The charge is negative so the particle leaving E must first clear the potential barrier at D before arriving at B (if the charge were positive, then it would have to climb over the potential energy barrier at C). The kinetic energy required to clear the D potential energy barrier is (31 + 10) × 20 = 820 µJ. (This is the same as the “activation energy” needed by two chemical reactants to clear a potential energy barrier before taking part in an exothermic reaction.) The kinetic energy of the particle when it arrives at point B is ΔK = −ΔU = −qΔV = −20 ×(23− 31) = 160µJ .
