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 dielectric 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 :

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

Answer 1

Answer: (A)

$40.5$ $pF$

The capacitance with air is $C_0=\frac{A{ϵ}_0}{d}=9pF$

After inserting dielectric, there are two capacitors $C_1,C_2$ and they are in series.

here, $C_1=\frac{A{K}_1{ϵ}_0}{d/3}=\frac{9A{ϵ}_0}{d}=9\times 9=81pF$

and $C_2=\frac{A{K}_2{ϵ}_0}{2d/3}=\frac{9A{ϵ}_0}{d}=9\times 9=81pF$

Thus new capacitance $C_{eq}=\frac{C_1{C}_2}{C_1+C_2}=\frac{81\times 81}{81+81}=40.5pF$