A) \[\frac{3K{{\varepsilon }_{0}}A}{4d}\]
B) \[\frac{4K{{\varepsilon }_{0}}A}{4d}\]
C) \[\frac{\left( K+1 \right){{\varepsilon }_{0}}A}{2d}\]
D) \[\frac{K\left( K+3 \right){{\varepsilon }_{0}}A}{2\left( K+1 \right)d}\]
Correct Answer: D
Solution :
[d] \[{{c}_{1}}=\frac{\left( A/2 \right){{\varepsilon }_{0}}}{d/2}=\frac{A{{\varepsilon }_{0}}}{d},\] \[{{c}_{2}}=K\frac{A{{\varepsilon }_{0}}}{d},{{c}_{3}}=K\frac{A{{\varepsilon }_{0}}}{2d}\] \[\therefore {{c}_{eq}}=\frac{{{c}_{1}}\times {{c}_{2}}}{{{c}_{1}}+{{c}_{2}}}+{{c}_{3}}=\frac{\left( 3+K \right)KA{{\varepsilon }_{0}}}{2d\left( K+1 \right)}\] (\[\therefore {{C}_{1}}\]and \[{{C}_{2}}\]are in series and resultant of these two in parallel with \[{{C}_{3}}\])You need to login to perform this action.
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