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0 Figure 1: 1. The potential energy U for figure 1A is Q2 = -9 23=4 Qu= 4 us 12=dris =v2 d d k @:Q = -2K9 True, False 2. The

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1. True , u= - 2K Since the deposits and vad U= -2Kia 2. True U= - ukat + 12 Kg - 12 kg Here we have 12 pairs of charges (one

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