Electric fields equalling zero

Click For Summary

Homework Help Overview

The discussion revolves around determining the point where the electric field from two charges, Q1 and Q2, is zero. The problem involves understanding the configuration of the charges and the implications of their signs on the location of this point.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the distances from the charges to the point where the electric field is zero, questioning how to express these distances in terms of each other. There is also a discussion about the implications of the charges being like or opposite in sign.

Discussion Status

Some participants have provided insights regarding the placement of the zero electric field point based on the nature of the charges. However, there is no explicit consensus on the correct approach or solution, and the discussion remains open with various interpretations being explored.

Contextual Notes

It is noted that both charges are positive, which influences the discussion about the location of the zero electric field point. There is also a mention of the need to consider the magnitudes of the electric fields rather than simply adding them as forces.

mrpbll
Messages
2
Reaction score
0
Trying to work this one out.
If you have two charges Q1 and Q2 separated by a distance of x then how do you show what distance from the charge Q1 is the total electric from the two charges zero.

I know that Enet =E1 +E2 and if i use coulombs law i get:
KQ1/x^2 +KQ2/x^2
the K cancles and you get Q1/x^2 + Q2/x^2 = 0 but that doest seeem right. anyone with any ideas?
 
Physics news on Phys.org
Are the charges like or opposite in sign? If theyre like in sign, the point will be inbetween them, but if theyre opposite in sign, the point will be outside and closer to the weaker charge.

Lets say the charges are like. That means that the distance from the spot to one charge + the distance from the spot to the other charge = x. The trick is to write one distance in terms of the other. The point will not be rigt inbetween the 2 unless the charges are equal, and then its not much of a question.

Now let's say the charges are opposite, one charge will be a distance r from the point and the second charge will be a distance r + x
 
both charges are positive
 
I know that Enet =E1 +E2 and if i use coulombs law i get:
KQ1/x^2 +KQ2/x^2
You didn't add forces here: you merely added their magnitudes.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
Replies
18
Views
5K