There are plenty of circuits available that will block the application of 400 VDC, typically using a MOSFET gated OFF until a "trigger" is received. The trigger, and associated logic, then allows the 400 VDC to be applied to subsequent circuit loads seven milliseconds (0.007s) after the trigger occurs. It is your design choice as to whether this occurs after the leading edge, or after the trailing edge, or at a particular voltage level of the trigger.

If the source of the 400 VDC is a capacitor, and the subsequent circuit load is a resistance, then the voltage across the resistance increases abruptly to the capacitor voltage, 400 VDC, at the end of the 7 ms delay period when the MOSFET is gated ON, and decreases exponentially after that according to this equation:

Eload = (400 V) (e^-(t/RC) )

where Eload is the voltage across the resistance load as well as the capacitor voltage, e is the base of natural logarithms or approximately 2.718281828459. ..., the caret ^ represents exponentiation, t is the elapsed time in seconds after the capacitor is connected to the load resistance, R is the load resistance in ohms, and C is the capacitance value in farads.

Neglected in this analysis is the small, decreasing, voltage drop from drain-to-source across the MOSFET when it is turned ON after the 7 ms delay.

Normally, a capacitor is considered to be completely discharged after six time constants, which is the product of R and C in seconds. Thus, if R = 1 megohm and C = 1 microfarad, the RC time constant is one second and the capacitor is considered discharged at the end of six seconds. This is nonsense of course, because the exponential factor never does equal zero, no matter how long t is. However, after a sufficiently long period of time, the only voltage remaining on the capacitor is random thermal noise.