Potential Difference and Electric Potential

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SUMMARY

The discussion centers on the effects of a uniform electric field of 375 N/C on an electron initially at rest, specifically addressing the work done, change in potential energy, and velocity of the electron after moving 3.2 cm. The correct calculations yield a work done of 1.92 x 10^-18 joules, a change in potential energy of -1.92 x 10^-18 joules, and a velocity of -2.05 x 10^6 m/s. Clarifications were made regarding the direction of movement for negative charges in an electric field, the implications of negative work, and the relationship between potential energy and kinetic energy in electric fields.

PREREQUISITES
  • Understanding of electric fields and forces
  • Familiarity with the work-energy theorem
  • Knowledge of kinetic energy equations
  • Concept of electric potential energy
NEXT STEPS
  • Study the implications of electric field direction on charged particles
  • Learn about the work-energy theorem in the context of electric fields
  • Explore the relationship between potential energy and kinetic energy in electric systems
  • Review the equations for work done by electric fields, specifically W = qEd
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone seeking to understand the dynamics of charged particles in electric fields.

johnnyies
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Homework Statement


A uniform electric field of magnitude 375 N/C pointing in the positive x-direction acts on an electron, which is initially at rest. After the electron has moved 3.2 cm, what is

a. the work done by the field on the electron
b. change in potential energy associated with the electron
c. velocity of the electron

Homework Equations


W = -qEd
KE = 1/2 mv^2

The Attempt at a Solution


As far as math goes, I got all the answers, but I do not have the signs correct.

answers are:
a. 1.92 x 10^-18 joules
b. -1.92 x 10^-18 joules
c. -2.05 x 10^6 m/s

I got those numbers, but the signs are entirely wrong. My book doesn't explain things very clearly. But what I'm confused about, is when an E field is pointing to the right, doesn't that mean the electron should move to the right? What does negative work mean? How does the electron lose potential energy?

Thanks.
 
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johnnyies said:

Homework Equations


W = -qEd
KE = 1/2 mv^2

Actually, W = +qEd. (The force is qE, and W=Fd.)
 
Okay, understood the equation mistake, but can someone answer my questions at the bottom?
 
johnnyies said:
But what I'm confused about, is when an E field is pointing to the right, doesn't that mean the electron should move to the right?

No. By definition, the direction of the electric field at a point is the direction in which a positive test charge would move if it were placed at that point. A negative charge would therefore move in the opposite of the field direction.


johnnyies said:
What does negative work mean?

If force and the displacement are in the opposite direction, then it is clear from the definition of work that the work done will be negative. The force acts to impede the existing motion, decelerating the particle. By the work-energy theorem, the work done is equal to the change in kinetic energy. Therefore, if negative work is done, the kinetic energy decreases. That's not what is happening in this situation though. In this situation, positive work is done and the charge is accelerated.

johnnyies said:
How does the electron lose potential energy?

Just as a mass raised to a certain height in a gravitational field has a certain potential energy, a charge located at a certain position within an electric field results in a system having a certain amount of electric potential energy. Conceptually, this potential energy can be thought of as the work that would have had to be done against the field in order to get the charge to that location. Just as when the mass is dropped and accelerates under gravity it loses potential energy (converted to kinetic) so too does charge lose potential energy (converted to kinetic) when accelerated by a field in the direction that it "wants" to go.
 

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