question_answer

In the figure shown below, the charge on the left plate of the \[10\mu F\]capacitor is \[-30\,\mu C\]. The charge on the right plate of the \[6\,\mu F\]capacitor is [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

A) \[-18\,\mu C\]                            

B) \[-12\,\mu C\]

C) \[+12\,\mu C\]  

D)                  \[+18\,\mu C\]

Correct Answer: D

Solution :

Let \[{{q}_{1}}\]and \[{{q}_{2}}\]be the charge on \[6\mu F\]and\[4\mu F\]respectively. \[{{q}_{1}}+{{q}_{2}}=q\]                                         ...(i) Also, \[\frac{{{q}_{1}}}{{{C}_{1}}}=\frac{{{q}_{2}}}{{{C}_{2}}}\]                                       ?(ii) [\[\because \]\[{{C}_{1}}\] and \[{{C}_{2}}\] are in parallel combination]       \[\Rightarrow \]\[{{q}_{2}}=\frac{{{C}_{2}}}{{{C}_{1}}}{{q}_{1}}=\frac{4}{6}{{q}_{1}}\]                            ?(iii) Using (i) and (iii),                                        \[\frac{10}{6}{{q}_{1}}=q\Rightarrow q=\frac{5}{3}{{q}_{1}}\Rightarrow {{q}_{1}}=\frac{3}{5}(30)=18\mu C\]