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Two charges 7 and -4 are placed at (-7 cm,0,0) and (7 cm,0,0) respectively. Given, 0=8.85 10-12 C2 N-1 m-2, the electrostatic potential energy of the charge configuration is :
(4) -1.8 J
[figure]
r=7-(-7)=14 cm=0.14 m; keep the signs of the charges.
U= 14 0 q1q2r=9 109 (7 10-6)(-4 10-6)0.14
U=-1.8 J (negative: the charges attract)
Answer: (4)
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