Question

Find the percentage increase in the capacitance of a parallel plate capacitor after it is half-filled with a dielectric of dielectric constant $5$ , if it is known that its capacitance with air as the medium is $C$ .

• A. $400\%$
• B. $66.6\%$
• C. $33.3\%$
• D. $200\%$

Answer 1

Answer: (B)

When it is half filled with dielectric of $K=5$ , its capacitance becomes $C_{s}$ which is series combination of
$C_{1}=\frac{K \epsilon _{0} A}{d / 1}=\frac{\epsilon _{0} A}{d / 2 K}=\frac{\epsilon _{0} A}{d / 10}$ and $C_{2}=\frac{\epsilon _{0} A}{d / 2}$
$\therefore \, \frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{d / 10}{\left(\epsilon \right)_{0} A}+\frac{d / 2}{\left(\epsilon \right)_{0} A}=\frac{\left(\right. 3 d / 5 \left.\right)}{\left(\epsilon \right)_{0} A}$
$\Rightarrow \, \, C_{s}=\frac{5}{3}\frac{\left(\epsilon \right)_{0} A}{d}=\frac{5 C}{3} \, \, \left(\right.using \, \left(\right.i\left.\right)\left.\right)$
Percentage increase in capacitance $=\frac{C_{s} - C \, }{C}\times 100$
$=\frac{\frac{5}{3} C - C}{C}\times =\frac{2}{3}\times 100=66.6\%$