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Practice Problem 18.3 Constants Now lets look at the potential energy of a charge that is in the vicinity of two other charges. Suppose two electrons are held in place 10.0 cm apart. Point a is midway between the two electrons, and point b is 12.0 cm directly above point a. (a) Calculate the electric potential at point a and at point b. (b) A third electron is released from rest at point b. What is the speed of this electron when it is far from the other two electrons? The mass of an electron is me -9.11 x 10-31 kg. Figure 1 of 13.4 cm 12.0 cm 6.0 cm^6.0 cmSOLUTION SETUP (Figure 1) shows our sketch. Point b is a distance rb - 1/(12.0 cm)2 (6.0cm) 13.4cm from each electron. 92 SOLVE Part(a): The electric potential V at each point is the sum of the electric potentials of each electron: V q1-e. At point a, r1-r2 Ta-0.060m, so V + ½ k + k with 2(8.99x109 N-m2/C2)(1.60x10-19 c) 0.060m ra -4.8 × 10-8 v = At point b, r1 r20.134m, so 2(8.99x109 N-m2/C2)(1.60x10-19 c) 0.134m Tb 2.1 × 10-8 v Part (b): Remember that our equation for potential assumes that U is zero when r = oo. Thus, when the third electron is far from the other two (at a location we designate c), we can assume that U 0. To find the electrons speed at point c, we use conservation of energy We solve for Kc. First we use Uq V and qe to rewrite the preceding expression as We know that V, = 0 because V, = kg/re and re is very large. Also, Kb = 0 because the electron is at rest before it is released. Then -eV,--(1.60 × 10-19 C)(-2.2 × 10-8 V) - +3.68 x 10-27J K. = SOSo 2Kc me 2(3.68x10-27 J) 9.11x10-31 kg 90m/s REFLECT Remember that electric potential is a scalar quantity. We never talk about components of V; there is no such thing. When we add potentials caused by two or more point charges, the operation is simple scalar addition, not vector addition. But the sign of V, determined by the sign of the q that produces V, is important. Note that at point a the electric fields of the two electrons are equal in magnitude and opposite in direction and sum to zero. But the potentials for the electrons are both negative and do not add to zero. Make sure you understand this distinction. When the third electron (the one that moves) is at point b, the electric potential energy is Ub-eVi +3.68 x 10-27 J. At point c, the potential energy is zero. All of the initial electric potential energy has been converted to kinetic energy because of the positive work done on it by the repulsive forces of the other two electrons. The negatively charged electron gains kinetic energy when it moves from a lower-potential point to a higher-potential point-in this case, from Vs =-2.1 × 10-8 V to V, = 0 Note that the net force on the third electron decreases as that electron moves away from point b. Its acceleration is not constant, so constant-acceleration equations cannot be used to find its final speed. But conservation-of-energy principles are easy to apply ▼ Part A-Practice Problem The electron at point b is replaced with a proton (mass mp 1.67 x 10kg) that is released from rest and accelerates toward point a. What is the speed of the proton when it reaches point a? Express your answer in meters per second to two significant figures. m/s Submit Request Answer

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