question_answer

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

A)  6                  

B)  8

C)  10               

D)  12

Correct Answer: B

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\[{{C}_{1}}={{C}_{2}},\] \[d=d+x-t+\frac{t}{k};\] \[t-x=\frac{t}{K}\] \[K=\frac{t}{t-x}=\frac{4\times {{10}^{-3}}}{4\times {{10}^{-3}}-3.5\times {{10}^{-3}}}=\frac{4}{4-3.5}=8\]