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Random helpful notes for your experiment:

1. Extending the analogy from mechanical to electrical systems that we started last week, we now look at the electrical analogue of gravitational potential energy. This analogue is simply the electrical potential energy of a charge and is equal to the amount of work required to move the charge against the electric forces that are present. Consider the following situation: suppose I have a electric field that goes from point B to point A: B+|-------->|-A. Now, if I have a positive charge at point A and I move it to point B, then I have to move it against the Electric Force (which is directed along Electric Field Line) all the way. That means I do work on it all that way, too. This work is stored as electrical potential energy when the charge reaches point B. Thus, pulling a charged object against the electric-field force that acts against it stores electrical potential energy in that charged object. This is exactly analogous to the mechanical case where we store gravitational potential energy in an object by lifting it against the pull of gravity (the gravitational force).
   
   As we see in Equation (1) below, the electrical force is F = qE. Thus, as the charge moves in the direction of the force, we do an amount of work W = Fd. Just as in the mechanical case, you can think of the electrical potential energy that was stored in it by virtue of it's position in the electrical field being converted to kinetic energy. The potential difference is defined simply as the work per unit positive charge; for a simple case with a constant electric field and a charge moving along the line of force (i.e., the electric field line), this reduces simply to V = Ed (the electric field is constant as long as both Voltage and the Distance between subsequent equipotential lines change in the same proportion). Thus, voltage is simply the electrical potential energy per unit charge. This means that, as in Equation (2) below, Work can also be defined as W = qDV.
   
   Equipotential lines are simply lines of constant potential. Since the potential difference is defined as the work per unit charge, we see that no work is done along an equipotential line (since the potential difference between 2 points at the same potential is 0). That means a moving charge can move along an equipotential line without having any work done on it. This can only be if the equipotential lines are perpendicular to the lines of force (since Work is defined as force times displacement in the direction of the force). Thus, the lines of force (i.e., the Electric Field Lines) are perpendicular to the equipotential lines.
   
   So the 3-step algorithm is:

   Same V        ===>  DV = 0
   DV = 0        ===>  W = 0
   W = 0         ===>  F is perpendicular to d
   

   
   
   To (maybe) help make this a little clearer, think of a gravitational analogue: suppose a ball is rolling along a table. The Earth exerts a gravitational field all around us and the top of the table can then be considered an equipotential line since anything placed there will be at the same energy; in order to move the ball up or down, I have to do work on the ball. If the electric field, E, is not perpendicular to the equipotential lines, then it does work so E lines are always perpendicular to equipotential lines.
2. When you're mapping out the equipotential lines, try to get about 7 points so you can see what the lines look like (see diagrams below for what it should look like (ignore the crappy drawing and look for the general idea)). Also, when you're mapping the well, make sure the bucket part is your negative terminal and try to use the following increments: 0.1, 0.5, 1, 3, 5, 7, and 8.5 Volts (this is for the 2nd template). Also, when figuring out the electric field in the last part, here's a hint: the important quantity is distance (biggest distance --> smallest E; see equations below). Finally, when you extend the sides of the bucket (in the last part), remember that the included terminal will tend to weaken the electric field inside the bucket.
   
   
3. Some helpful equations:

   
          (1)  F = qE    ==> W = FDd = qEDd
          (2)  DV = EDd  ==> W = qDV = qEDd
          (3)  E = DV/Dd = F/q
          

1. Please ignore Appendix A; we'll be using a digital power supply instead. All you need to remember is to turn it on, plug in your leads, and set the Voltage. Don't mess with any of the other buttons!

Required Materials:

1. Laboratory Manual (SGM 407)
2. Laboratory Answer Book
3. Calculator with statistical functions
4. Paper to draw on
5. Ruler

1. This lab should also be very quick; nothing too complicated or dangerous (next week's lab should fix that!).