Quantifying Bending of Water Stream by Charge ``` Name: Velson Status: teacher Grade: 9-12 Country: United Kingdom Date: Summer 2012 ``` Question: Bending of water stream by static electricity. This is a common experiment, but is never quantified. I can calculate the force applied by measuring the displacement. I should like to get students to estimate the charge generated by rubbing. Can you help by providing the attractive force generated by the charge in the adjacent stream? Replies: Velson, This is a problem that lends itself well to some experiments and successive refinements in your approach. To start with a “first cut” analysis, I would assume that the water and the charged element (bending the water) are two electrodes of a capacitor. A basic capacitor is two electrode plates separated by a distance d. If you place a voltage (V) across the plates, the amount of charge (in Coulombs) is given by Q=CV (total capacitor charge) For a simple two-plate capacitor, each plate having an area of A (m^2), the capacitance is approximated by: C= eps*A/d Where eps=the permittivity of air (between the plates) = 8.85x10-12 (Farads/m) The total energy (U) in a capacitor is U=CV^2/2 (Joules) We can thus determine the energy stored based on this simple geometry: U=(Q^2)d/(2eps*A) Note that the energy depends on the distance between the plates (d). We can calculate the force between the plates as the change in energy with small changes in plate separation: F=-dU/dx=-Q^2/(2*eps*A) (Newtons) So, now we have a force based on a simple geometry and the net charge on the attractive plate. This simple model should provide a starting point for charge calculations. It won’t be exact, but it should get you in the ballpark. Now, a few things to consider: 1. The electrodes (water and whatever you have charged) are not metal as in typical capacitors. Charge can distribute unevenly across an insulator (say a balloon) and uneven distribution will change the force. Interesting to play with would be to take a balloon, rub it locally to charge it, then see if bending changes as you rotate the charged spots away from the water. Most balloons are made of insulative material, so charge will not move around readily on the surface. 2. The capacitor equation uses the area seen by the two electrodes and assumes the plates are flat with the same sizes. If you use a stream, you are creating a rounded surface so that you may need to correct for this surface geometry. You could try looking for a capacitor model closer to your geometry and go through the same calculation. A cylinder (the stream) next to a flat plane (the attractive surface) may be the next step. There are many capacitive calculations to help with this. 3. Given that the capacitor equation assumes a voltage difference across two plates, you need to think about what the potential difference (V) is between the water stream and the attracting element. Water is somewhat conducting, and it usually can be considered to be a big “sink” for charges, especially if there is a connection to the earth (copper pipes, for example). The potential of the water, in this case, can be assumed to be zero. If the water source “floats” electrically, its potential may not be zero, and the force will change. What happens, for example, if you charge two balloons and hang them from insulating strings? What is the force if you charge just one, or charge them both? What if you paint one with conductive aluminum paint and tie it to ground? To get a more precise calculation of the force, you would need to do a field analysis of the particular problem. A correct analysis would likely be fairly involved. I think, though, you could start applying refinements starting with this simple model to get a good feel for charge, capacitances, and forces. Check back if you have questions or if you end up doing some experiments. Kyle Bunch, PhD, PE Click here to return to the Physics Archives

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