Four metallic plates each with a surface area of one side A are placed at a distance d from each other. The plates are connected as shown in the circuit diagram. Then the capacitance of the system between a and b is

Question

Four metallic plates each with a surface area of one side A are placed at a distance d from each other. The plates are connected as shown in the circuit diagram. Then the capacitance of the system between a and b is (A) (3 ε0 A/d) (B) (2 ε0 A/d) (C) (2 ε0 A/3 d) (D) (3 ε0 A/2 d)

Tardigrade Admin · Accepted Answer

The given circuit can be redrawn as follows <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/3fb2e5ad1fc403cd473002f9bb4e71f2-.png" /> ⇒ Ce q=(3 C/2)=(3 ε0 A/2 d)

Q. Four metallic plates each with a surface area of one side $A$ are placed at a distance $d$ from each other. The plates are connected as shown in the circuit diagram. Then the capacitance of the system between $a$ and $b$ is

$\frac{3 ε _{0} A}{d}$

$\frac{2 ε _{0} A}{d}$

$\frac{2 ε _{0} A}{3 d}$

$\frac{3 ε _{0} A}{2 d}$

The given circuit can be redrawn as follows

$\Rightarrow C_{e q} = \frac{3 C}{2} = \frac{3 ε _{0} A}{2 d}$

Q. Four metallic plates each with a surface area of one side A are placed at a distance d from each other. The plates are connected as shown in the circuit diagram. Then the capacitance of the system between a and b is

d3ε0​A​

d2ε0​A​

3d2ε0​A​

2d3ε0​A​