Question

A dielectric of thickness $5$ cm and a dielectric constant $10$ is introduced between the plates of a parallel plate capacitor having plate area $500sq.$ cm and separation between the plates $10cm$. The capacitance of the capacitor with the dielectric slab is ${\epsilon }_0=8.8\times {10}^{-12}{C}^2/N-m^2$

• A. $4.4pF$
• B. $6.2pF$
• C. $8pF$
• D. $10pF$

Answer 1

Answer: (C)

We know
$C=\frac{k{l}_{0}A}{d}$
effective capacitors is series combination of $\frac{k{l}_{0}A}{5cm}and\frac{l_{0}A}{5cm}$
$\therefore \frac{1}{C_{eff}}=\frac{5cm}{k{l}_{0}A}+\frac{5cm}{k{l}_{0}A}$

$\Rightarrow C_{eff}=\frac{k{l}_{0}A}{(5+k5)\times {10}^{-2}}$

$\Rightarrow C_{eff}=\frac{10\times 8.8\times {10}^{-12}\times 500\times {10}^{-4}}{55\times {10}^{-2}}$
$\Rightarrow C_{eff}=8pF$