Micro/macro collection between electrostatic

  • Thread starter Thread starter 743344
  • Start date Start date
  • Tags Tags
    Electrostatic
Click For Summary

Homework Help Overview

The problem involves calculating the fraction of electrons removed from a charged copper ball with a specified diameter and charge. The subject area pertains to electrostatics and atomic structure.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between charge and the number of electrons, with one participant questioning the meaning of the variable 'e'. Another participant suggests determining the initial number of electrons in the copper ball using its density and atomic properties.

Discussion Status

The discussion is ongoing, with participants exploring different methods to approach the problem. Some guidance has been offered regarding the calculation of initial electron count, but no consensus has been reached on the best method to solve the problem.

Contextual Notes

There is a mention of needing to know the initial number of electrons in the copper ball, which may involve assumptions about its density and atomic structure. The original poster expresses uncertainty about the problem setup.

743344
Messages
19
Reaction score
0
i have question, if A 2.00 mm-diameter copper ball is charged to + 70.0 nC. What fraction of its electrons have been removed?

I know i have to use the relation, q = (Nprotons - Neletrons)e, but i can't figure out the answer.
what does e stand for.
please help
thankyou
 
Physics news on Phys.org
e is the charge of a proton.
 
thank
i still need help for the question though
 
It's easy to figure out how many electrons were lost. But if they want to know the fraction of electrons removed, then you'll have to know how much it started with. Using the density of copper and its atomic number and weight from the periodic table, one could figure out how many electrons there would be in a neutral 2mm diameter ball.

I'm not sure if that's the best way to do this problem though.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
Replies
12
Views
3K
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
2
Views
2K
Replies
7
Views
7K
  • · Replies 9 ·
Replies
9
Views
11K
Replies
7
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K