Question

A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness d1 and dielectric constant K1 and the other has thickness d2 and dielectric constant K2 as shown in figure. This arrangement can be through as a dielectric slab of thickness d(=d1+d2) and effective dielectric constant K. The K is

• A. $\frac { \mathrm { K } _ { 1 } \mathrm {~d} _ { 1 } + \mathrm { K } _ { 2 } \mathrm {~d} _ { 2 } } { \mathrm {~d} _ { 1 } + \mathrm { d } _ { 2 } }$
• B.
• C.
• D. $\frac { 2 \mathrm {~K} _ { 1 } \mathrm {~K} _ { 2 } } { \mathrm {~K} _ { 1 } + \mathrm { K } _ { 2 } }$

Answer 1

Answer: (C)

Answer 2

: The capacities of two individual condensers are

and

The arrangement is equivalent to two capacitors joined in series.

$\therefore$Equivalent capacitance,

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

or, …(i)

Also …(ii)

From (i) and (ii),

$\varepsilon _ { 0 } \mathrm {~A} \left( \frac { \mathrm { K } _ { 1 } \mathrm {~K} _ { 2 } } { \mathrm {~d} _ { 2 } \mathrm {~K} _ { 1 } + \mathrm { d } _ { 1 } \mathrm {~K} _ { 2 } } \right) = \varepsilon _ { 0 } \mathrm {~A} \left( \frac { \mathrm { K } } { \mathrm { d } _ { 1 } + \mathrm { d } _ { 2 } } \right)$

$\therefore \mathrm { K } = \frac { \mathrm { K } _ { 1 } \mathrm {~K} _ { 2 } \left( \mathrm {~d} _ { 1 } + \mathrm { d } _ { 2 } \right) } { \mathrm { d } _ { 2 } \mathrm {~K} _ { 1 } + \mathrm { d } _ { 1 } \mathrm {~K} _ { 2 } }$