question_answer

In the given figure each plate of capacitance C has partial value of charge                                              [MP PMT 2003]

A)            CE

B)            \[\frac{CE{{R}_{1}}}{{{R}_{2}}-r}\]

C)            \[\frac{CE{{R}_{2}}}{{{R}_{2}}+r}\]

D)            \[\frac{CE{{R}_{1}}}{{{R}_{1}}-r}\]

Correct Answer: C

Solution :

           In steady state current drawn from the battery \[i=\frac{E}{({{R}_{2}}+r)}\] In steady state capacitor is fully charged hence No current will flow through line (2) Hence potential difference across line (1) is  \[V=\frac{E}{({{R}_{2}}+r)}\times {{R}_{2}}\], the same potential difference appears across the capacitor, so charge on capacitor \[Q=C\times \frac{E{{R}_{2}}}{({{R}_{2}}+r)}\]