A) \[C\,E\]
B) \[\frac{C\,E\,{{R}_{1}}}{{{R}_{1}}+r}\]
C) \[\frac{C\,E\,{{R}_{2}}}{{{R}_{2}}+r}\]
D) \[\frac{C\,E\,{{R}_{1}}}{{{R}_{2}}+r}\]
Correct Answer: C
Solution :
From the figure, we see that the current in the circuit would be equal to \[I({{R}_{2}}+r)=E\] As no current flows through the capacitance in a DC connection Thus, we get \[I=\frac{E}{({{R}_{2}}+r)}\] Thus, the potential across the resistance \[{{R}_{2}}\]is equal to \[V=I{{R}_{2}}=\frac{E{{R}_{2}}}{({{R}_{2}}+r)}\]and thus the charge on the capacitor would be equal to\[Q=CV=\frac{E{{R}_{2}}V}{({{R}_{2}}+r)}\]
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