (D)

$C=\frac{{\epsilon }_{0}A}{d}=9\times 10^{-12}F$
With dielectric, $C=\frac{{\epsilon }_{0}KA}{d}$
$C_1=\frac{{\epsilon }_{0}A\cdot 3}{d/3}=9C$;
$C_2=\frac{{\epsilon }_{0}A\cdot 6}{2d/3}=9C$
$\therefore C_{\text{total }}=\frac{C_1{C}_2}{C_1+C_2}$
as they are in series.
$C_{\text{total}}=\frac{9C\times 9C}{18C}=\frac{9}{2}\times C$
$=\frac{9}{2}\times 9\times 10^{-12}F$
$=40.5pF$