lecture3 - 3. Lecture 3 3.1 Dipole and quadrupole The law...

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Unformatted text preview: 3. Lecture 3 3.1 Dipole and quadrupole The law of addition of vectors allows us to find the electric field produced by two charges. If the two charges are of equal magnitude but opposite sign (and separated by a distance so that they do not cancel each other) then we have what is called a dipole. The electric lines can be found in fig.10. In the laboratory practice you will be able to use a program that draws these lines for you. Try several configurations. For example arrange four charges so that they look as two dipoles of opposite orientation and see what happens. In fig. 11 and 12 we illustrate typical outputs of the program. Besides the electric lines it also shows equipotential surfaces everywhere perpendicular to the electric field. We discuss what they mean below. − + Figure 10: Electric field of a dipole obtained by adding (vectorially) the electric fields of each charge. A simple rule for the lines of electric field is that they only start from positive charges and end in negative ones. They can never end “in the middle of the air” so to say. Before, we mentioned that electric field can exist independently of charges, in that case the lines of electric field have to close on themselves since there is nowhere for them to end. We will see how this works later. 3.2 Electrostatic energy To motivate the idea of energy we use the Kelvin water dropper (fig.13), a simple – 17 – Figure 11: Typical output from the computer program you use in the lab. In this case illustrating the electric field and equipotentials of a dipole. Figure 12: Illustrating the electric field and equipotentials of a quadrupole (meaning four poles). device that generates electricity. The way it works is illustrated in fig.14. Assuming that one metallic jar is positively charged and the other negative, opposite charges will be induced in the streams of water falling through the respective rings. The water breaks in droplets which then fall into the jars accumulating charge. This is because of the cross connection between jars and rings which makes the positive droplets fall into the positively charged jar and the negative ones in the negatively charged one. If the water were not to break in droplets we would have a conducting circuit from the jars to the water tank, the charge will be repelled and would not accumulate in the jar. It is the same effect as when we discussed induction that we needed to cut the connection – 18 – to ground and remove the rod. Ingeniously here the connection breaks simply because the water stream breaks into droplets. When enough charge is accumulated the electric field is strong enough to push the charge through the bulbs lighting them up which simultaneously neutralizes the system. Charge will build up again by repeating the procedure. The initial charge imbalance appears just randomly, the important effect is that any random charge fluctuation will be amplified by the device. The idea we wanted to illustrate is also that we were able to generate electric power and so the energy has to come from somewhere. It might not be apparent initially where it is coming from. To figure that out one has to run the system and see when it will stop and what I need to replace to make it work again. Obviously many electric devices have batteries and we know they are the source of energy because when they run out we need to replace them or the device won’t work anymore. Here it is clear that after a while the tank runs out of water. So we need to take the water from the jars and put it back up into the tank. To do that we need to lift water against gravity and that is where we put energy back into the system. Therefore, gravitational energy is being converted into electric energy. One way of seeing that is that, since the jars repel the water droplets, gravity is necessary to push them down into their respective jars allowing the accumulation of charge. Having said that let us now look at how we understand energy when there are electric fields present. From Newtonian mechanics we know that forces can be derived from potentials. Basically a particle feels a force that tries to make it move in the direction where it can decrease its potential energy faster. The same occurs in the case of electrostatics. Consider a positive charge Q which is fixed at some point and take another charge q that we are going to move. Let us say that the other charge is negative so there is an attractive force. If we want to move the charge q further apart by an amount ∆r we need to do some work because there is a force that opposes us. The amount of work needed is ￿ W = |F |∆r = ∆U = Ufinal − Uinitial (3.1) where ∆U is the change in potential energy of the system. A couple of comments. When computing the work we should use only the component of the force in the direction of the motion. Since here we move parallel to the force the formula is OK. If not we need to multiply by cos θ where θ is the angle between the force and the direction of motion. Second comment is that the formula is valid for constant force. Since the force depends on r we are going to consider ∆r ￿ r and therefore to consider the force constant is a – 19 – Figure 13: Kelvin invented this device to generate electricity. I find it of great interest since it is basically just water dropping from a tank and nevertheless it is able to generate electricity enough to light a few neon bulbs. Essentially illustrates how, by understanding how nature works one can create something interesting out of almost nothing. good approximation. With this in mind we compute 1 Qq ∆U = Ufinal − Uinitial = − δr 4πε0 r2 (3.2) The minus sign is because we should have ∆U > 0 as discussed but the product of the charges is negative since they have opposite sign. We claim now that the potential energy is 1 Qq U= (3.3) 4πε0 r Indeed the difference in potential energy between the initial and final situation is 1 Qq 1 Qq 1 Qq ∆U = Ufinal − Uinitial = − =− ∆r 4πε0 r + ∆r 4πε0 r 4πε0 r(r + ∆r) (3.4) If we use now that ∆r ￿ r we find that ∆U = − 1 Qq δr 4πε0 r2 (3.5) as we computed from the force. This verifies our expression (3.3) for the potential energy. – 20 – water tank + + −− −− −− −− + + − − ++ ++ ++ + − − − − − rings + − − − − − − − − + − + − − − − − − −−− −− − − + + + + + − + + + + + + + + +++++ metallic jars neon lamps Figure 14: The key is the cross connection of the rings and the jars. This amplifies any charge difference accumulating opposite charges in the two metallic jars. Another case in which we can compute the potential energy is that of a constant ￿ uniform electric field. Namely E is independent of the position, that is a probe charge feels the same force no matter where it is located. Of course this is an idealized situation but in many cases is a good approximation for the electric field in certain regions where it does not change very much. In such case, let us say that the electric field points in the direction x. If we move a charge in directions, y or z , perpendicular to x, the force ˆ ˆ ˆ ˆ does not oppose or help the motion so we do not need to do any work. If we move it – 21 – in direction x however, since we move in the direction of the force we extract work, or, ˆ we make negative work: ￿ ￿ W = −|F |∆x = −q |E |∆x = δ U = Ufinal − Uinitial = U (x + ∆x) − U (x) (3.6) The final energy of the system is therefore smaller than the initial one. We see that the equation is satisfied if the potential energy is simply given by ￿ U = −q |E |x (3.7) Another way to figure out the sign is to notice that the charge will move in the direction of decreasing energy. So, if q > 0 then U has to decrease toward the right and therefore the minus sign. 3.3 Electrostatic potential We see, as a consequence of the force being proportional to the charge, that so is the potential energy. For that reason we define the electrostatic potential V (x, y, z ) such that is we put a charge at position (x, y, z ) the energy of the system changes by U = qV (x, y, z ) (3.8) For a charge Q the electrostatic potential is given by V= 1Q , 4πε0 r where, as always, r= ￿ x2 + y 2 + z 2 (3.9) On the other hand, for a constant electric field in direction x it is ˆ ￿ V = −|E |x (3.10) By analogy with the equation that determines the work done in terms of the potential energy we see that if we move by a distance ∆x in a direction parallel to the electric field the change in electrostatic potential is ￿ ∆ V = −|E |∆ x (3.11) ￿ Equivalently the component of E in direction x is Ex = − ∆V ∆x (3.12) So, the electric field can be computed from the potential by moving in a direction x by an amount ∆x. The change in potential ∆V determines the electric field through – 22 – eq.(3.12). This actually works if we move in any direction, it always give the component ￿ ￿ of E in that particular direction (or equivalently the projection of E along the direction ￿ of motion). For example, if we move in a direction perpendicular to E then there is no change in the potential. This gives rise to the notion of equipotential surfaces. This means surfaces of the same (equal) potential, that is surfaces where V has a constant value. From what we just said, such surfaces should be perpendicular to the electric field. For example, for a single charge they are spheres concentric with the charge. For ￿ a constant electric field they are planes perpendicular to E . In other cases, for example for the dipole, they are more complicated but one can have some idea by drawing the electric field and then drawing surfaces perpendicular to it (see fig. 11). An important property of equipotential surfaces is that if we move a charge along such surface we do not do any work. Another important property is that to be in a static situation, namely with charges not moving, the surface of a conductor has to be an equipotential surface. This is because on the surface of the conductor the electric field is perpendicular to it, otherwise if there were a component parallel to the surface, charges would move until they cancel the electric field. Similarly, if charges are not moving then inside a conductor the electric field is zero, implying that all the conductor has the same value of the potential. We should emphasize that this refers to the static situation. If charges are moving then the electric field inside a conductor need not be zero and the conductor need not be all at the same potential. The principle of superposition also applies to the electrostatic potential. So if we have several charges, the total electrostatic potential is the sum of the electrostatic potentials due to each of them. It becomes apparent the usefulness of the potential, since the potential add as numbers as opposed to the electric field which add as vectors. Finally the potential is measured in N m/C since the electric field is measured in N/C and ∆x in meters. Since this is a very common unit it has received the special name of Volt (V). So we have Nm 1V = 1 (3.13) C The potential difference many times is referred simply as “voltage” Another demo illustrates the existence of this potential by measuring directly the potential of a charged sphere and another by showing how putting a fluorescent light in the electric field produces a discharge because of the potential difference between the two ends of the tube (fig.15). – 23 – Figure 15: A voltmeter and a fluorescent bulb make manifest the existence of a potential difference (and electric field). – 24 – ...
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