How Does Charge Movement Affect Electric Potential and Kinetic Energy?

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SUMMARY

The discussion focuses on the movement of a test charge -q0 between two charged rods with net charge -Q0, analyzing the electric force, work done, and electric potential difference. The work done on the charge is determined to be negative, as the charge moves in the direction of the electric field. The electric potential difference ΔVx→y is assessed, with the conclusion that it is negative, indicating that point Y is at a lower potential than point X. Finally, the speed of the test charge at point Y is greater than its initial speed v0 due to the acceleration from the electric force acting on it.

PREREQUISITES
  • Understanding of electric forces and fields
  • Familiarity with the concept of electric potential and potential difference
  • Knowledge of vector addition in physics
  • Basic principles of kinetic energy and work-energy theorem
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  • Learn about the work-energy theorem in the context of electric forces
  • Explore the concept of equipotential surfaces and their implications
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Students of physics, educators teaching electromagnetism, and anyone interested in understanding the dynamics of charged particles in electric fields.

jlmccart03
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Homework Statement


Two charged rods, each with net charge -Q0 are held in place as shown in the top view diagram below.

upload_2017-2-14_10-24-42.png


a. A small test charge -q0 travels from point X to point Y along the circular arc shown.
i. Draw an arrow on the diagram at each point to show the direction of the electric force on the test charge at that point. Explain why you drew the arrows as you did.

ii. Is the work done on the charge by the electric field positive, negative, or zero? Explain.

iii.Is the electric potential difference ΔVx→y positive, negative, or zero? Explain.

b. The test charge is launched from point X with an initial speed v0 and is observed to pass through point Y. Is the speed of the test charge at point Y grater than, less than, or equal to v0? Explain.

Homework Equations


W = -ΔVq
ΔV = Vf-Vi
Vector addition

The Attempt at a Solution


For the first part (i) as shown in the diagram I drew the arrows based on the two forces and did vector addition to get the resulting vector.

For the second part (ii) I say it is negative work since you are moving the charge in a direction it wants to go so it is negative work done.

For the third part (iii) I really don't know what I am supposed to look at here besides that the electric potential must be positive (possibly) I really don't understand how to look at a diagram and figure this piece out.

The final part (b) I say the speed is greater than the speed v0 since the charge is going where it wants to so it has a force to go that direction increasing its acceleration.
 
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jlmccart03 said:
A small test charge -q0 travels from point X to point Y along the circular arc shown.
What does this tell you about the force acting on the test charge?
 
cnh1995 said:
What does this tell you about the force acting on the test charge?
The force is different at all points along the arc since they are all at different distances from the two other -Q charges correct?
 
jlmccart03 said:
drew the arrows based on the two forces and did vector addition
Ok, but you could get the directions a little more accurate by taking into account the proximities of the charges when drawing the force vector lengths.
jlmccart03 said:
negative work since you are moving the charge in a direction it wants to go
That may be true towards the end of the path, but it is certainly not true at the start. Can you think of a way to answer it just by considering the start and end points, not worrying about the path?
jlmccart03 said:
For the third part (iii) I really don't know what I am supposed to look at here
This is really the same as part ii). Which of X and Y is at the higher potential?
cnh1995 said:
What does this tell you about the force acting on the test charge?
Nothing. There is no suggestion that the motion is purely in response to the electric field. Indeed, it could not be.
 

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