Similar Question
A particle with a charge of + 4.20nC is in a uniform electric field E directed to the left. It is released from rest and moves to the left; after it has moved 6.00 cm, its kinetic energy is found to be + 1.50*10^ -6 J.
What work was done by the electric force?
What is the potential of the starting point with respect to the endpoint?
What is the magnitude of E?
Answer
The work done gave the particle the kinetic energy change it acquired:
W = Δ(KE) = 1.50×10^-6 joule
q.ΔU = W
ΔU = W / q = 1.50×10^-6 / 4.20×10^-9 = 357 volts
E = U/d = 357 / 6.00×10^-2 = 5,952 V/m
Similar Question
A particle with a charge of 4.30 nC is in a uniform electric field E directed to the left. It is released from rest and moves to the left; after it has moved 8.00 cm, its kinetic energy is found to be 1.50×10−6 J.
Part A.
What work was done by the electric force?
W= J
Part B.
What is the potential of the starting point with respect to the endpoint?
ΔV= V
Part C.
What is the magnitude of E?
E= V/m
Answer
The work done gave the particle the kinetic energy change it acquired:
W = Δ(KE) = 1.50×10^-6 joule
q.ΔU = W
ΔU = W / q = 1.50×10^-6 / 4.30×10^-9 = 349 volts
E = U/d = 349 / 6.00×10^-2 = 5816.66 V/m
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