Sure. The equation relating capacitance, charge and voltage is: V = Q/C where V = voltage, Q = charge and C = capacitance Taking the time differential of both sides we get: dV/dt = (1/C)*dQ/dt where dV/dt = voltage change/time = slew rate, dQ/dt = change in charge/time = current = I So: dV/dt = I/C, or C*dV/dt = I For my example, dV/dt = 30 V/microsecond, C = 100 pf, plugging everything in gives I = 3 mA for 1600 volts peak-to-peak. I used to be a physics major in a previous life. ) The reason you got a different answer for slew rate is that you are calculating for a full-power, 20 kHz signal. However, music does not contain full-power signals at 20 kHz. As Nelson Pass and Peter Baxandall found, the music power spectrum rolls off above around 5-6 kHz at a rate of approximately 6 dB/octave. Because of this, for a 100 watt amp that can swing 80 volts peak-to-peak the fastest slew rate with a music signal is 1.5 volts/microsecond up to clipping, whereas a sine wave 20 kHz signal at clipping would slew at 5 volts/microsecond for the same amp. The fact is that music signals are not that fast compared to some test signals. Multiply those results by 20 for a 1600 volt peak-to-peak signal at 20 kHz gets you to 100 volts/microsec, which is what you calculated, or 30 volts/microsecond for music signals, which is what I got. You'll get the same answer as me if you plug in a full-power signal at 6 kHz, which is what Baxandall said was needed to reproduce music. Now, the calculations in the second post assumed a signal of 800 volts peak-to-peak, which is 400 volts peak, whereas your calculation was for 800 volts peak, hence your answer is twice as high as mine. My number also includes the 1.5 mA current into the capacitative load of the headphone, and because that current is 90 degrees out of phase with the current to the resistor, the overall answer 8.1 mA (rounded off) rather than 8 mA for the resistor load alone.