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(D) The arrangement consists of three capacitors connected in series.For a parallel plate capacitor with a dielectric slab of thickness $t$ and dielec

Vedclass · Accepted Answer

$\frac{A \varepsilon_0}{\left[\frac{t_1}{k_1} + \frac{t_2}{k_2} + \frac{t_3}{k_3}\right]}$

Vedclass · Answer

(D) The arrangement consists of three capacitors connected in series.For a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $k$,the capacitance is $C = \frac{k \varepsilon_0 A}{t}$.Since the slabs are placed in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.Substituting the values,we get $\frac{1}{C_{eq}} = \frac{t_1}{k_1 \varepsilon_0 A} + \frac{t_2}{k_2 \varepsilon_0 A} + \frac{t_3}{k_3 \varepsilon_0 A}$.Taking $\frac{1}{\varepsilon_0 A}$ as a common factor,$\frac{1}{C_{eq}} = \frac{1}{\varepsilon_0 A} \left[\frac{t_1}{k_1} + \frac{t_2}{k_2} + \frac{t_3}{k_3}\right]$.Therefore,the resultant capacity is $C_{eq} = \frac{A \varepsilon_0}{\left[\frac{t_1}{k_1} + \frac{t_2}{k_2} + \frac{t_3}{k_3}\right]}$.

Two parallel plates with dielectric slabs placed between them are as shown in the figure. The resultant capacity of the capacitor will be: [$A$ = area of plate,$t_1, t_2, t_3$ are thicknesses of dielectric slabs,$k_1, k_2, k_3$ are dielectric constants.]

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