Why impulse response of analog components is not FIR

hi everyone ,i don't understand physically why the imopulse response of analog components like capaciter is not FIR .
Mathematically it is proved by laplace transform but physically there is no explanation known to me.
Any response would be helpful.

The reason I can cite is:

Mathematically its very easy to explain the infinite response of capacitor, because here we are considering the ideal case where there is no resistive component. But in practical stystem, always there will be some resistive component accompanied with the capacitive component. And hence there will be dissipation of energy as heat from the resistive component.

For your example of a capacitor, the answer is simple in my mind. An FIR lacks "memory", that is the resulting output (say voltage on the plates) is the result of the input at the current time (current) and some limited number of inputs from the past.

An IIR has "memory" or some "internal" variable that is a result of the "accumulation" of past inputs (all of them from time zero). In your example the internal variable is the charge on the plates. So if we send an impulse of current into a capacitor it will physically accumulate a charge on the plates and the capacitor will (in an ideal world) maintain a voltage proportional to the accumulated charge forever. The response is not finite in duration.

In the practical world, you end up drawing the charge off the plates through time (whether the charges travel through the air or through the dielectic in the capacitor). The capacitor "leaks" the accumulated charge. I suppose you could model this as an FIR, but modeling as an ideal capacitor (IIR) in parallel with a resistor is more convient.

-jonathan