Four identical metal plates are located in air at equal distance $d$ from each other. The area of each plate is $S$. If the outer most plates are connected by a conducting wire as shown in the figure the capacitance between points $A$ and $B$ will be

• A

  $\frac{\varepsilon_0 S}{d}$
• Correct answer
  B

  $\frac{3}{2} \frac{\varepsilon_0 S}{d}$
• C

  $\frac{1}{2} \frac{\varepsilon_0 S}{d}$
• D

  $\frac{2}{3} \frac{\varepsilon_0 S}{d}$

Solution

The equivalent arrangement of given combination is shown in the following figure,
we know that the capacitance, $C=\frac{\varepsilon_0 S}{d}$ where, $S$ is the area of each plate and $d$ is distance between two plates. So, equivalent capacitance between points $A$ and $B$, $\begin{aligned} C_{A B} & =\frac{C \times C}{C+C}+C \\ & =\frac{C}{2}+C=\frac{3 C}{2}=\frac{3}{2} \cdot \frac{\varepsilon_0 S}{d}\end{aligned}$

Hence, the correct answer is option: 'B'.