Re: hmmm... let me think...
the current through a capacitor = C dv/dt which when solved for a sinusoidal wave Magnitude(V) = Magnitude(i)/ (2*pi*f*C) thus for very low frequencies applying a voltage across the capacitor yields very small current which makes the capacitor acts as a very high impedance (or open circuit at d.c). Similarly for very high frequencies applying a voltage across the capacitor yields very high current which makes the capacitor act as a short circuit at high frequencies.
thus for high pass filter at low frequencies the capacitor isolates the input voltage and the resistance has no current in it thus producing an output voltage =0
at high requencies the capacitor acts as a short ciruit connecting the input and the output voltages
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