A) \[{{V}_{0}}\frac{({{C}_{1}}-{{C}_{2}})}{{{C}_{1}}}\]
B) \[{{V}_{0}}\frac{{{C}_{1}}}{({{C}_{1}}-{{C}_{2}})}\]
C) \[{{V}_{0}}\frac{{{C}_{1}}}{({{C}_{1}}+{{C}_{2}})}\]
D) \[{{V}_{0}}\frac{({{C}_{1}}+{{C}_{2}})}{{{C}_{1}}}\]
Correct Answer: C
Solution :
Key Idea: Total amount of charge in an isolated system remains constant. When a conductor is given a charge, its potential rises in proportion to the charge given. Thus, if a charge Q raises the potential of the conductor by \[{{V}_{o}},\]then \[Q={{C}_{1}}{{V}_{0}}\]where \[{{C}_{1}}\]is capacitance of first capacitor. When the charging battery is removed and capacitor is connected to an uncharged capacitor \[{{C}_{2}},\]this charge \[{{C}_{1}}{{V}_{0}}\]is now shared across \[{{C}_{1}}\]and \[{{C}_{2}}\]. From law of conservation of charge, we have \[{{C}_{1}}{{V}_{0}}={{C}_{1}}V+{{C}_{2}}V\] \[\Rightarrow \] \[V={{V}_{0}}\frac{{{C}_{1}}}{{{C}_{1}}+{{C}_{2}}}\]
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