Electric field and positive and a negative charge

In summary, the question is about determining the direction of the electric field at point P, which is equidistant from a positive and negative charge placed on the x-axis. The answer from the textbook explains that the contributions to the field at P from both charges are equal in magnitude and are directed along the line joining P to the charges. However, while the contribution from the positive charge is directed away from it, the contribution from the negative charge is directed towards it. This results in the resultant electric field at P being in the positive x-direction. The reason for this convention is that the electric field points away from positive charges and towards negative charges. The questioner is struggling to understand why this convention was chosen and suggests that the fields from both
  • #1
John Marow
1
0
This is not a homework question, but a question I am stuck on in my textbook. I have the answer but struggling to understand it.

Homework Statement


Picture below with answer.

A positive and a negative charge of equal magnitude are placed at a distance 's' from each other on the x-axis as shown in Figure 1.38. Determine the direction of the electric field at point P, which is equidistant from both charges. ** I have attached the answer that is a diagram to this post** I do not understand it.

Homework Equations


N/A

The Attempt at a Solution



Answer from textbook
The contributions to the field at P due to thepositive and negative charges are shown in Figure 1.60.Their magnitudes are the same, because the magnitudes of the two charges are the same, as are their distances from P.The directions are along the line joining P to the chargesand away from the positive charge but towards thenegative charge. Thus, the two contributions make equal angles above and below the positive x-direction. The resultant field at P is therefore in the positive x-direction.


What I don't understand

Why is the elecric field from the negative charge and positive charge towards the right on x-axis but in opposite y directions. The charge of P is not known so why is it to the right, I would think it would be to the right if it was a positive charge, because +ve q would repel P and -ve q would attract P, thus both going in same direction. But if P is negative, the vectors should going left?

My way of thinking is the fields from the -ve and +ve particles propogate from their center around them and cancel each other out at P because the particles are of the same charge mag but opposite, therefore same field magnitude at P and equal distance from P. Therefore no resultant.
 

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  • #2
The electric field points, by convention, away from + charges and towards - charges.

It could have been the reverse, but that's not how it was chosen. But it has to be away from one and towards the other.
 

Related to Electric field and positive and a negative charge

1. What is an electric field?

An electric field is a force field that surrounds a 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, either positive or negative, which generates a force that can act on other charged particles in its surroundings.

3. What is the difference between a positive and a negative charge?

A positive charge has an excess of protons, while a negative charge has an excess of electrons. These charges have opposite forces and will attract or repel each other depending on their polarity.

4. How do positive and negative charges interact in an electric field?

In an electric field, positive and negative charges will either attract or repel each other, depending on their polarity. Opposite charges will attract, while like charges will repel.

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

The strength of an electric field is measured by its electric field intensity, which is the force exerted on a unit charge within the field. It is typically measured in units of newtons per coulomb (N/C).

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