question_answer

In the given circuit, charge \[{{Q}_{2}}\] on the \[2\mu F\] capacitor changes as C is varied from \[1\mu F\] to \[3\mu F.{{Q}_{2}}\] as a function of ‘C’ is given properly by : (figures are drawn schematically and are not to scale)  [JEE MAIN 2015]

A)

B)

C)

D)

Correct Answer: D

Solution :

[d]

\[{{C}_{eq}}=\frac{3C}{C+3}\]

\[Q=\frac{3C\,{{E}_{0}}}{C+3}\]

\[{{Q}_{2}}=\frac{2}{3}\left( \frac{3C\,{{E}_{0}}}{C+3} \right)=\frac{2C{{E}_{0}}}{C+3}\]

\[C=1\Rightarrow {{Q}_{2}}=\frac{2{{E}_{0}}}{4}=\frac{{{E}_{0}}}{2}\]

\[C=3\Rightarrow {{Q}_{2}}=\frac{2\times 3\times {{E}_{0}}}{2\times 3}={{E}_{0}}\]

\[{{Q}_{2}}=2{{E}_{0}}\left[ \frac{C}{C+3} \right]\]

\[\frac{d{{Q}_{2}}}{dC}=2{{E}_{0}}\left[ \frac{\left( C+3 \right)\times 1-\left( C \right)\times 1}{{{\left( C+3 \right)}^{2}}} \right]\]

\[=2{{E}_{0}}\frac{\left( C+3-C \right)}{{{\left( C+3 \right)}^{2}}}\]

\[d{{Q}_{2}}=\frac{6{{E}_{0}}}{{{\left( C+3 \right)}^{2}}}\]slope decreases