Electronic – Is power the rate of energy flowing in and out a capacitor

"I want to transfer a few Joules in a few nanoseconds."

This is going to pose some major obstacles.

In order to reduce the discharge time the capacitor should be as small as possible. To increase de level of energy storage I'd set the maximum voltage at 1000V, which is already hazardous.

C = 2E/V² or 2*5/(1000*1000) = 10μF

When loaded by resistor R, a capacitor could be considered fully discharged after 5*R*C secs. R = t/(5*C) = 5.10𐄐⁹/(5*10.10𐄐⁶) = 0.1mΩ.

This is about the resistance of 10cm of 15AWG copper wire and far below the ESR of a physical capacitor.

One of your equations is incorrect: the energy in a capacitor is \$ \frac{1}{2}QV\$. Then, the power is

\$P = \frac{dw}{dt} = \frac{d}{dt}(\frac{1}{2}QV) = \frac{1}{2}[Q\frac{dV}{dt} + V\frac{dQ}{dt}]\$

Then, since V = \$\frac{Q}{C}\$, \$\frac{dV}{dt} = \frac{1}{C}\frac{dQ}{dt}\$, so

\$ P = \frac{1}{2}[\frac{Q}{C}\frac{dQ}{dt} + V\frac{dQ}{dt}] = \frac{1}{2}[V\frac{dQ}{dt} + V\frac{dQ}{dt}] = IV \$

with no assumptions about constant voltage.

I haven't thought about your second question, but I think the same idea will help out there too.