Question

A charge Q1 = 1.13 μC is at rest and is located 2.20 cm away from another fixed charge Q2 = 1.75 μC. The first charge is then released. Calculate the kinetic energy of charge Q1 when it is 5.70 cm away from charge Q2.

Answer 1

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q1 = 1.13 μC
q2 = 1.75 μC

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Initial Potential Energy = k*q1q2/r
Pin = (8.9*10^9 * 1.13 * 10^-6 * 1.75 * 10^-6)/(2.20 * 10^-2)
Pin = 0.799989 J

Inititial Kinetic Energy = 0

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Final Kinetic Energy , Kf =?
Final Potential Energy, Pf = k*q1q2/r
Pf = (8.9*10^9 * 1.13 * 10^-6 * 1.75 * 10^-6)/(5.7 * 10^-2)
Pf = 0.308768 J

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Now, Using Energy Conservation

K.E in + P.Ein = K.E_fin + P.Efin
0 + 0.799989 = K.E_fin + 0.308768
K.E_fin = 0.49122 J

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kinetic energy of charge Q1 when it is 5.70 cm away from charge Q2 = 0.49122 J

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