Two charge of 4 micro coulomb and 7 micro coulomb are placed 4.5 cm apart. Find potential energy of the system.
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Answer:
5600 J
Explanation:
given
q1 = 4mC = 4 x 10^-6 C
q2 = 7mC = 7 x 10^-6 C
and r12 = 4.5 cm = 0.045 m
therefore
potential energy U = Kq1q2/r12
here k=9×10^-9
Placing the values in equation we get
U = 4 x 10^-6 C × 7 x 10^-6 C x 9×10^-9/0.045
therefore U = 5600J
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