Question

Capacitance of a capacitor becomes $ \frac{7}{6} $ time its original value if a dielectric slab of thickness t= $ \frac{2}{3} $ d is introduced in between the plates d. is the separation between the plates. The dielectric constant of the dielectric slab is:

• A. $ \frac{14}{11} $
• B. $ \frac{11}{14} $
• C. $ \frac{7}{11} $
• D. $ \frac{11}{7} $

Answer 1

Answer: (A)

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(1-\frac{1}{k})}=\frac{{{\varepsilon }_{0}}A}{d-\frac{2}{3}d(1-\frac{1}{k})} $ ?(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} $