A) 40.5 pF
B) 20.25 pF
C) 1.8 pF
D) 45 pF
Correct Answer: A
Solution :
\[{{C}_{0}}=9pF=\frac{{{\varepsilon }_{0}}A}{d}\,\,;\,\,\,\frac{1}{C}=\frac{d/3}{{{\varepsilon }_{0}}A{{\kappa }_{1}}}+\frac{2d/3}{{{\varepsilon }_{0}}A{{\kappa }_{2}}}\] \[{{\kappa }_{1}}=3,\,{{\kappa }_{2}}=6\] \[\therefore \]\[C=\frac{9}{2}\frac{{{\varepsilon }_{0}}A}{2}=\frac{9}{2}\left( 9pF \right)=40.5\,\,pF\]You need to login to perform this action.
You will be redirected in
3 sec