# Two isolated conducting spheres S1 and 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   S1 andS2 are respectively   Option: 1 Option: 2 Option: 3   Option: 4

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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