A) 15.6 \[\mu F\]
B) 11.2\[\mu F\]
C) 19\[\mu F\]
D) 22.4 \[\mu F\]
Correct Answer: B
Solution :
\[C=\frac{{{\varepsilon }_{0}}A}{d}\propto \frac{1}{d}\] Hence,\[\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{{{d}_{2}}}{{{d}_{1}}}\] or \[\frac{2}{{{C}_{2}}}=\frac{0.2}{0.4}\] or \[{{C}_{2}}=\frac{0.4\,\times 2}{0.2}=4\mu F\] Final capacity is \[C=4\times k=4\times 2.8=11.2\,\mu F\]
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