Question

When a dielectric slab of thickness $4$ cm is introduced between the plates of parallel plate condenser, it is found that the distance between the plates has to be increased by $3$ cm to restore the capacity to its original value. The dielectric constant of slab is
[A] $\dfrac{3}{4}$
[B] $4$
[C] $3$
[D] $1$

Answer 1

Hint : In order to solve this question, we will understand the fact that whenever a dielectric is placed in an electric field it gets polarized and charges of opposite nature induced on the surface and volume of dielectric. So when dielectric is placed between parallel plates of capacitor it gets polarized and due to the same capacitance value increases by k if thickness of dielectric is same as distance between parallel plates of capacitor.

Complete Step By Step Answer:
Consider a parallel plate capacitor of plate area A and separation d.
Capacitance of such system is given by $C = \dfrac{{{ \in _0}A}}{d}$
Where ${ \in _0}$ is the permittivity of free space having value $8.854 \times {10^{ - 12}}k{g^{ - 1}}{C^2}{s^2}{m^{ - 3}}$
When a dielectric slab of thickness t and dielectric constant k is introduced between plates of capacitor then it is given by
$C' = \dfrac{{{ \in _0}A}}{{(d' - t) + (\dfrac{t}{k})}}$
According to Question $t = 4cm$ , $d' = (d + 3)cm$
And to have same capacitance $C' = C$
Hence putting values we get
$C' = C$
$\dfrac{{{ \in _0}A}}{{(d + 3 - 4) + (\dfrac{4}{k})}} = \dfrac{{{ \in _0}A}}{d}$
$d = (d - 1) + (\dfrac{4}{k})$
$d - d = - 1 + \dfrac{4}{k}$
$0 = - 1 + \dfrac{4}{k}$
$k = 4$
So from here we get $k = 4$
So the correct option is [B] $4$ .

Note :
It should be remembered that after inserting dielectric of thickness t we have to either decrease d or increase A so to put capacitance the same. One must put new increased d in capacitance formula, Also charges induced on dielectric on its surface area of opposite sign so electric field inside dielectric will be opposite in direction of external field