Two metallic plates form a parallel plate capacitor. The distance between the plates is 'd'. A metal sheet of thickness ( d /2) and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor ?

Question

Two metallic plates form a parallel plate capacitor. The distance between the plates is 'd'. A metal sheet of thickness ( d /2) and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor ? (A) 2: 1 (B) 1:2 (C) 1: 4 (D) 4: 1

Tardigrade Admin · Accepted Answer

C 1=(ϵ0 A / d ) C 2=(ϵ0 A /( d )2+( d / 2) propto=(2 ϵ0 A / d ) ( C 2/ C 1)=(2/1)

$2 : 1$

$1 : 2$

$1 : 4$

$4 : 1$

$C_{1} = \frac{ϵ _{0} A}{d}$
$C_{2} = \frac{ϵ _{0} A}{\frac{d}{2} + \frac{d /2}{\propto}} = \frac{2 ϵ _{0} A}{d}$
$\frac{C _{2}}{C _{1}} = \frac{2}{1}$

Q. Two metallic plates form a parallel plate capacitor. The distance between the plates is 'd'. A metal sheet of thickness 2d​ and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor ?

2:1

1:2

1:4

4:1

C1​=dϵ0​A​ C2​=2d​+∝d/2​ϵ0​A​=d2ϵ0​A​ C1​C2​​=12​

$2 : 1$

$1 : 2$

$1 : 4$

$4 : 1$

$C_{1} = \frac{ϵ _{0} A}{d}$
$C_{2} = \frac{ϵ _{0} A}{\frac{d}{2} + \frac{d /2}{\propto}} = \frac{2 ϵ _{0} A}{d}$
$\frac{C _{2}}{C _{1}} = \frac{2}{1}$