Input Leakage Current
To determine your resistors voltage drop from the gate you need to use the leakage current from the datasheet. Microchip specifies an "Input Leakage Current" on their datasheets. The [datasheet that I have looked up][1] specifies an input leakage current of 1uA. This could cause a .1V or 100mV, which is only double what Robert calculated, probably not a problem on your signal.
Now remember, if you are dividing a 30V signal down to 30/11 (2.7v) volts full read then the 100mV is added to this, causing up to 3% error on your 30V signal.
If you need a resolution of 1V, divide that by 11 and then add the 100mV. This 100mV could be larger than the 1V signal.
Input Capacitance
Robert is correct, there will be a capacitance, but this really specifies an amount of time that is needed to take the ADC measurement. This also, combined with the input resistance you chose, creates a low pass filter; if you were wanting to measure signals with a higher frequency, you are not going to be able to capture them.
Reducing the error
The easiest way is to either reduce your resistance on your divider, or to buffer your signal. When you buffer the signal you will replace the PIC's leakage current with your op-amps leakage current which you can get quite low.
This 1uA is a worst case, unless it costs you a large amount to make minor changes to the design, fab your design and test how bad it is for you.
Please let me know if there is anything I can do to make this easier to read.
Actually the capacitors acts as an impedance which influences the circuit.
Impedance, the vector sum of reactance and resistance, describes the phase difference and the ratio
of amplitudes between sinusoidally varying voltage and sinusoidally varying current at a given frequency.
Fourier analysis allows any signal to be constructed from a spectrum of frequencies,
whence the circuit's reaction to the various frequencies may be found.
The reactance of a capacitor is given by:
$$X_c=\frac{1}{2\pi{} f C}$$
$$\pi\approx3.14$$
$$f=frequency $$
$$C=Capacitance$$
As the frequency increases the reactance decreases.
like this capacitance influences the circuit.
Best Answer
Those high impedances suggest that you are interested in voltage rather than current. For voltage, your measuring instrument loads down the source, reducing measured voltage. A Thevenin equivalent is used to model the source.
For current, the source model is a Norton equivalent, and its current is divided between the 100M source and the 1M instrument. Most of the current in this case goes into the instrument.
There are other effects. If your 100M source is connected to lock-in and/or spectrum analyzer with coax, you must consider cable capacitance in determining effective bandwidth and phase response. A large instrument input-Z results in a low-pass filter whose pass band is small. A lower instrument input-Z results in a smaller voltage signal, but extends the pass band. In some older lock-ins, signal harmonics contribute to measured signal. Your phase and amplitude response may change with cable length.
Since these instruments are often used to measure signals where noise is significant, the instrument loading effect may make the instrument noise floor dominate noise from your source. Lock-in noise is usually spec'd in terms of volts-per-root-Hertz, like 6 nV/(rt Hz). You can measure very small noise levels by extending its time constant, and pay the penalty of extending your measurement time.
Like lock-ins, spectrum analyzers can measure smaller signals by reducing the resolution bandwidth, but sweep times must be reduced, again extending measurement time.