Question

Three different dielectrics are filled in a parallel plate capacitor as shown. What should be the value of dielectric constant of a material , which when fully filled between the plates produces same capacitance ? (Asssume terminals are connected between $AC$ and $BD$).

• A. $5$
• B. $6$
• C. $9$
• D. $4$

Answer 1

The correct option is A $5$

From the given arrangement of dielectric slabs we can infer that it forms three capacitors in total, out of which two are in series and the third one $(k_1=6)$ is in parallel combination with them as shown below.


Here for capacitor $C_1$:
$A_1=\frac{A}{2},d_1=d$

Thus,
$C_1=\frac{k_1{A}_1{\epsilon }_0}{d_1}$

$\Rightarrow C_1=\frac{6\left(\frac{A}{2}\right){\epsilon }_0}{d}=\frac{3A{\epsilon }_0}{d}$

Similarly for capacitors $C_2$ & $C_3$,
$A_2=\frac{A}{2},A_3=\frac{A}{2}$
$d_2=\frac{d}{2},d_3=\frac{d}{2}$
Hence, $C_2=\frac{k_2{A}_2{\epsilon }_0}{d_2}=\frac{3\left(\frac{A}{2}\right){\epsilon }_0}{\left(\frac{d}{2}\right)}=\frac{3A{\epsilon }_0}{d}$

And, $C_3=\frac{k_3{A}_3{\epsilon }_0}{d_3}=\frac{6\left(\frac{A}{2}\right){\epsilon }_0}{\left(\frac{d}{2}\right)}=\frac{6A{\epsilon }_0}{d}$

Let $C_1=\frac{3A{\epsilon }_0}{d}=C$

Thus, $C_2=C$ & $C_3=2C$


For the series combination:

$C_{eq}=\frac{2C\times C}{2C+C}=\frac{2C^2}{3C}=\frac{2C}{3}$

For the above parallel arrangement,
$C_{net}=C+\frac{2C}{3}$

$\Rightarrow C_{net}=\frac{5C}{3}$

$\Rightarrow C_{net}=\frac{5}{3}\left(\frac{3A{\epsilon }_0}{d}\right)=\frac{5A{\epsilon }_0}{d}....\left(i\right)$

Let, the material having dielectric constant $k^{\prime }$ fills the given capacitor plates fully. Then

$C^{\prime }=\frac{k^{\prime }A{\epsilon }_0}{d}....\left(ii\right)$

Comparing equations (i) & (ii) we get,

$\frac{k^{\prime }A{\epsilon }_0}{d}=\frac{5A{\epsilon }_0}{d}$

$\therefore k^{\prime }=5$

Hence, option (c) is the correct answer.

Why this question? In problems involving more than two dielectric slabs, always focus on observing area of plate & separation between them for individual capacitors then segregate them for series & parallel grouping.