Question

• Construct a capacitor by filling a paper cup with water. What is the analog of voltage in this case, and what is the capacitance?
• Poke a small hole on the side of the paper cup near the bottom and let the water drain. How is this similar to an RC circuit?
• Decrease the R by making the hole slightly larger in diameter and re-run the demonstration. How is the rate of water flow consistent with the new RC value?
• Time how long it takes the cup to drain in each case and estimate the drainage rate (ounces or milliliters per second), make note of it, and compare the two.

Answer 1

A capacitor holds electric charge across itself which leads to an electric potential getting developed between the two plates (or surfaces). The plastic cup holds water in it whose volume/mass can be compared to the electrical charge. This water causes a development of potential between surface of water-air and the bottom of the cup. The height can be considered as an analog of voltage as Potential difference is directly proportional to height.

Capacitance = Charge/Voltage (C =Q/V)

Hence in our example, C = Volume of water/Height of water = Area of bottom of the cup.

Hence the cross sectional area of the cup is the capacitance in this case. In an RC circuit, the charge of the capacitor is released as current flows across the resistor. Here water flows out of the cup while reducing water level in the cup. Flow of water is analog to the flow of current. As the hole is a conductor for water out of the cup, the conductivity increases with the size of the hole. Hence R decreases, Current increases.

For a capacitor: I(t) = (Q_o/RC) e^-t/RC = I_o e^-t/RC

Similarly here, the flow is dependent on the size of the hole (Increases if resistance or capacitance decreases) while a decrease in RC also means that the flow rate decreases at a faster rate (due to faster drainage and reduction of potential). The impact of capacitance is due to lower surface area meaning greater reduction of potential per volume of water released.