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 \[q=30\,\,\mu C,\] the 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 :

 When switch S is closed, the charge on sphere A will be redistributed on A and C in the ratio of their capacitances, \[\frac{{{q}_{1}}}{q-{{q}_{1}}}=\frac{4\pi {{\varepsilon }_{0}}\,\left( \frac{ab}{b-a} \right)}{4\pi {{\varepsilon }_{0}}C}\] \[{{q}_{1}}=\frac{q\,ab}{ab+bc-ca}\] Taking \[q=30\,\mu C\] \[a=1\,cm,\]     \[b=2\,cm,\]     \[c=1\,cm\] We get, \[{{q}_{1}}=20\,\,\mu C\] \[\therefore \]    \[q-{{q}_{1}}=30-20=10\,\mu C\]