question_answer

Capacitance of a parallel plate capacitor becomes 6/5 times its original value, if a dielectric slab of thickness t = d/3 is inserted between the plates (d is the original separation between the plates) what will be the value of dielectric constant of the slab?

Answer:

K = 2 Capacitance of a parallel plate capacitor \[C=\frac{{{\varepsilon }_{0}}A}{d}\] When a dielectric slab of thickness t is inserted between the plates, then Capacitance \[C'=\frac{{{\varepsilon }_{0}}A}{d-t+(t/K)}\] [where, K is the value of dielectric constant of the slab] Here,     \[C'=\frac{6}{5}C\Rightarrow t=\frac{d}{3}\] \[\therefore \]      \[\frac{6}{5}C=\frac{{{\varepsilon }_{0}}A}{d-\frac{d}{3}+\frac{d/3}{K}}\] \[\frac{6}{5}C=\frac{{{\varepsilon }_{0}}A}{d\left( \frac{1+2\,K}{3K} \right)}=C\left( \frac{3K}{1+2K} \right)\] i.e.        \[\frac{3K}{1+2K}=\frac{6}{5}\Rightarrow K=2\]