Question

Find the capacitances of the capacitors shown in figure (31-E24). The plate area is A and the separation between the plates is d. Different dielectric slabs in a particular part of the figure are of the same thickness and the entire gap between the plates is filled with the dielectric slabs.
Figure

Answer 1

The overall capacitance is equal to the sum of individual capacitance of the circuit.

Explanation:

$C_{1}=\frac{\varepsilon_{0} A k_{1}}{d / 2}$

$C_{2}=\frac{\varepsilon_{0} A k_{2}}{d / 2}$

Where C = Capacitance

A = Area of the plates

d = distance between plates

k= dielectric constant

$C=\frac{C_{1} C_{2}}{C_{1}+C_{2}}$

   = $\frac{\frac{2 \varepsilon_{0} A k_{1}}{d} \times \frac{2 \varepsilon_{0} A k_{2}}{d}}{\frac{2 \varepsilon_{0} A k_{1}}{d}+\frac{2 \varepsilon_{0} A k_{2}}{d}}$

$=\frac{\frac{\left(2 \varepsilon_{0} A\right)^{2} k_{1} k_{2}}{d^{2}}}{\left(2 \varepsilon_{0} A\right) \frac{k_{1} d+k_{2} d}{d^{2}}}=\frac{2 k_{1} k_{2} \varepsilon_{0} A}{d\left(k_{1}+k_{2}\right)}$

Similarly,

When capacitors connected in series the total equivalent capacitance of the circuit is equal to the sum of the individual capacitance of the circuit. By representing the above statement into the formula the total circuit capacitance is

$\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}$

$=\frac{\frac{1}{3 \varepsilon_{0} A k_{1}}}{d}+\frac{\frac{1}{3 \varepsilon_{0} A k_{2}}}{d}+\frac{1}{\frac{3 \varepsilon_{0} A k_{3}}{d}}$

$=\frac{d}{3 \varepsilon_{0} A}\left[\frac{1}{k_{1}}+\frac{1}{k_{2}}+\frac{1}{k_{3}}\right]$

$=\frac{d}{3 \varepsilon_{0} A}\left[\frac{k_{2} k_{3}+k_{1} k_{3}+k_{1} k_{2}}{k_{1} k_{2} k_{3}}\right]$

$\therefore C=\frac{3 \varepsilon_{0} A k_{1} k_{2} k_{3}}{d\left(k_{1} k_{2}+k_{2} k_{3}+k_{1} k_{3}\right)}$

Hence the total capacitance of the circuit is equal to the sum of the individual capacitance of the circuit.