1. Consider Example 25.4 then answer the following: Two pieces of dust, 3 ug each, are both charged with 2.0 nC of charge and held 1 cm apart before being let go. If the two dust particles zip off in opposite directions, how fast are they going when they get really far away from each other? Neglect gravity and air resistance. 2. Write down the equations for the force, electric field, potential energy, and potential (that is voltage) a distance r away from a point charge. Then write down the general equations which relate these equations together. 3. Take a look at Example 25.10: The potential of a ring charge. Notice that the process here is very similar to determining the electric field due to a continuous charge distributions Identify a key difference which makes the potential easier to deal with than the electric field, in general.EXAMPLE 25.4 | Launching an electron Three electrons are spaced 1.0 mm apart along a vertical line. The SOLVE a. The center electron is in equilibrium exactly in the center outer two electrons are fixed in position. because the two electric forces on it balance. But if it moves a a. Is the center electron at a point of stable or unstable equilibrium? little to the right or left, no matter how little, then the horizontal b. If the center electron is displaced horizontally by a small distance, components of the forces from both outer electrons will push the what will its speed be when it is very far away? center electron farther away. This is an unstable equilibrium for horizontal displacements, like being on the top of a hill. MODEL Energy is conserved. The outer two electrons don't move, so we don't need to include the potential energy of their interaction. b. A small displacement will cause the electron to move away. If the displacement is only infinitesimal, the initial conditions are VISUALIZE FIGURE 25.12 shows the situation. (12)i = (123)i = 1.0 mm and v; = 0. "Far away" is interpreted as FIGURE 25.12 Three electrons. - co, where U = 0. There are now two terms in the potential energy, so conservation of energy K + Uf = K; + U, gives Before: After: my/ +0+0=0+ Kq192 Kq23 (rza) (fiz) = 1.0 mm Ke2 Ke2 Vi=0 = ( riz )i (23)i 2 (rza) = 1.0 mm This is easily solved to give 2 Ke2 Ke2 V = 1000 m/s m (12): (r23)i.EXAMPLE 25.10 | The potential of a ring of charge A thin, uniformly charged ring of radius R has total charge Q. Find solving strategy. We've chosen a coordinate system in which the the potential at distance z on the axis of the ring. ring lies in the xy-plane and point P is on the z-axis. We've then MODEL Because the ring is thin, we'll assume the charge lies along divided the ring into A small segments of charge AQ, each of a circle of radius R. which can be modeled as a point charge. The distance r; between VISUALIZE FIGURE 25.30 on the next page illustrates the problem- segment i and point P is Continued 706 CHAPTER 25 The Electric Potential ri = VR2+ 22 SOLVE The potential V at P is the sum of the potentials due to each segment of charge: Note that r, is a constant distance, the same for every charge segment. FIGURE 25.30 Finding the potential of a ring of charge. V = > Mz 1 AQ 1 ATTEO ATTED VR' + 23 EAQ Segment i with charge We were able to bring all terms involving z to the front because z AO is a constant as far as the summation is concerned. Surprisingly, we don't need to convert the sum to an integral to complete this calculation. The sum of all the AO charge segments around the ring R is simply the ring's total charge, E(AQ) = 0; hence the electric R+z potential on the axis of a charged ring is Vring on axis 4TEO VR' + 22 ASSESS From far away, the ring appears as a point charge O in the What is the potential here? .... distance. Thus we expect the potential of the ring to be that of a point charge when z > R. You can see that Vring = 0/476oz when z >> R, which is, indeed, the potential of a point charge