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  4. If a slab of insulating material 4× 10-3 m thick is introduced between the plates of a parallel plate capacitor, the separation between plates has to be increased by 3.5× 10-3 m to restore the capacity to original value. The dielectric constant of the material will be

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Q. If a slab of insulating material $ 4\times {{10}^{-3}} $ m thick is introduced between the plates of a parallel plate capacitor, the separation between plates has to be increased by $ 3.5\times {{10}^{-3}} $ m to restore the capacity to original value. The dielectric constant of the material will be

AMUAMU 1999

Solution:

: Capacity of a parallel plate capacitor $ =\frac{{{\varepsilon }_{0}}KA}{d} $ Initially, $ {{C}_{1}}=\frac{{{\varepsilon }_{0}}A}{d} $ Let separation be increased by $ x $ .
Finally, $ {{C}_{2}}=\frac{{{\varepsilon }_{0}}A}{(d+x-t)+\frac{t}{K}} $ Since $ {{C}_{1}}={{C}_{2}}, $ $ d=d+x-t+\frac{t}{k}; $ $ t-x=\frac{t}{K} $ $ K=\frac{t}{t-x}=\frac{4\times {{10}^{-3}}}{4\times {{10}^{-3}}-3.5\times {{10}^{-3}}}=\frac{4}{4-3.5}=8 $

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