The experiment you describe (in the question and the comments) will almost certainly give you no measurable result; the capacitance of the setup is much too small. Let's do the math:
1 cm spacing, 8 cm x 8 cm area, air gap.
$$C = \frac{\epsilon_0 A}{d} = \frac{8.8 \cdot 10^{-12} \cdot 0.08 \cdot 0.08}{0.01} = 5.6\cdot 10^{-12} F$$
5.6 pF is not a very large capacitance. When you put 4 V across that, you have 20 pC of charge. If your voltmeter has an impedance of 1 MOhm, the RC time constant will be 5.6 µs.
There is no chance that you can observe such a short blip with the naked eye.
You might be able to observe something with a fast oscilloscope, assuming that you do the experiment in vacuum (or at least VERY dry air): the moment you disconnect the voltage source from your capacitor, charge will leak away. Since there is very little charge, a tiny leakage current will be sufficient to remove the charge very quickly.
I addressed the question of charge leakage in this earlier answer - you might find it useful.
You could measure the capacitance of your setup using a (high) frequency generator, and determining that the voltage observed across the capacitor "rolls off" as you increase the frequency. Again, for this you would need a high quality oscilloscope and a variable frequency oscillator. You can look at the phase of the voltage across the capacitor, and you will see that as you reach frequencies corresponding to $\frac{1}{RC}$, the observed phase of the voltage across the capacitor begins to lag.
Your plate has capacitance. Your experiment will not show it.
