Electric field from a charge q1

Click For Summary
SUMMARY

The discussion centers on the electric field generated by a point charge q1 = -10 * 10^9 C located at coordinates (3, 4) m. The calculated electric field vector at the origin is (-2160i -2880j) N/C. It is established that adding another charge q2 at the origin does not alter the electric field produced by q1 at that point, as the electric field at a location excludes any charge present at that location. The formula used for calculating the electric field, E = k(q/r²), is confirmed to be correct for this scenario.

PREREQUISITES
  • Understanding of electric fields and point charges
  • Familiarity with Coulomb's law and the formula E = k(q/r²)
  • Knowledge of vector notation in physics
  • Basic concepts of charge interactions and field calculations
NEXT STEPS
  • Study the principles of superposition in electric fields
  • Learn about electric field lines and their representation
  • Explore the effects of multiple charges on electric fields
  • Investigate the concept of electric field strength and its applications
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in understanding electric field interactions and calculations involving point charges.

happyparticle
Messages
490
Reaction score
24
Homework Statement
Electric fields vector
Relevant Equations
Electric fields vector
Hi,
I have a charge q1 = -10 * 10^9. The the coordinatesare (3,4)m.
I found the electric field vector that is (-2160i -2880j) n/c.

My questions is if I add a charge q2 to the the coordinates(0,0) is the electric field stay the same?
 
Physics news on Phys.org
EpselonZero said:
Homework Statement:: Electric fields vector
Relevant Equations:: Electric fields vector

Hi,
I have a charge q1 = -10 * 10^9. The the coordinatesare (3,4)m.
I found the electric field vector that is (-2160i -2880j) n/c.

My questions is if I add a charge q2 to the the coordinates(0,0) is the electric field stay the same?

Do you mean the electric field at the origin?
 
Actually, I'm not sure. I had to find the electric field without the charge q2 and then I have a second question that ask what happen to the field calculated if we add a charge q2 at the origin.
 
EpselonZero said:
Actually, I'm not sure. I had to find the electric field without the charge q2 and then I have a second question that ask what happen to the field calculated if we add a charge q2 at the origin.

Where is the charge ##q_1## and where are you calculating the electric field?
 
q1 is at the point (3m,4m) I calculated the electric field at the origin, but without the charge q2.
 
EpselonZero said:
q1 is to the point (3m,4m) I calculated the electric field at the origin, but without the charge q2.

The electric field at a point excludes any charge at that point. If you tried to apply the formula for the electric field of a point charge at the point itself, then you have undefined division by zero.

If you consider the point charge as an infinitesimal symmetric distribution of charge, then the field at the centre would be zero. That's how you should treat the situation here.
 
I'm not sure to understand.
I used E = k(q/r²) to find the electric field from (3m,4m) to the origin by the charge q1.

Is that correct?
 
EpselonZero said:
I'm not sure to understand.
I used E = k(q/r²) to find the electric field from (3m,4m) to the origin by the charge q1.

Is that correct?

Yes. What would the field be at the point ##(3m, 4m)##?

The answer is that it has to be zero, as a point charge cannot exert a net force on itself.
 
If I have a charge q2 at the origin the electric field by q1 is the same?
 
  • #10
EpselonZero said:
If I have a charge q2 at the origin the electric field by q1 is the same?
I'm not sure I understand that. The electric field at the origin does not change if you put a charge at the origin.
 
  • Like
Likes   Reactions: happyparticle
  • #11
Alright! That was my question sorry. I wasn't sure about that. Thanks!
 

Similar threads

  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 6 ·
Replies
6
Views
997
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
2K
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
3K
  • · Replies 68 ·
3
Replies
68
Views
8K
Replies
6
Views
2K