Question

Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area $\frac{'A'}{3}$ and the separation between the plates is ‘$d$’, ‘$2d$’ and ‘$3d$’ respectively. The equivalent capacity of combination is ($∈_0 =$ absolute permittivity of free space)

• A. $\frac{7∈_{0}A}{18\,d}$
• B. $\frac{11∈_{0}A}{18\,d}$
• C. $\frac{13∈_{0}A}{18\,d}$
• D. $\frac{17∈_{0}A}{18\,d}$

Answer 1

Answer: (B)

Capacitance of parallel plate capacitor $C =\frac{ A\epsilon _{ o }}{ d }$
where $A$ is the area of each plate and $d$ is the separation betwee the plates.
Thus capacitance of first capacitor $C _{1}=\frac{ A\epsilon _{ o }}{3 d }$
Capacitance of second capacitor $C _{2}=\frac{ A\epsilon _{ o }}{6 d }$
Capacitance of second capacitor $C _{3}=\frac{ A\epsilon _{ o }}{9 d }$
Equivalent capacitance of capacitors connected in aprallel $C _{ eq }= C _{1}+ C _{2}+ C _{3}$
$\therefore C _{ eq }=\frac{ A \epsilon_{ o }}{3 d }+\frac{ A \epsilon_{ o }}{6 d }+\frac{ A \epsilon_{ o }}{9 d }$
$\Rightarrow C _{ eq }=\frac{11 A \epsilon_{ o }}{18 d }$