A slab of copper of thickness d / 2 is introduced between the plates of a parallel plate capacitor where d is the seperation between its two plates. If the capacitance of the capacitor without copper slab is C and with copper slab is C' then (C'/C) is :

Question

A slab of copper of thickness d / 2 is introduced between the plates of a parallel plate capacitor where d is the seperation between its two plates. If the capacitance of the capacitor without copper slab is C and with copper slab is C' then (C'/C) is : (A) √2 (B) 2 (C) 1 (D) (1/√2)

Tardigrade Admin · Accepted Answer

Without copper C =(ϵ0 A/d) with copper C 1=(ϵ0 A/d-t(1-(1)ϵr)) for copper ϵr=∞ and t=(d/2) Hence C1=(ϵ0 A/d-(d)2)=2 (ϵ0 A/d) (C1/C)=2

Q. A slab of copper of thickness $d /2$ is introduced between the plates of a parallel plate capacitor where $d$ is the seperation between its two plates. If the capacitance of the capacitor without copper slab is $C$ and with copper slab is $C^{'}$ then $\frac{C ^{'}}{C}$ is :

$2$

$2$

$1$

$\frac{1}{2}$

Without copper $C = \frac{ϵ _{0} A}{d}$
with copper $C^{1} = \frac{ϵ _{0} A}{d - t ( 1 - \frac{1}{ϵ _{r}} )}$
for copper $ϵ_{r} = \infty$ and $t = \frac{d}{2}$
Hence $C^{1} = \frac{ϵ _{0} A}{d - \frac{d}{2}} = 2 \frac{ϵ _{0} A}{d}$
$\frac{C ^{1}}{C} = 2$

Q. A slab of copper of thickness d/2 is introduced between the plates of a parallel plate capacitor where d is the seperation between its two plates. If the capacitance of the capacitor without copper slab is C and with copper slab is C′ then CC′​ is :

2​

2

1

2​1​

Without copper C=dϵ0​A​ with copper C1=d−t(1−ϵr​1​)ϵ0​A​ for copper ϵr​=∞ and t=2d​ Hence C1=d−2d​ϵ0​A​=2dϵ0​A​ CC1​=2

Q. A slab of copper of thickness $d /2$ is introduced between the plates of a parallel plate capacitor where $d$ is the seperation between its two plates. If the capacitance of the capacitor without copper slab is $C$ and with copper slab is $C^{'}$ then $\frac{C ^{'}}{C}$ is :

$2$

$2$

$1$

$\frac{1}{2}$

Without copper $C = \frac{ϵ _{0} A}{d}$
with copper $C^{1} = \frac{ϵ _{0} A}{d - t ( 1 - \frac{1}{ϵ _{r}} )}$
for copper $ϵ_{r} = \infty$ and $t = \frac{d}{2}$
Hence $C^{1} = \frac{ϵ _{0} A}{d - \frac{d}{2}} = 2 \frac{ϵ _{0} A}{d}$
$\frac{C ^{1}}{C} = 2$