Question

Four metal plates numbered $1,2,3$ and $4$ are arranged as shown below. The area of each plate is $A=10{\text{cm}}^2$ and separation between adjacent plates is $d=1.5\text{mm}$. The capacitance of the arrangement is

• A. $35.4\text{pF}$
• B. $70.8\text{pF}$
• C. $17.7\text{pF}$
• D. $8.85\text{pF}$

Answer 1

The correct option is C $17.7\text{pF}$


Plate $1$ is connected to plate $3$ and plate $2$ is connected to plate $4$.

Thus, there are three capacitors in parallel, each of capacitance
$C,$ as shown below.


So, the equivalent capacitance between points $\text{A}$ and $\text{B}$ will be

$C_{eq}=3C=\frac{3{ϵ}_{0}A}{d}$

Here, $A={10}^{-3}{\text{m}}^2;d=1.5\times {10}^{-3}\text{m}$

$C_{eq}=\frac{3\times 8.85\times {10}^{-12}\times {10}^{-3}}{1.5\times {10}^{-3}}$

$\therefore C_{eq}=17.7\text{pF}$

Hence, option $\left(c\right)$ is the right answer.