If a slab of insulating material 4times 10-5-m thick is introduced between the plates of a parallel plate capacitor- the distance between the plates has to be increased by 3-5times 10-5-m to restore the capacity to original value- Then the dielectric constant of the material of slab is -861210

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Toppr Admin · Accepted Answer

If K be the dielectric constant- the distance increased due to introduce dielectric is x-t-x2212-tK-t-1-x2212-1K-xA0-where t- thickness of dielectric-According to question- x-3-5-xD7-10-x2212-5-x21D2-t-1-x2212-1K-3-5-xD7-10-x2212-5

If a slab of insulating material 4×10−5m thick is introduced between the plates of a parallel plate capacitor, the distance between the plates has to be increased by 3.5×10−5m to restore the capacity to original value. Then the dielectric constant of the material of slab is :861210

If K be the dielectric constant, the distance increased due to introduce dielectric is x=t−tK=t(1−1K) where t= thickness of dielectric.According to question, x=3.5×10−5⇒t(1−1K)=3.5×10−5

If $K$ be the dielectric constant, the distance increased due to introduce dielectric is $x = t - \frac{t}{K} = t (1 - \frac{1}{K})$
where $t =$ thickness of dielectric.
According to question, $x = 3.5 \times 10^{- 5}$
$\Rightarrow t (1 - \frac{1}{K}) = 3.5 \times 10^{- 5}$