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Electric Field

A particle with a charge of 4.30 nC is in a uniform electric field vec {rm E} directed to the left. It is released from rest and moves to the left; after it hasmoved 8.00 {rm 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 respectto the endpoint? What is the magnitude of vector E?
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Answer #1
Application of energy conservation:

Energy gain in kinetic = energy loss in potential
?KE = ?U

Electric potential energy (U) = qV
The amount of energy loss in potential is ?U = q?V
While in an uniform electric field V = Ed
?V = E?d (d is distance between two point and is 0.06mfor this problem)
Therefore ?U = q(E?d)

Since ?KE = ?U
1.5x10-6J = 4.2x10-9 x E x 0.06
so E = 5952.38 N/C
answered by: sofia
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