B Why does an inductor discharge?

[versine]

A capacitor will discharge because once a path is created there is a potential difference $V=Q/C$. But for an inductor it's $L\frac{di}{dt}$. Why would an inductor want to discharge once you connect it to a capacitor?

 

[Baluncore]

If there was no current flowing through the inductor, then there is no energy stored, and no magnetic field.
E = ½·L·i² ;

If a current is flowing through the inductor when it is connected to a zero volt capacitor, the current will begin to charge the capacitor.
Q = I·t ; v = Q / C ; E = ½·C·v² ;

As the capacitor voltage increases, the inductor voltage will also increase, that will reduce the inductor current, until it stops, reverses, and repeats, circulating the energy between L and C.
v = L·di/dt ;

 

[vanhees71]

Just write down the equation from Kirchhoff's rule:
$$L \dot{i}+Q/C=0$$
Since $i=\dot{Q}$ this means
$$L \ddot{Q}+Q/C=0$$
The general solution of this "harmonic-oscillator equation of motion" is
$$Q(t)=A \cos(\omega t) + B \sin(\omega t) \quad \text{with} \quad \omega=\sqrt{\frac{1}{LC}}$$
and
$$i(t)=-A\omega \sin(\omega t) + B \omega \cos(\omega t).$$
For $Q(0)=Q_0$ and $i(0)=0$, i.e., having some charge on the capacitor and then at $t=0$ connect it to the coil, then you get $A=Q_0$ and $B=0$. So the final solution reads
$$Q(t)=Q_0 \cos(\omega t), \quad i(t)=-Q_0 \omega \sin(\omega t).$$