Question

Five identical conducting plates $1,2,3,4$ and $5$ are fixed parallel to and equidistant from each other as shown in figure. The equivalent capacitance between $\text{A}$ and $\text{B}$ is ($A=$ area of each plate)

• A. $\frac{5{\epsilon }_{0}A}{3d}$
• B. $\frac{3{\epsilon }_{0}A}{2d}$
• C. $\frac{5{\epsilon }_{0}A}{4d}$
• D. $\frac{3{\epsilon }_{0}A}{5d}$

Answer 1

The correct option is A $\frac{5{\epsilon }_{0}A}{3d}$

We can redraw the given arrangement of plates as follows,




$\therefore$ The equivalent capacitance between $\text{A}$ and $\text{B}$ is,

$C_{AB}=C+\frac{2C}{3}=\frac{5C}{3}$

$C_{AB}=\frac{5{\epsilon }_{0}A}{3d}[\because C=\frac{{\epsilon }_{0}A}{d}]$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (a) is the correct answer.