Question

The plates in a parallel plate capacitor are separated by a distance d with air as the medium between the plates. In order to increase the capacity by 66% a dielectric slab of dielectric constant 5 is introduced between the plates. What is the thickness of the dielectric slab?

• A. $\frac{d}{4}$
• B. $\frac{d}{2}$
• C. $\frac{5d}{8}$
• D. $d$

Answer 1

Answer: (B)

The capacity in air $C=\frac{{\epsilon }_{0}A}{d}$ The capacity when dielectric slab of dielectric constant 5 is introduced between the plates $C=\frac{{\epsilon }_{0}A}{d-t+\frac{t}{5}}$ $\therefore$ $\frac{C}{C}=\frac{d-t+\frac{t}{5}}{d}$ $C=\frac{166}{100}C$ $\frac{100}{166}=\frac{d-t+\frac{t}{5}}{d}=\frac{d-\frac{4t}{5}}{d}$ or $100d=166d-166\left(\frac{4t}{5}\right)$ $166\left(\frac{4t}{5}\right)=66d$ $t=\frac{66d\times 5}{166\times 4}=\frac{d}{2}$