Electric Potential Energy & Speed of Charge: Q=+85.5 µC, q=-2.73 µC

  • Thread starter Thread starter wr1015
  • Start date Start date
  • Tags Tags
    Charge Speed
Click For Summary

Homework Help Overview

The discussion revolves around the calculation of electric potential energy and the speed of a charge in an electric field. The problem involves two point charges, one fixed and one movable, with specific values for charge and mass.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss methods for calculating the change in electric potential energy as the second charge moves from one position to another. There is an exploration of the relationship between potential energy and speed, with attempts to apply relevant formulas.

Discussion Status

Some participants are attempting to clarify their calculations regarding the change in potential energy and its impact on the speed of the second charge. There is an ongoing exploration of how to correctly compute the final potential energy and the differences involved.

Contextual Notes

Participants are working under the constraints of the problem, including specific charge values and distances. There is an emphasis on ensuring the correct application of formulas related to potential energy and kinetic energy without reaching a definitive conclusion.

wr1015
Messages
53
Reaction score
0
A point charge Q = +85.5 µC is held fixed at the origin. A second point charge, with mass m = 0.0526 kg and charge q = -2.73 µC, is placed at the location (0.323 m, 0).

(a) Find the electric potential energy of this system of charges.

(b) If the second charge is released from rest, what is its speed when it
reaches the point (0.125 m, 0)?


ok so i found part (a) -6.496597059 J but how do i calculate (b)?? I've tried subtracting the electric potential energy of the distance traveled from .323 to .125 (.198 m) from the electric potential energy of the system to find the change in electric potential energy. Then i divided that by my second charge (-2.73E-6) to get my change in potential so i could plug that into [tex]v= \sqrt{2q\Delta V/m}[/tex] and am getting wrong answers... any guess on where I'm going wrong??
 
Last edited:
Physics news on Phys.org
wr1015 said:
A point charge Q = +85.5 µC is held fixed at the origin. A second point charge, with mass m = 0.0526 kg and charge q = -2.73 µC, is placed at the location (0.323 m, 0).

(a) Find the electric potential energy of this system of charges.

(b) If the second charge is released from rest, what is its speed when it
reaches the point (0.125 m, 0)?


ok so i found part (a) -6.496597059 J but how do i calculate (b)?? I've tried subtracting the electric potential energy of the distance traveled from .323 to .125 (.198 m) from the electric potential energy of the system to find the change in electric potential energy. Then i divided that by my second charge (-2.73E-6) to get my change in potential so i could plug that into [tex]v= \sqrt{2q\Delta V/m}[/tex] and am getting wrong answers... any guess on where I'm going wrong??
[itex]v = \sqrt{2U/m}[/itex] where U is the potential energy change.
I get the change in potential energy to be 10.3 J which gives a speed of 19.8 m/sec.

AM
 
Last edited:
wr1015 said:
(b)?? I've tried subtracting the electric potential energy of the distance traveled from .323 to .125 (.198 m) from the electric potential energy of the system to find the change in electric potential energy.

Initially, the charges are separated by 0.323m and have the potential energy that you calculated in part (a). Finally, the charges are separated by 0.125m and have potential energy = ?. You need to calculate the final potential energy, and then the difference between the initial and final potential energies.
 
thank you all for your help
 

Similar threads

Replies
5
Views
3K
Replies
16
Views
5K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
5
Views
2K
Replies
2
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K