A) 4
B) 2
C) 3
D) 1
Correct Answer: C
Solution :
\[{{C}_{air}}=\frac{{{\varepsilon }_{0}}A}{d}\]With dielectric slab inserted between the plates, the capacitance \[C=\frac{{{\varepsilon }_{0}}A}{\left( d-t+\frac{t}{k} \right)}\] Here, \[C=\frac{3}{2}C\] \[\Rightarrow \] \[\frac{{{\varepsilon }_{0}}A}{d-t+\frac{t}{k}}=\frac{3}{2}\frac{{{\varepsilon }_{0}}A}{d}\] \[\Rightarrow \] \[\frac{1}{d\left( 1-\frac{1}{2}+\frac{1}{2k} \right)}=\frac{3}{2d}\] \[\Rightarrow \] \[\frac{1}{2}+\frac{1}{2k}=\frac{2}{3}\] \[\Rightarrow \] \[\frac{1}{2k}=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\] \[\Rightarrow \] \[k=3\]
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