A) 6
B) 8
C) 4
D) 2
Correct Answer: D
Solution :
Original capacity \[{{C}_{0}}=\frac{{{\varepsilon }_{0}}A}{d}\] On introducing dielectric slab of thickness \[\frac{d}{2}\] the capacity becomes \[C=\frac{{{\varepsilon }_{0}}A}{\left( d-\frac{d}{2} \right)+\frac{d}{2K}}=\frac{{{\varepsilon }_{0}}A}{\frac{d}{2}\left( 1+\frac{1}{K} \right)}\] As \[C=\frac{4}{3}{{C}_{0}}\] \[\therefore \] \[\frac{{{\varepsilon }_{0}}A}{\frac{d}{2}\left( 1+\frac{1}{K} \right)}=\frac{4}{3}\frac{{{\varepsilon }_{0}}A}{d}\] Or \[1+\frac{1}{K}=\frac{3}{2}\]or \[K=2\]
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