Electric Field and Work on a Moving Charge

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SUMMARY

The discussion focuses on the physics of electric fields and work done on a moving charge. The force acting on a charge +q in a constant electric field ε V/m is calculated using the formula F = εNq, directed positively. The work done by this force when the charge moves a distance d is expressed as W = Fd. The voltage at position x = d is derived from the relationship ε = -dv/dx, leading to Vd = -εd. The relationship between electric potential energy loss and work done is confirmed through energy conservation principles, indicating that the kinetic energy (KE) of the charge at x = d equals the loss in electric potential energy.

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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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