Discussion Forum

 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  $(\epsilon _{0}=$ absolute permitivity of free space)

Answer:

Explanation:

 According to question, capacitance of parallel plate capacitor  is given by $C=\frac{\epsilon_{0 A}}{d}$

 For first capacitor  , $C_{1}=\frac{\epsilon_{0 A}}{3d}$

For second capacitor = $C_{2}=\frac{\epsilon_{0 A}}{6d}$

For third capacitor , $C_{3}=\frac{\epsilon_{0 A}}{9d}$

 Again they are arranged in parallel 

combination , so equivalent capacitor is given by

    $C_{eq}= C_{1}+C_{2}+C_{3}$

 $=\frac{\epsilon_{0 A}}{d}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{9}\right)=\frac{\epsilon_{0 A}}{d}\times\frac{11}{18}=\frac{11\epsilon_{0 A}}{18d}$