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Addendum:
I have just read Strict(er) adherence to homework guidelines? and think that this is the reason why there have been three votes to close this topic. Notice, though, that I am not asking for a solution, as I already know how to solve this (multiplying by a possible $\pm1=c$ mentioned further on).

Two fixed identical metallic spheres $A$ and $B$ of radius $R=50\,\mathrm{cm}$ each are placed on a non-conducting plane at a very large distance from each other and they are connected by a coil of inductance $L=9\,\mathrm{mH}$ as shown in figure.

One of the spheres (say $A$) is imparted an initial charge and the other is kept uncharged. The switch $S$ is closed at $t=0\,\mathrm{s}$. After what minimum time $t$ does the charge on the first sphere decrease to half of its initial value ? figure

Setting up the differential equation of the loop law, and eliminating $Q_B$, I get: $$\frac{kQ_A(t)}{R} - \frac{k(Q_0-Q_A(t))}{R} = Li'(t)$$ I know that $i'(t)>0$ since the left hand side is positive, and thus $i(t)=Q'(t)$ should be growing, i.e $Q''(t)>0$, but I don't see how I can verify this for either of the charges as they are not given in the form of an explicit function.

Any idea? Choosing $Q=cQ_A$, for example, works, since I can determine $c$ to be either $1$ or $-1$ during my identifying the variables in the solution to the differential equation, but is there a way to actually know directly without solving further and seeing how to chose $c$? What's the physically correct way?

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    $\begingroup$ Note: With the switch closed, this acts like an LC circuit with no resistance. $\endgroup$
    – R.W. Bird
    Commented May 5, 2020 at 18:16

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The setup of the exercise is ok, now you only have to choose which is the positive direction of the current $i(t)$: e.g. if you choose that $i(t)>0$ when it flows from $A$ to $B$ then you will have that -in that case- it is $Q_B(t)$ that increases, so $$i(t) = Q_B'(t) = -Q_A'(t) $$ At this point you can substitute this in your equation and get a second-order differential equation for $Q_A(t)$.

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  • $\begingroup$ I have already chosen that the charge flows from A to B! Please reread the before last paragraph. $\endgroup$
    – GDGDJKJ
    Commented May 5, 2020 at 14:46
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    $\begingroup$ The direction of the current that you choose is conventional, you only need it to solve the exercise: if you choose the "wrong" direction you will simply get a minus sign in the solution. In this problem you can have a physical intuition of where the current is flowing to and choose that, but this is not always possible in more complex configurations. Also consider that the direction of the current itself is conventional, as it is physically produced of the flow of electrons in the opposite direction. $\endgroup$
    – Luca M
    Commented May 5, 2020 at 15:04

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