Question

Two parallel plates of area A are separated by two different dielectric. The net capacitance is

• A. $\frac{\varepsilon_{0} A}{d}$
• B. $\frac{7}{20} \frac{\varepsilon_{0} A}{d}$
• C. $\frac{20}{7} \frac{\varepsilon_{0} A}{d}$
• D. $\frac{2 \varepsilon_{0} A}{d}$

Answer 1

Answer: (C)

This is a series combination because there is separation charges and in parallel combination area changes
$\frac{1}{ C _{ eq }}=\frac{1}{ C _{1}}+\frac{1}{ C _{2}} $
$\Rightarrow \frac{1}{ C _{ eq }}$
$=\frac{ d _{1}}{ K _{1} \varepsilon_{ 0 } A }+\frac{ d _{2}}{ K _{2} \varepsilon_{0} A }$
$\therefore \frac{1}{ C _{ eq }}=\frac{ d }{2 \varepsilon_{ 0 } A }\left[\frac{1}{2}+\frac{1}{5}\right]$
$\therefore \frac{1}{ C _{ eq }}=\frac{ d }{2 \varepsilon_{ 0 } A }\left(\frac{5+2}{10}\right) $
$\Rightarrow \frac{1}{ C _{ eq }}$ $=\frac{7 d }{20 \varepsilon_{ 0 } A }$
$C _{ eq }=\frac{20 \varepsilon_{ 0 } A }{7d }$