Calculating Charge Q in an Electric Field at Point P: 50 cm from Source Charge Q

  • Thread starter Thread starter NIT14
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves calculating the magnitude of a charge Q based on the electric field strength at a point P, which is 50 cm away from the source charge. The context is within electrostatics, specifically relating to electric fields generated by point charges.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss Coulomb's law and its application to find the electric field strength. There are comments on the constants involved, such as the permittivity of free space and the Coulomb constant. Some participants question the assumptions regarding the medium in which the charge is situated.

Discussion Status

The discussion is ongoing, with participants providing insights into the relevant equations and constants. There is an acknowledgment of different interpretations regarding the use of constants in various media, but no consensus has been reached on the specific approach to solving the problem.

Contextual Notes

There is a mention of the potential confusion for the original poster regarding the use of epsilon versus the Coulomb constant, suggesting a possible gap in understanding the underlying concepts.

NIT14
How can I solve this problem?...

An electron is placed at point P in the electric field set up by a source charge, Q. Point P is located 50 cm from Q and has an electric field strength of 1.08x10^5 N/C directed away from Q. What is the magnitude of charge Q?
 
Physics news on Phys.org
Coulombs law:

[tex]E(r) = \frac{Q}{4 \pi \epsilon r^2}[/tex]

gives the electric field strength E(r) at a distance r from a point charge Q. The electrical permittivity [tex]\epsilon[/tex] can be found in a tablebook. For vacuum or air it is approx. 8.85e-12 F/m.
 
Last edited:
Just a comment:
the fraction[tex]\frac{1}{4 \pi \epsilon}[/tex]
is equal to the coulomb constant "k" (8.99 E9)
 
Originally posted by Chi Meson
Just a comment:
the fraction[tex]\frac{1}{4 \pi \epsilon}[/tex]
is equal to the coulomb constant "k" (8.99 E9)
Only when:

[tex]\epsilon = \epsilon_0[/tex]

If you are in a different medium there is a relative permeability [tex]\epsilon_r[/tex] in which case:

[tex]\epsilon = \epsilon_0\epsilon_r[/tex]

And K is different.
 
Agreed.

It was my assumption that the person asking the question was not yet at that level, and might have been taken aback by the use of epsilon when the textbook uses "k."
 

Similar threads

Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
23
Views
6K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K