If we don't have enough energy to kill the insulator what will happen? The resistor limits the maximum current, so high power and temperature won't occur.
The constant and increased DC leakage can cause permanent damage? I especially interested about gate-oxide capacitors, where the insulator is SiO2, the plates are polySi and doped Si, the overvoltage is not transient peak, so about DC condition.
I assume this will result in problems.
You don´t need much energy to destroy insulation. The stored energy in the capacitor is sufficient, especially when there is a high voltage source charging the capacitor. It may be a long time process.
Yes, the capacitor charged in slower rate then when resistor is not connected but the maximum charge that will store in the capacitor will be same.
Because of above mentioned reason the capacitor will not work properly.
Why will be the same? The insulator has a breakdown voltage, which can't allow higher voltage than it. So charge and energy is limited.
I think if the generator's voltage is over the breakdown voltage the conduction of the capacitor will increase only. Charge and so energy in the capacitor won't.
The power is limited by the resistance of the voltage source so high current and high temperature can't appear. If we choose the series resistance correctly I think we can protect the device. https://www.allaboutcircuits.com/textbook/direct-current/chpt-12/insulator-breakdown-voltage/