Electric Field and Work on a Moving Charge

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In the discussion on electric fields and work on a moving charge, the force acting on a charge +q in a constant electric field ε is determined to be F = εNq, directed positively. The work done by this force when the charge moves a distance d is calculated using W = Fd. The voltage at position x = d is related to the electric field by Vd = -εd, reflecting the relationship between electric potential and field strength. The loss in electric potential energy is equated to the work done by the force, consistent with the conservation of energy principle. For the kinetic energy at position x = d, it can be shown that it equals the loss in electric potential energy through energy conservation or by calculating acceleration and velocity.
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Homework Statement



Given an electric field ε V/m that is constant in the x-direction and a charge +q C located at the origin and free to move in the x-direction. (a) What is the magnitude and direction of the force F acting on q? (b) If q moves under the influence of F a distance d m, how much work is done by F? (c) Assuming the voltage at the origin to be zero, what is the voltage Vd at x = d m, bearing in mind that ε = -dv/dx? (d) How is the loss in electric potential energy related to the work done by F? (e) Assuming the charge has a mass m kg and zero velocity at the origin, show that the KE of the charge at x = d is equal to the loss in electric potential energy.

Homework Equations





The Attempt at a Solution



a. F= εNq in the positive direction
b. W=Fd
c. I can guess that it's -εx but I don't know how to explain it properly
d. Equal by conservation of energy.
e. Not too sure how.

Thank you in advance.
 
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a-d: right.
For (e), I would use energy conservation. Alternatively, calculate the acceleration, and use this to determine the velocity there.

The question uses an ugly way to deal with units :(.
 
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