A) \[\frac{8{{\varepsilon }_{0}}A}{d}\]
B) \[\frac{8{{\varepsilon }_{0}}A}{3d}\]
C) \[\frac{26{{\varepsilon }_{0}}A}{35d}\]
D) \[\frac{12{{\varepsilon }_{0}}A}{35d}\]
Correct Answer: B
Solution :
Here three slabs are in series \[{{C}_{eq}}=\frac{{{\varepsilon }_{0}}A}{\left( \frac{d/4}{2}+\frac{d/7}{8/7}+\frac{d/2}{4} \right)}\] \[=\frac{{{\varepsilon }_{0}}A}{\left( \frac{d}{8}+\frac{d}{8}+\frac{d}{8} \right)}=\frac{8{{\varepsilon }_{0}}A}{3d}\]You need to login to perform this action.
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