Question

The following arrangement consists of five identical metal plates marked $1,2,3,4$ and $5$ parallel to each other. Area of each plate is $A$ and separation between the successive plates is $d$. The capacitance between $P$ and $Q$ is

• A. $5 \frac{\varepsilon_{0} A}{d}$
• B. $\frac{7}{3} \frac{\varepsilon_{0} A}{d}$
• C. $\frac{5}{3} \frac{\varepsilon_{0} A}{d}$
• D. $\frac{4}{3} \frac{\varepsilon_{0} A}{d}$

Answer 1

Answer: (C)

The arrangement will work as a system of capacitors shown in figure (a) and (b). Each capacitor in figure (a) is. $C=\varepsilon_{0} A / d$. Total capacitance between $P$ and $Q$ is

$C' = C + \frac{C \times 2C}{C + 2C} = \frac{5}{3} C = \frac{5}{3} \frac{\varepsilon_0A}{d}$