Question

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 $d / 3$ while the other one has dielectric constant $K_{2}=6$ and thickness $2 d / 3 .$ 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)

$C=\frac{\varepsilon_{0} A}{d}=9 \times 10^{-12} F$
With dielectric, $C=\frac{\varepsilon_{0} K A}{d}$
$C_{1}=\frac{\varepsilon_{0} A \cdot 3}{d / 3}=9 C$;
$ C_{2}=\frac{\varepsilon_{0} A \cdot 6}{2 d / 3}=9 C$
$\therefore C_{\text {total }}=\frac{C_{1} C_{2}}{C_{1}+C_{2}}$
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 .5\,pF$