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Alexphysics edited
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+### Statement
+
+$11.4.20.$ [Insert the problem statement]
+
+### Solution
+
+Let q be the charge that flows from the positive plate of$ C_0$ toward C. Then the charge on$ C_0$ is $Q_0 - q $(with $ Q_0 = C_0 V_0$) and the charge on C is q. The voltages are
+
+$V_{C0} = \frac{Q_0 - q}{C_0}, \qquad V_C = \frac{q}{C}$
+
+The circuit equation is obtained from the mesh:
+
+$V_{C0} - V_C = L \frac{dI}{dt} = L \frac{d^2q}{dt^2}$
+
+Substituting the above expressions,
+
+$\frac{Q_0}{C_0} - \frac{q}{C_0} - \frac{q}{C} = L \ddot{q}$
+
+Rearranging, we obtain the differential equation of the forced harmonic oscillator:
+
+$L \ddot{q} + q\left(\frac{1}{C_0} + \frac{1}{C}\right) = V_0$
+
+The natural angular frequency of the circuit is
+
+$\omega = \frac{1}{\sqrt{L\,C_{\text{eq}}}} = \sqrt{\frac{C+C_0}{L\,C\,C_0}}$
+
+where
+$ C_{\text{eq}} = \dfrac{C\,C_0}{C+C_0}$ is the equivalent series capacitance.
+
+The general solution with the initial conditions $q(0)=0 $(capacitor C starts uncharged) and$ I(0)=\dot{q}(0)=0$ (the inductor prevents abrupt changes in current) is
+
+$q(t) = C_{\text{eq}} V_0 \bigl(1 - \cos\omega t\bigr)$
+
+Therefore, the voltage on capacitor C evolves as
+
+$V_C(t) = \frac{q(t)}{C}
+ = \frac{C_{\text{eq}}}{C}\,V_0 \bigl(1 - \cos\omega t\bigr)
+ = \frac{V_0}{1 + \dfrac{C}{C_0}} \bigl(1 - \cos\omega t\bigr)$
+
+The maximum value that V_C can reach is
+
+$V_{C,\max} = \frac{2V_0}{1 + \dfrac{C}{C_0}}$
+
+If the breakdown voltage V is greater than this maximum, capacitor C never breaks down.
+Otherwise $(V < V_{C,\max})$ breakdown occurs at the instant $\tau $when $V_C(\tau) = V$
+
+Solving for the cosine and then for time:
+
+$\cos\omega\tau = 1 - \left(1 + \frac{C}{C_0}\right)\frac{V}{V_0}$
+
+$\boxed{\tau = \frac{1}{\omega} \arccos\!\left[1 - \left(1 + \frac{C}{C_0}\right)\frac{V}{V_0}\right]}$
+
+#### Answer
+
+[Insert a concise answer or boxed result]