A parallel plate capacitor having a separation between the plates d, plate area A and material with dielectric constant K has capacitance C0. Now one-third of the material is replaced by another material with dielectric constant 2K, so that effectively there are two capacitors one with area (1/3) A, dielectric constant 2K and another with area (2/3) A and dielectric constant K. If the capacitance of this new capacitor is C then (C/C0) is

Question

A parallel plate capacitor having a separation between the plates d, plate area A and material with dielectric constant K has capacitance C0. Now one-third of the material is replaced by another material with dielectric constant 2K, so that effectively there are two capacitors one with area (1/3) A, dielectric constant 2K and another with area (2/3) A and dielectric constant K. If the capacitance of this new capacitor is C then (C/C0) is (A) 1 (B) (4/3) (C) (2/3) (D) (1/3)

Tardigrade Admin · Accepted Answer

C0 =(k ∈0 A/d) C=(k ∈0 2 /3 d)+(2k ∈0 A/3d)=(4k∈0 A/3 d) ∴ (C/C0) = ((4/3)(k ∈0 A/d)/(k ∈0 A)d) =(4/3)

$1$

$\frac{4}{3}$

$\frac{2}{3}$

$\frac{1}{3}$

$C_{0} = \frac{k \in _{0} A}{d}$
$C = \frac{k \in _{0} 2}{3 d} + \frac{2 k \in _{0} A}{3 d} = \frac{4 k \in _{0} A}{3 d}$
$∴ \frac{C}{C _{0}} = \frac{\frac{4}{3} \frac{k \in _{0} A}{d}}{\frac{k \in _{0} A}{d}} = \frac{4}{3}$

1

34​

32​

31​

C0​=dk∈0​A​ C=3dk∈0​2​+3d2k∈0​A​=3d4k∈0​A​ ∴C0​C​=dk∈0​A​34​dk∈0​A​​=34​

$1$

$\frac{4}{3}$

$\frac{2}{3}$

$\frac{1}{3}$

$C_{0} = \frac{k \in _{0} A}{d}$
$C = \frac{k \in _{0} 2}{3 d} + \frac{2 k \in _{0} A}{3 d} = \frac{4 k \in _{0} A}{3 d}$
$∴ \frac{C}{C _{0}} = \frac{\frac{4}{3} \frac{k \in _{0} A}{d}}{\frac{k \in _{0} A}{d}} = \frac{4}{3}$