A) \[\frac{mx}{{{(m+1)}^{2}}}\]
B) \[\frac{mx}{{{(m-1)}^{2}}}\]
C) \[\frac{{{(m+1)}^{2}}}{m}x\]
D) \[\frac{{{(m-1)}^{2}}}{m}x\]
Correct Answer: A
Solution :
\[{{C}_{1}}={{K}_{1}}\frac{{{\varepsilon }_{0}}\frac{A}{2}}{d}=\frac{{{\varepsilon }_{0}}A}{d}\frac{{{K}_{1}}}{2}\] \[{{C}_{2}}={{K}_{2}}\frac{{{\varepsilon }_{0}}\frac{A}{2}}{d/2}={{K}_{2}}\frac{{{\varepsilon }_{0}}A}{d}\] \[{{C}_{3}}={{K}_{3}}\frac{{{\varepsilon }_{0}}\frac{A}{2}}{d/2}={{K}_{3}}\frac{{{\varepsilon }_{0}}A}{d}\] \[{{C}_{2}}\] and \[{{C}_{3}}\] are in series. \[{{C}_{23}}={{\varepsilon }_{0}}\frac{A}{d}\left[ \frac{{{K}_{2}}{{K}_{3}}}{{{K}_{2}}+{{K}_{3}}} \right]\] Now, \[{{C}_{1}}\] and \[{{C}_{23}}\] are in parallel. \[\therefore \] \[{{C}_{123}}=\frac{{{\varepsilon }_{0}}A}{d}\frac{{{K}_{1}}}{2}+\frac{{{\varepsilon }_{0}}A}{d}\left[ \frac{{{K}_{2}}{{K}_{3}}}{{{K}_{2}}+{{K}_{3}}} \right]\] \[=\frac{{{\varepsilon }_{0}}A}{d}\left[ \frac{{{K}_{1}}}{2}+\frac{{{K}_{2}}{{K}_{3}}}{{{K}_{2}}+{{K}_{3}}} \right]\]
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