A) \[{{q}_{A}}=275\,\mu C,{{q}_{B}}=3.15\,\mu C\]
B) \[{{q}_{A}}=1.09\,\mu C,{{q}_{B}}=1.53\,\mu C\]
C) \[{{q}_{A}}={{q}_{B}}=5.5\,\mu C\]
D) None of the above
Correct Answer: C
Solution :
When touched charge flows from higher potential to lower potential till it equalises on the two discs. Given, \[{{q}_{1}}={{10}^{-6}}C,{{q}_{2}}={{10}^{-5}}C\] \[\therefore \] \[q=\frac{{{q}_{1}}+{{q}_{2}}}{2}=\frac{{{10}^{-6}}+{{10}^{-5}}}{2}=5.5\mu C\]
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