A) 6
B) 4
C) 3
D) 5
Correct Answer: D
Solution :
\[{{C}_{1}}=\frac{{{\varepsilon }_{0}}A}{3}\,\] before \[{{C}_{1}}=\frac{k{{\varepsilon }_{0}}A}{3}\,+\frac{{{\varepsilon }_{0}}A}{2.4}\] after \[\frac{{{\varepsilon }_{0}}A}{3}=\frac{k\frac{{{\varepsilon }_{0}}A}{3}.\,\frac{{{\varepsilon }_{0}}A}{2.4}}{k\frac{{{\varepsilon }_{0}}A}{3}+\frac{{{\varepsilon }_{0}}A}{2.4}}\] \[3k=2.4k+3\] \[0.6\,k=3\Rightarrow \,k=\frac{3}{0.6}\] \[k=\frac{30}{6}=5\]You need to login to perform this action.
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