Question

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

• A. $\frac { 1 } { K } = \frac { 1 } { K _ { 1 } } + \frac { 1 } { K _ { 2 } } + \frac { 1 } { 2 K _ { 3 } }$
• B. $\frac { 1 } { K } = \frac { 1 } { K _ { 1 } + K _ { 2 } } + \frac { 1 } { 2 K _ { 3 } }$
• C. $K = \frac { K _ { 1 } K _ { 2 } } { K _ { 1 } + K _ { 2 } } + 2 K _ { 3 }$
• D. $K = K _ { 1 } + K _ { 2 } + 2 K _ { 3 }$

Answer 1

Answer: (B)

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

Answer 2

The effective capacitance is given by

$\frac { 1 } { C _ { e q } } = \frac { d } { \varepsilon _ { 0 } A } \left[ \frac { 1 } { 2 K _ { 3 } } + \frac { 1 } { \left( K _ { 1 } + K _ { 2 } \right) } \right]$

The capacitance of capacitor with single dielectric of dielectric constant K is $C = \frac { K \varepsilon _ { 0 } A } { d }$

According to question $C _ { e q } = C$ i.e.,

$\frac { \varepsilon _ { 0 } A } { d \left[ \frac { 1 } { 2 K _ { 3 } } + \frac { 1 } { K _ { 1 } + K _ { 2 } } \right] } = \frac { K \varepsilon _ { 0 } A } { d }$

⇒ $\frac { 1 } { K } = \frac { 1 } { 2 K _ { 3 } } + \frac { 1 } { K _ { 1 } + K _ { 2 } }$.