A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constant K1, K2 and K3 as shown in 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

We have C1=(A/2)ε0K1(d/2)=Aε0K1dC2=(A/2)ε0K2(d/2)=Aε0K2d and C3=Aε0K2(d/2)=2Aε0K3dThe capacitors C1 and C2 are in parallel and their equivalent capacitance is C′=C1+C2=Aε0dK1+K2 This combination is in series with C3 .Hence the net capacitance is 1C′′=1C′+1C3=dAε0K1+K2+d2Aε0K3=dε0A1K1+K2+12K3 or C′′=Aε0Kd where 1K=1K1+K2+12K3

A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constant K1, K2 and K3 as shown in 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

We have C1=(A/2)ε0K1(d/2)=Aε0K1dC2=(A/2)ε0K2(d/2)=Aε0K2d and C3=Aε0K2(d/2)=2Aε0K3dThe capacitors C1 and C2 are in parallel and their equivalent capacitance is C′=C1+C2=Aε0dK1+K2 This combination is in series with C3 .Hence the net capacitance is 1C′′=1C′+1C3=dAε0K1+K2+d2Aε0K3=dε0A1K1+K2+12K3 or C′′=Aε0Kd where 1K=1K1+K2+12K3

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constant K₁, K₂ and K₃ as shown in 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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{1}{K} = \frac{1}{K_{1}} + \frac{1}{K_{2}} + \frac{1}{2 K_{3}}$

$K = \frac{K_{1} K_{2}}{K_{1} + K_{2}} + 2 K_{3}$

$K = K_{1} + K_{2} + K_{3}$

$\frac{1}{K} = \frac{1}{K_{1} + K_{2}} + \frac{1}{2 K_{3}}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

We have $C_{1} = \frac{(A / 2) ε_{0} K_{1}}{(d / 2)} = \frac{{Aε}_{0} K_{1}}{d}$
$C_{2} = \frac{(A / 2) ε_{0} K_{2}}{(d / 2)} = \frac{{Aε}_{0} K_{2}}{d} and C_{3} = \frac{{Aε}_{0} K_{2}}{(d / 2)} = \frac{2 {Aε}_{0} K_{3}}{d}$
The capacitors C₁ and C₂ are in parallel and their equivalent capacitance is $C^{'} = C_{1} + C_{2} = \frac{{Aε}_{0}}{d} (K_{1} + K_{2})$ This combination is in series with C₃ .
Hence the net capacitance is $\frac{1}{C^{''}} = \frac{1}{C^{'}} + \frac{1}{C_{3}} = \frac{d}{{Aε}_{0} (K_{1} + K_{2})} + \frac{d}{2 {Aε}_{0} K_{3}}$
$= \frac{d}{ε_{0} A} [\frac{1}{(K_{1} + K_{2})} + \frac{1}{2 K_{3}}] or C^{''} = \frac{{Aε}_{0} K}{d} where \frac{1}{K} = \frac{1}{(K_{1} + K_{2})} + \frac{1}{2 K_{3}}$