question_answer

A spherical capacitor of inner radius a = 1 cm and outer radius b = 2 cm is earthed as shown. It can be connected to an isolated metallic sphere of radius c = 1 cm through a switch S and a very long conducting wire. If initial charge on inner sphere is\[q=30\mu C\], then charge on the sphere of radius c, when switch S is closed, will be

A)  \[10\mu C\]                      

B)  \[20\mu C\]

C)  \[25\mu C\]                      

D)  \[30\mu C\]

Correct Answer: A

Solution :

 Idea This question is judging the knowledge of Electrostatics of conductor and Electric potential. Here after connecting two spheres, their potential will be same. When switch S is closed, the charge on sphere A will be redistributed on A and C in the ratio of their capacitances, \[\frac{{{q}_{1}}}{q-{{q}_{1}}}=\frac{4\pi {{\varepsilon }_{0}}\left( \frac{ab}{b-a} \right)}{4\pi {{\varepsilon }_{0}}c}\] \[{{q}_{1}}=\frac{qab}{ab+bc-ca}\] Taking\[q=30\mu C\] \[a=1cm,b=2cm,c=1cm\] We get, \[{{q}_{1}}=20\mu C\] \[\therefore \]\[q-{{q}_{1}}=30-20=10\mu C\] TEST Edge Similar questions where two concentric shells are connected with conducting wires and all the charge on inner shell would flow on the outer shell could be asked. Here, one should remember that positive charge will flow towards lower potential and vice-versa.