Electric field from a charge q1

In summary, the electric field at the origin is zero, but the field at the point (3m, 4m) would be different if a charge q2 was present.
  • #1
happyparticle
400
20
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?
 
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  • #2
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?
 
  • #3
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.
 
  • #4
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?
 
  • #5
q1 is at the point (3m,4m) I calculated the electric field at the origin, but without the charge q2.
 
  • #6
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.
 
  • #7
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?
 
  • #8
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.
 
  • #9
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 happyparticle
  • #11
Alright! That was my question sorry. I wasn't sure about that. Thanks!
 

1. What is an electric field?

An electric field is a physical force field that surrounds an electrically charged particle and exerts a force on other charged particles within its vicinity.

2. How is an electric field created?

An electric field is created by a charged particle, such as an electron or proton. The charged particle exerts a force on other particles in its surroundings, creating an electric field.

3. How is the strength of an electric field measured?

The strength of an electric field is measured by the force it exerts on a test charge placed in the field. The unit of measurement for electric field strength is volts per meter (V/m).

4. How does the electric field from a charge q1 affect other charges?

The electric field from a charge q1 exerts a force on other charges within its vicinity. The direction and magnitude of the force depend on the charges of the particles and their distance from q1.

5. How does the distance from a charge q1 affect the strength of the electric field?

The strength of the electric field decreases as the distance from q1 increases. This is because the electric field spreads out as it moves away from the source charge, resulting in a weaker force on other charges.

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