Use the graph to determine the time to ‘half-max’.
t_1/2 =  s

Calculate the capacitance (C) of the capacitor.
C =  µF

Determine the percent error between the calculated capacitance and the value on the capacitor (330 µF).

% error =

× 100% =  %

What is the maximum charge for the capacitor in this experiment (approximately)?

Charge equals capacitance multiplied by voltage, or 0.000330 F × 4.00 V, or 0.00132 C or 1320 µC.Charge equals capacitance multiplied by voltage, or 330 F × 4.00 V, or 1.32 C.     Charge equals capacitance divided by voltage, or 0.000330 F / 4.00 V, or 0.00132 C or 1320 µC.Charge equals capacitance multiplied by the inverse of the voltage, or 0.000330 F × 1/4.00 V, or 0.0000825 C or 82.5 µC.