A parallel- plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constant $k_{1}, k_{2}, k_{3}$ and $k_{4}$ as shown in the figure below. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by

$k = k_{1} + k_{2} + k_{3} + k_{4}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$k = \frac{2}{3} (k_{1} + k_{2} + k_{3}) + k_{4}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{2}{k} = \frac{3}{k_{1} + k_{2} + k_{3}} + \frac{1}{k_{4}}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$\frac{1}{k} = \frac{1}{k_{1}} + \frac{1}{k_{2}} + \frac{1}{k_{3}} + \frac{3}{2 k_{4}}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$\frac{2}{k} = \frac{3}{k_{1} + k_{2} + k_{3}} + \frac{1}{k_{4}}$

$k_{1}, k_{2}, k_{3}$ are in parallel so Arithmetic mean.

$k_{e q} = \frac{k_{1} + k_{2} + k_{3}}{3}$

$k_{e q}$ is in series with $k_{4}$ .So harmonic mean.

$\Rightarrow \frac{2}{k} = \frac{1}{k_{e q}} + \frac{1}{k_{4}} \Rightarrow \frac{2}{k} = \frac{3}{k_{1} + k_{2} + k_{3}} + \frac{1}{k_{4}}$

Suggest Corrections

Similar questions

Q. A parallel-plate capacitor of area

A

, plate separation 𝑑, and capacitance

C

is filled with four dielectric material having dielectric constants

k_{1}, k_{2}, k_{3}

and

k_{4}

as shown in the figure. If a single dielectric material is to be used to have the same capacitance

C

in this capacitor, then its dielectric constant

k

is given by

A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants $K_{1}, K_{2} a n d K_{3}$ as shown in Fig. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant K is given by