A) \[\frac{14}{11}\]
B) \[\frac{11}{14}\]
C) \[\frac{7}{11}\]
D) \[\frac{11}{7}\]
Correct Answer: A
Solution :
The capacitance of a capacitor is given by \[{{C}_{1}}=\frac{{{\varepsilon }_{0}}A}{d}\] ? (i) If dielectric slab of constant k of thickness \[t=\frac{2}{3}d\] is introduced, then capacitance becomes \[{{C}_{2}}=\frac{{{\varepsilon }_{0}}A}{d-t\left( 1-\frac{1}{k} \right)}\] \[=\frac{{{\varepsilon }_{0}}A}{d-\frac{2}{3}d\left( 1-\frac{1}{k} \right)}\] ?. (ii) Given, \[{{C}_{2}}=\frac{7}{6}{{C}_{1}}\] Eq. (i) becomes \[{{C}_{2}}=\frac{{{\varepsilon }_{0}}A}{d\left[ \left( 1-\frac{2}{3} \right)+\frac{2}{3k} \right]}\] \[\Rightarrow \] \[{{C}_{2}}=\frac{{{C}_{1}}}{\left( \frac{1}{3}+\frac{2}{3k} \right)}\] \[\Rightarrow \] \[\frac{7}{6}{{C}_{1}}=\frac{{{C}_{1}}}{\frac{1}{3}+\frac{2}{3k}}\] \[\Rightarrow \] \[\frac{1}{3}+\frac{2}{3k}=\frac{6}{7}\] \[\Rightarrow \]\[\frac{2}{3k}=\frac{6}{7}-\frac{1}{3}=\frac{11}{21}\] Or \[33k=21\times 2\] \[k=\frac{42}{33}=\frac{14}{11}\]
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