question_answer

Three dielectric slabs of dielectric constants \[{{k}_{1}},\,{{k}_{2}}\] and \[{{k}_{3}}\] have been put in between the plates of a capacitor as shown in figure. Determine the equivalent capacitance. Dielectric slabs

Answer:

The given arrangement can be represented as For each portion, we use             \[C=\frac{{{\varepsilon }_{0}}A}{d-{{t}_{1}}-{{t}_{2}}+\frac{{{t}_{1}}}{{{K}_{1}}}+\frac{{{t}_{2}}}{{{K}_{2}}}}\] For parallel combination,\[{{C}_{p}}={{C}_{1}}+{{C}_{2}}\] \[\therefore \]      \[\frac{K{{\varepsilon }_{0}}A}{d}=\frac{{{\varepsilon }_{0}}\frac{A}{2}}{d-\frac{d}{2}-\frac{d}{2}+\frac{d}{2{{K}_{1}}}+\frac{d}{2{{K}_{3}}}}\]             \[+\frac{{{\varepsilon }_{0}}(A/2)}{d-\frac{d}{2}-\frac{d}{2}+\frac{d}{2{{K}_{1}}}+\frac{d}{2{{K}_{3}}}}\] \[\frac{K{{\varepsilon }_{0}}A}{d}=\frac{{{\varepsilon }_{0}}A}{\frac{d}{{{K}_{1}}}+\frac{d}{{{K}_{2}}}}+\frac{{{\varepsilon }_{0}}A}{\frac{d}{{{K}_{1}}}+\frac{d}{{{K}_{3}}}}\]             \[K=\frac{{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}}+\frac{{{K}_{1}}{{K}_{3}}}{{{K}_{1}}+{{K}_{3}}}\]