Question

Given that q = +1.70 µC and d = 20.0 cm, find the electric potential energy of the three charge system.

9,--2.09 93 = +3.0 q

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Answer #1


potentiual energy of the system is U = -ke[(q1*q2/r12)+(q2*q3/r23)+(q3*q1/r31)]

r12 = 20 cm = 0.2 m = r32

r31 = 20+20 = 40 cm =0.4m


U = (-9*10^9)*(((-1.7*2*1.7*10^-12)/0.2) + (-2*1.7*3*1.7*10^-12)/(0.2) + (1.7*3*1.7*10^-12)/(0.4))) = 0.845325 J

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