- #26

- 562

- 164

At this point, ignore ##Q_0##. I think that, up until now, we are clear on$$\frac{kQ_A(t)}{r} - \frac{k(Q_0-Q_A(t))}{r} = L\frac{di}{dt}$$See

various thoughts crossing my mind-

1. I have four unknowns ( ##Q_0## ,##Q_A## ,##Q_B## and i ) and only 3 equations

2. Even if manage to solve the equations and find ##Q_A##(t)/ whatever, how will i get the time in which charge flow occurs

could you give me a birds eye view of the sum once again and then let's proceed with equation solving:)

We need to find the appropriate expression for ##i(t)##; either ##Q'_B(t)##, or ##Q'_A(t)##.

Before losing half of its initial charge, it should be clear that ##Q_A(t)\geq Q_B(t)##, since ##Q_B(t)## starts from being zero coulombs and grows by taking the charge lost from A's.

Thus

$$\frac{kQ_A(t)}{r} - \frac{k(Q_0-Q_A(t))}{r}\geq0$$

as long as the charge in A is greater than half of its initial value.

Now, we have$$\frac{kQ_A(t)}{r} - \frac{k(Q_0-Q_A(t))}{r} = L\frac{di}{dt}$$

or, in other terms, ##\alpha=\beta##. If ##\alpha>0##, then what can you infer about ##\beta##?