A) 3q, 6q
B) 6q, 3q
C) 4.5q, 4.5q
D) 5q, 4q
Correct Answer: B
Solution :
Let charges is\[{{q}_{1}}\]and\[{{q}_{2}}\]after closing key S then \[V=\frac{{{q}_{1}}}{6C}=\frac{{{q}_{2}}}{3C}\] \[\Rightarrow \] \[\frac{{{q}_{1}}}{2}={{q}_{2}}\] \[\Rightarrow \] \[{{q}_{1}}=2{{q}_{2}}\] ...(i) Charge is always conserved \[\therefore \] \[{{q}_{1}}+{{q}_{2}}=3q+6q\] or \[{{q}_{1}}+{{q}_{2}}=9q\] ...(ii) From Eqs. (i) and (ii) \[2{{q}_{2}}+{{q}_{2}}=9q\] \[\Rightarrow \] \[3{{q}_{2}}=9q\] From Eq. (i) \[{{q}_{1}}=2\times 3q\] \[\Rightarrow \] \[{{q}_{1}}=6q\] \[(\because {{q}_{1}}=6q,{{q}_{2}}=3q)\]You need to login to perform this action.
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