A)
B)
C)
D)
Correct Answer: D
Solution :
\[{{C}_{eq}}=\frac{3C}{C+3}\] \[Q=\frac{3C\,{{E}_{0}}}{C+3}\] \[{{Q}_{2}}=\frac{2}{3}\left( \frac{3C\,{{E}_{0}}}{C+3} \right)=\frac{2C{{E}_{0}}}{C+3}\] \[C=1\Rightarrow {{Q}_{2}}=\frac{2{{E}_{0}}}{4}=\frac{{{E}_{0}}}{2}\] \[C=3\Rightarrow {{Q}_{2}}=\frac{2\times 3\times {{E}_{0}}}{2\times 3}={{E}_{0}}\] \[{{Q}_{2}}=2{{E}_{0}}\left[ \frac{C}{C+3} \right]\] \[\frac{d{{Q}_{2}}}{dC}=2{{E}_{0}}\left[ \frac{\left( C+3 \right)\times 1-\left( C \right)\times 1}{{{\left( C+3 \right)}^{2}}} \right]=2{{E}_{0}}\left( C+3-C \right){{\left( C+3 \right)}^{2}}\] \[d{{Q}_{2}}=\frac{6{{E}_{0}}}{{{\left( C+3 \right)}^{2}}}\]slope decreasesYou need to login to perform this action.
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