Question

The figure shows an arrangement of identical metal plates placed parallel to each other. The diagram also shows the variation of potential between the plates. Using the details given in the diagram, the equivalent capacitance connected across the battery is given as $\frac{{x\epsilon }_{0}A}{5l}$. Find $x$ :

(separation between the consecutive plates is equal to $l$ and the cross-sectional area of each plate is $A$).

Answer 1

From the graph , $V_1=0V,V_2=10V,V_3=0V,V_4=-10V,V_5=-10V,V_6=0,V_7=10V$
Using this data we will get 5 active capacitors with capacitance $C=\frac{A{ϵ}_0}{d}$, which are shown in the equivalent circuit.
For left branch , $C_{eq1}=C+C+C=3C$
for right branch, $C_{eq2}=C+C=2C$
$\therefore C_{eq}=\frac{C_{eq1}{C}_{eq2}}{C_{eq1}+C_{eq2}}=\frac{\left(3C\right)\left(2C\right)}{3C+3C}=\frac{6C}{5}=\frac{6A{ϵ}_0}{5d}$
thus $x=6$