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 $C_1=\frac{{\epsilon }_{0}A}{d}$ ?(i) If dielectric slab of constant k of thickness $t=\frac{2}{3}d$ is introduced, then capacitance becomes $C_2=\frac{{\epsilon }_{0}A}{d-t(1-\frac{1}{k})}=\frac{{\epsilon }_{0}A}{d-\frac{2}{3}d(1-\frac{1}{k})}$ ?(ii) Given: $C_2=\frac{7}{6}{C}_1$ Eq.(i) becomes $C_2=\frac{{\epsilon }_{0}A}{d\left[\left(1-\frac{2}{3}\right)+\frac{2}{3k}\right]}$ $\Rightarrow$ $C_2=\frac{C_1}{\left(\frac{1}{3}+\frac{2}{3k}\right)}$ $\Rightarrow$ $\frac{7}{6}{C}_1=\frac{C_1}{\frac{1}{3}+\frac{2}{3k}}$ $\Rightarrow$ $\frac{1}{3}+\frac{2}{3k}=\frac{6}{7}\Rightarrow \frac{2}{3k}=\frac{6}{7}-\frac{1}{3}=\frac{11}{21}$ or $33k=21\times 2$ $k=\frac{42}{33}=\frac{14}{11}$