A particle with charge+4.20 nC is in a uniform electric field E^vector directed

Question

A particle with charge+4.20 nC is in a uniform electric field E^vector directed to the left. The charge is released from rest and moves to the left: after it has moved 6.00 cm, its kinetic energy is +2.20 times 10^-6 J. Part A What is the work done by the electric force? Express your answer with the appropriate units W = __________ Part B What is the potential of the starting point with respect to the end point? Express your answer with the appropriate units Delta V = _____ Part C What is the magnitude of E^vector? Express your answer with the appropriate units. |E| = ___________

Explanation / Answer

Here ,

let the electric field is E

part A)

work done by the electric force = change in kinetic energy

work done by the electric force = 2.2 *10^-6 - 0

work done by the electric force = 2.2 *10^-6 J

B)

as the charge is positive ,

it moves from higher potential to lower potential

potential of the starting point = 2.2 *10^-6/(4.2 *10^-9)

potential of the starting point = 523.8 V

the potential is 523.8 V

c)

electric field = potential/distance

electric field = 523.8/.06

electric field = 8730 N/C

the electric field E is 8730 N/C

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