64. (a)When the electron is released, its energy is K+ U= 3.0 eV − 6.0 eV (the latter value is inferred from the graph along with the fact that U = qVand q = − e). Because of the minus sign (of the charge) it is convenient to imagine the graph multiplied by a minus sign so that it represents potential energy in eV. Thus, the 2 V value shown at x= 0 would become –2 eV, and the 6 V value at x =4.5 cm becomes –6 eV, and so on. The total energy (−3.0 eV) is constant and can then be represented on our (imagined) graph as a horizontal line at −3.0 V. This intersects the potential energy plot at a point we recognize as the turning point. Interpolating in the region between 1.0 cm and 4.0 cm, we find the turning point is at x = 1.75 cm 1.8 cm.≈(b) There is no turning point towards the right, so the speed there is nonzero. Noting that the kinetic energy at x= 7.0 cm is
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