Question

Two charge of 4 micro coulomb and 7 micro coulomb are placed 4.5 cm apart. Find potential energy of the system.

Answer 1

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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