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Step-by-Step Solution
1. Understanding the Scenario
A capacitor of capacitance $C$ is initially charged to a voltage $V_1$. It is then connected to an ideal inductor of inductance $L$. Once connected, the capacitor starts discharging through the inductor and the system undergoes oscillations of charge and current. We need to find the current through the inductor at the instant when the potential difference across the capacitor has fallen to $V_2$.
2. Expressing Charge on the Capacitor
The maximum charge (initially) on the capacitor is given by:
$$ Q_0 = C \, V_1. $$
Once the circuit is connected, the charge on the capacitor varies with time, often expressed as
$$ q(t) = Q_0 \cos(\omega t), $$
where $\omega = \frac{1}{\sqrt{LC}}$ is the angular frequency of the LC oscillations.
3. Relating Charge to Voltage
At any instant, the voltage across the capacitor is
$$ V(t) = \frac{q(t)}{C}. $$
We are interested in the moment when the capacitor’s voltage is $V_2$. This corresponds to
$$ V_2 = \frac{q(t)}{C}. $$
If $q(t) = C \, V_2$, then from the assumed cosine dependence:
$$ \cos(\omega t) = \frac{q(t)}{Q_0} = \frac{C \, V_2}{C \, V_1} = \frac{V_2}{V_1}. $$
4. Expression for the Current
The current $I$ through the inductor is the time derivative of the charge:
$$ I(t) = \frac{dq(t)}{dt} = \frac{d}{dt} \bigl( Q_0 \cos(\omega t) \bigr)
= -\,Q_0 \,\omega \,\sin(\omega t). $$
Taking the magnitude,
$$ |I| = Q_0\, \omega \left|\sin(\omega t)\right|. $$
5. Finding the Sine Term
Using $\sin^2 x + \cos^2 x = 1$, we have
$$ \sin(\omega t) = \sqrt{1 - \cos^2(\omega t)}
= \sqrt{1 - \left(\frac{V_2}{V_1}\right)^2}. $$
Hence,
$$ |I| = \bigl(C\,V_1\bigr)\left(\frac{1}{\sqrt{LC}}\right) \sqrt{1 - \left(\frac{V_2}{V_1}\right)^2}. $$
6. Substituting and Simplifying
Factor out the terms and simplify:
$$ |I| = V_1 \sqrt{\frac{C}{L}} \sqrt{1 - \frac{V_2^2}{V_1^2}}
= V_1 \sqrt{\frac{C}{L}} \,\sqrt{\frac{V_1^2 - V_2^2}{V_1^2}}
= \sqrt{\frac{C\,(V_1^2 - V_2^2)}{L}}. $$
7. Final Answer
The current through the inductor (in magnitude) when the potential difference across the capacitor is $V_2$ is
$$ \boxed{ I = \sqrt{\frac{C\bigl(V_1^2 - V_2^2\bigr)}{L}} }. $$
