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Charge A has qA = 2.2 μC, charge B has qB = -1.6 μC, and charge...

Charge A has qA = 2.2 μC, charge B has qB = -1.6 μC, and charge C has qC = -7.5 μC. A is at the origin, B is at coordinates (10.0 cm, 0 cm), and C is at coordinates (0, 7.8 cm). Calculate the electrostatic potential energy of this charge configuration.

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Answer #1

We know that potential energy of the system

   (r is the distance between the charges)

here

U=kq1q2/r12+kq2q3/r23+kq3q1/r13 ...1

here k=9*109 , q1=qA=2.2*10-6 C , q2=qB=-1.6*10-6C , q3 =qc=-7.5*10-6C , r12=10cm=0.10m  . r13 =7.8 cm=0.078 m

put these value in equation 1

U=9*109*2.2*10-6*(-1.6*10-6)/0.10+9*109*(-1.6*10-6)(-7.5*10-6)/0.1268+9*109*2.2*10-6*(-7.5*10-6)/0.078

U=-0.3168+0.8517-1.9034

U=-1.3685 J Answer

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