question_answer

Two metal spheres of radii 0.01 m and 0.02 m are given a charge of \[15mC\]and \[45mC\]respectively. They are then connected by a wire. The final charge on the first sphere is ____\[\times {{10}^{-3}}C\]

A) \[40\]    

B) \[30\]    

C) \[20\]    

D)  \[10\]

Correct Answer: C

Solution :

: The situation is as shown in the figure. When they are connected by a wire the charge flows and both the spheres attain the common potential which is given by \[V=\frac{{{Q}_{1}}+{{Q}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\frac{{{Q}_{1}}+{{Q}_{2}}}{4\pi {{\varepsilon }_{0}}({{R}_{1}}+{{R}_{2}})}\] \[\therefore \] Final charge on the first sphere is \[Q_{1}^{}={{C}_{1}}V=\frac{4\pi {{\varepsilon }_{0}}\,{{R}_{1}}({{Q}_{1}}+{{Q}_{2}})}{4\pi {{\varepsilon }_{0}}({{R}_{1}}+{{R}_{2}})}\]  (Using (i)) \[=\frac{{{R}_{1}}({{Q}_{1}}+{{Q}_{2}})}{{{R}_{1}}+{{R}_{2}}}\] Here, \[{{R}_{1}}=0.01m,\,\,{{R}_{2}}=0.02m\] \[{{Q}_{1}}=15mC=15\times {{10}^{-3}}C\] \[{{Q}_{2}}=45mC=45\times {{10}^{-3}}C\] \[\therefore \] \[Q_{1}^{}=\frac{0.01(15\times {{10}^{-3}}+45\times {{10}^{-3}})}{0.01+0.02}\] \[=\frac{0.01\times 60\times {{10}^{-3}}}{0.03}=20\times {{10}^{-3}}C\]