Question

In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be : (Given area of plate = A) 

 

• A. $\frac{25}{6}\frac{K\epsilon_0 A}{d}$
• B. $\frac{15}{34}\frac{K\epsilon_0 A}{d}$
• C. $\frac{9}{6}\frac{K\epsilon_0 A}{d}$
• D. $\frac{15}{6}\frac{K\epsilon_0 A}{d}$

Hint:

When dielectric slabs completely fill the space and are stacked between the plates (dividing the distance), they are in series. If they are placed side-by-side (dividing the area), they are in parallel.

Answer 1

The Correct Option is B

The arrangement shows three dielectric slabs placed between the plates of a capacitor. Since the slabs are placed one after another, this configuration is equivalent to three capacitors connected in series.
Let the three capacitors be $C_1$, $C_2$, and $C_3$.
For $C_1$: Dielectric constant = K, thickness = d, area = A.
$C_1 = \frac{K \epsilon_0 A}{d}$
For $C_2$: Dielectric constant = 3K, thickness = 2d, area = A.
$C_2 = \frac{3K \epsilon_0 A}{2d}$
For $C_3$: Dielectric constant = 5K, thickness = 3d, area = A.
$C_3 = \frac{5K \epsilon_0 A}{3d}$
For capacitors in series, the reciprocal of the equivalent capacitance ($C_{eq}$) is the sum of the reciprocals of individual capacitances.
$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$
$\frac{1}{C_{eq}} = \frac{d}{K \epsilon_0 A} + \frac{2d}{3K \epsilon_0 A} + \frac{3d}{5K \epsilon_0 A}$
Factor out the common term $\frac{d}{K \epsilon_0 A}$:
$\frac{1}{C_{eq}} = \frac{d}{K \epsilon_0 A} \left( 1 + \frac{2}{3} + \frac{3}{5} \right)$
Find a common denominator for the terms in the parenthesis (which is 15):
$\frac{1}{C_{eq}} = \frac{d}{K \epsilon_0 A} \left( \frac{15}{15} + \frac{10}{15} + \frac{9}{15} \right)$
$\frac{1}{C_{eq}} = \frac{d}{K \epsilon_0 A} \left( \frac{15+10+9}{15} \right) = \frac{d}{K \epsilon_0 A} \left( \frac{34}{15} \right)$
Now, invert the expression to find $C_{eq}$:
$C_{eq} = \frac{15}{34} \frac{K \epsilon_0 A}{d}$