A) \[C=\frac{\lambda d}{\ln (1+{{K}_{o}}\lambda d)}{{C}_{o}}\]
B) \[C=\frac{\lambda }{d.ln(1+{{K}_{o}}\lambda d)}{{C}_{o}}\]
C) \[C=\frac{\lambda d}{ln(1+\lambda d/{{K}_{o}})}{{C}_{o}}\]
D) \[C=\frac{\lambda }{d.ln(1+{{K}_{o}}/\lambda d)}{{C}_{o}}\]
Correct Answer: C
Solution :
The value of dielectric constant is given as, \[K={{K}_{0}}+\lambda x\]And, \[V=\int\limits_{0}^{d}{Edr}\] \[V=\int\limits_{0}^{d}{\frac{\sigma }{K}dx}\] \[=\sigma \int\limits_{0}^{d}{\frac{1}{({{K}_{0}}+\lambda x)}dx}\] \[=\frac{\sigma }{\lambda }\left[ \ln \left( {{K}_{0}}+\lambda d \right)-\ln {{K}_{0}} \right]\] \[=\frac{\sigma }{\lambda }\ln \left( 1+\frac{\lambda d}{{{K}_{0}}} \right)\] Now it is given that capacitance of vacuum \[={{C}_{0}}.\] Thus, \[C=\frac{Q}{V}\]\[=\frac{\sigma .s}{v}\] (Let surface area of plates = s) \[=\frac{\sigma .s}{\frac{\sigma }{\lambda }\ln \left( 1+\frac{\lambda d}{{{K}_{0}}} \right)}\] \[=s\lambda .\frac{d}{d}\frac{1}{\ln \left( 1+\frac{\lambda d}{{{K}_{0}}} \right)}\](\[\because \] in vacuum e0 =1) \[c=\frac{\lambda d}{\ln \left( 1+\frac{\lambda d}{{{K}_{0}}} \right)}.{{C}_{0}}\]\[\left( \text{here,}\,{{C}_{0}}=\frac{s}{d} \right)\]
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