| A parallel palate capacitor with square plates is filled with four dielectrics of dielectric constants \[{{K}_{1,}}\]\[{{K}_{2,}}\]\[{{K}_{3,}}\] \[{{K}_{4,}}\] arranged as shown in the figure. The effective dielectric constant K will be: | ||||||||||||
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A) \[K=\frac{({{K}_{1}}+{{K}_{3}})\,({{K}_{2}}+K4)}{{{K}_{1}}+{{K}_{2}}+{{K}_{3}}+{{K}_{4}}}\]
B) \[K=\frac{({{K}_{1}}+{{K}_{2}})\,({{K}_{3}}+{{K}_{4}})}{2\,({{K}_{1}}+{{K}_{2}}+{{K}_{3}}+{{K}_{4}})}\]
C) \[K=\frac{({{K}_{1}}+{{K}_{2}})\,({{K}_{3}}+{{K}_{4}})}{{{K}_{1}}+{{K}_{2}}+{{K}_{3}}+{{K}_{4}}}\]
D) \[K=\frac{({{K}_{1}}+{{K}_{4}})\,({{K}_{2}}+{{K}_{3}})}{2\,({{K}_{1}}+{{K}_{2}}+{{K}_{3}}+{{K}_{4}})}\]
Correct Answer: A
Solution :
| \[{{C}_{1}}=\frac{{{\varepsilon }_{0}}{{K}_{1}}\frac{{{L}^{2}}}{2}}{\frac{d}{2}}+\frac{{{\varepsilon }_{0}}{{K}_{3}}\frac{{{L}^{2}}}{2}}{\left( \frac{d}{2} \right)}\]\[=\frac{{{\varepsilon }_{0}}{{L}^{2}}}{d}({{K}_{1}}+{{K}_{3}})\] |
| \[{{C}_{2}}=\frac{{{\varepsilon }_{0}}{{K}_{2}}\frac{{{L}^{2}}}{L}}{\frac{d}{2}}+\frac{{{\varepsilon }_{0}}{{K}_{4}}\frac{{{L}^{2}}}{2}}{\frac{d}{2}}\]\[=\frac{{{\varepsilon }_{0}}{{L}^{2}}}{d}({{K}_{2}}+{{K}_{4}})\] |
| \[\therefore \] \[\frac{1}{c}=\frac{1}{c}+\frac{1}{{{c}^{2}}}\] |
| \[\Rightarrow \] \[\frac{d}{{{\varepsilon }_{0}}K{{L}^{2}}}=\frac{d}{{{\varepsilon }_{0}}{{L}^{2}}({{K}_{1}}+{{K}_{3}})}+\frac{d}{{{\varepsilon }_{0}}{{L}^{2}}({{K}_{2}}+{{K}_{4}})}.\] |
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