Op-amp circuit oscillation
The concept of virtual ground cannot be applied to this circuit. It is applicable only to circuits where the net opamp circuit is in negative feedback mode and it is the negative feedback which ensures that \$V- = V+\$. Here the opamp circuit has positive feedback via R2 and R3. The output of the opamp is driven to saturation due to this positive feedback. Hence \$V-\$ need not always equal \$V+\$.
&quot;Because after the capacitor voltage reaches VTH, when it starts to decrease&quot;. The terminology &quot;starts to decrease&quot; can only be applied to a real opamp with finite slew rate. For an ideal opamp with infinite slew rate, the output instantly changes from \$V_s\$ to \$-V_s\$ (It doesn't go through zero voltage). So there is no time instant where the current is zero. Current also instantly changes sign.
Before switching
\$V^+ = +V_{TH} &gt; V_C = V^-\$. So \$V_o = +V_s\$. (consistent set of eqns)
current \$I = (V_s - V_C)/R\$. and \$V_C \rightarrow V_{TH}\$.
Instantly after switching
\$V_C = V^- &gt; V^+\$. So \$V_o = -V_s\$. So \$V^+ = -V_{TH}\$. (consistent set of eqns)
\$I = (-V_s - V_C)/R\$.
i.e. current is non zero always.
With ideal components which can have infinite slew rate and discontinuous outputs, the intuition breaks down and you have to rely on which set of equations are consistent and which are not. In your case current = 0 assumption leads to a set of inconsistent equations (as you have noticed). That means that current doesn't reach such a condition.