Two isolated conducting spheres Sand S2 of radius  \frac{2}{3}Rand \frac{1}{3}R have 12 \mu C and -3\mu C charges respectively and are at a large distance from each other. They are now connected by a conducting wire. A long time after this is done the charges on   SandS2 are respectively
 
Option: 1 + 4.5\mu C\; and\; + 4.5\mu C
Option: 2 + 4.5\mu C\; and\; - 4.5\; \mu C
Option: 3 + 3\mu C\; and\; +6\; \mu C  
Option: 4 + 6\mu C\; and\; +3\; \mu C

Answers (1)

After connecting with wires, there will be charge flow such that the potential of both spheres are equal.

Let the flow of charge be q from (1) to (2)

So, final charge on (1)=(12-q)\mu C

and, final charge on (2)=(-3+q)\mu C

So,

V_{1}=V_{2}

\frac{k(12-q)}{(2R / 3)}=\frac{k(q-3)}{(R / 3)}

\begin{aligned} 12-q &=2q-6 \\ 3 q=& 18 \\ \Rightarrow q=6 \mu C \\ So, \\ Q_{1}=& 12-6=6 \mu \mathrm{c} \\ Q_{2}=&-3+6=3 \mathrm{\mu C} \end{aligned}

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