Question

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

Answer 1

Answer: (B)

: Capacity of a parallel plate capacitor $=\frac{{\epsilon }_{0}KA}{d}$ Initially, $C_1=\frac{{\epsilon }_{0}A}{d}$ Let separation be increased by $x$ .
Finally, $C_2=\frac{{\epsilon }_{0}A}{(d+x-t)+\frac{t}{K}}$ Since $C_1=C_2,$ $d=d+x-t+\frac{t}{k};$ $t-x=\frac{t}{K}$ $K=\frac{t}{t-x}=\frac{4\times {10}^{-3}}{4\times {10}^{-3}-3.5\times {10}^{-3}}=\frac{4}{4-3.5}=8$