Is 30 pf per foot considered low?
If you just want the simple answer, just scroll down to the bottom. If you want he explanation, keep reading.
The potential problem with cable capacitance is that shunts away part of the signal from the circuit you're trying to send it to. This normally would not be a serious problem if the effect were frequency independent. But as shown below, it IS frequency dependent, affecting high frequencies more than low which leads to distortion of the signal. Here's how it works:
The effect depends on the length of the cable you're using. The capacitance will represent an impedance in parallel with the input impedance of the circuit the cable is feeding. The impedance of a capacitor is given by:
Z = 1/[2*pi*C*f]
Where pi = 3.14159, etc., C is the capacitance (in Farads, not pico Farads) and f is the frequency of the signal being transmitted. In your case the capacitance will be 30pF times the length of the cable. To give you a concrete example, I'll consider a 10 foot cable carrying a 10 K Hertz signal into a circuit with 10 K ohms input impedance.
Then in the above equation, f = 10000, and C = 300 pF, which is 300*10
-12 Farad. This gives a capacitive impedance of 159 K ohms. The parallel combination of this capacitance with the 10 K ohm input impedance reduces the total impedance to 9.4 K ohms; a 6% change, or less than 0.3 dB, which is a difference well below the ability of the ear to detect.
The change of impedance for a 100 Hertz signal would be, by a similar calculation, less than 0.1%; even less detectable.
So unless you have an unusual situation, such as a 100 foot or more cable, you surely are OK with 30pF per foot.