Question

Four identical metal plates are located in ain 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. $\frac{{\epsilon }_{0}S}{d}$
• B. $\frac{3}{2}\frac{{\epsilon }_{0}S}{d}$
• C. $\frac{1}{2}\frac{{\epsilon }_{0}S}{d}$
• D. $\frac{2}{3}\frac{{\epsilon }_{0}S}{d}$

Answer 1

Answer: (B)

The equivalent arrangement of given combination is shown in the following figure,

we know that the capacitance, $C=\frac{{\epsilon }_{0}S}{d}$
where, $S$ is the area of each plate and $d$ is distance between two plates.
So, equivalent capacitance between points $A$ and $B$,
$C_{AB}=\frac{C\times C}{C+C}+C$
$=\frac{C}{2}+C=\frac{3C}{2}=\frac{3}{2}\cdot \frac{{\epsilon }_{0}S}{d}$