Question

A parallel plate capacitor is made using three different dielectric materials as shown in figure. Separation between the plates of the capacitor is $d$. The equivalent capacitance between the plates is

• A. $\frac{{\epsilon }_{0}KA}{d}$
• B. $\frac{2{\epsilon }_{0}KA}{d}$
• C. $\frac{3{\epsilon }_{0}KA}{d}$
• D. $\frac{5{\epsilon }_{0}KA}{d}$

Answer 1

The correct option is B $\frac{2{\epsilon }_{0}KA}{d}$

As we know, in parallel plate capacitor, if dielectric divides the area then parts formed will be in parallel and if dielectric divides the distance between the plates the plates parts formed are in series. Hence we can redraw given combination of dielectrics as follows.


$C_1=\frac{{\epsilon }_0\frac{KA}{2}}{\frac{d}{2}}=\frac{{\epsilon }_{0}KA}{d}$

$C_2=\frac{2{\epsilon }_0\frac{KA}{2}}{\left(\frac{d}{2}\right)}=\frac{2{\epsilon }_{0}KA}{d}$

$C_3=\frac{3{\epsilon }_{0}KA}{\left(\frac{d}{2}\right)}=\frac{6{\epsilon }_{0}KA}{d}$

$C_1$ and $C_2$ are in parallel and $C_3$ is in series with this combination,

$C_{12}=C_1+C_2=\frac{{\epsilon }_{0}KA}{d}+\frac{2{\epsilon }_{0}KA}{d}$

$C_{12}=\frac{3{\epsilon }_{0}KA}{d}$

$C_{eff}=\frac{C_{12}\times C_3}{C_{12}+C_3}=\frac{\left(\frac{3{\epsilon }_{0}KA}{d}\right)\left(\frac{6{\epsilon }_{0}KA}{d}\right)}{\frac{3{\epsilon }_{0}KA}{d}+\frac{6{\epsilon }_{0}KA}{d}}$

$=\frac{18\left(\frac{{\epsilon }_{0}KA}{d}\right)}{9}$

$=\frac{2{\epsilon }_{0}KA}{d}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (b) is the correct answer.