Twoparticles with Q 1 = 40 µCand Q 2 = 88 µC are initially separated by adistanc
ID: 1666739 • Letter: T
Question
Twoparticles with Q1 = 40 µCand Q2 = 88 µC are initially separated by adistance of 2.8 m and thenbrought closer together so that the final separationis 1.4m. What is the change inthe electric potential energy?J
Twoparticles with Q1 = 40 µCand Q2 = 88 µC are initially separated by adistance of 2.8 m and thenbrought closer together so that the final separationis 1.4m. What is the change inthe electric potential energy?
J
Twoparticles with Q1 = 40 µCand Q2 = 88 µC are initially separated by adistance of 2.8 m and thenbrought closer together so that the final separationis 1.4m. What is the change inthe electric potential energy?
J
Twoparticles with Q1 = 40 µCand Q2 = 88 µC are initially separated by adistance of 2.8 m and thenbrought closer together so that the final separationis 1.4m. What is the change inthe electric potential energy?
J
Explanation / Answer
Charges q = 40 * 10 ^ -6 C
q’= 88 * 10 ^ -6 C
Initial separation r = 2.8 m
Final separation r ‘ = 1.4 m
Initial potential energy U = -Kqq’ / r
Where K = 8.99 * 10 ^ 9 N m^2 / C ^ 2
Plug the values we get U = -11.3 J
Final potential energy U ‘ = Kqq’ /r’
= -22.6 J
Change in potential energy = U ‘ – U = -11.3J
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