Electric field between two charges in 1D

In summary, the question asks for the magnitude and direction of the electric field at point A, which is 0.25 m to the right of the left-hand charge. Using the equation E = kQ/r^2, it is necessary to subtract the charge at 0.75 m from the charge at 0.25 m to get the correct answer. The direction of the electric field is determined by the direction of the force on a positive test-charge, with a positive charge to the right pushing left and a positive charge to the left pushing right. By adding the fields from the two charges at 0.25 m and 0.75 m, the final magnitude and direction of the electric field at point A can be
  • #1
Tekee
20
0

Homework Statement



Two identical positive charges of +Q are 1 m apart. What is the magnitude and direction of the electric field at point A, 0.25 m to the right of the left-hand charge?

Homework Equations



E = kQ/r^2

The Attempt at a Solution



I summed up kQ+/(.25^2) and kQ+/(.75^2) and what I got was one of the answer choices, but apparently it was wrong. Evidently you have to subtract the .75 charge from the .25 to get the right answer. I also am unsure of which direction the electric field points.
 
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  • #2
okay tekee, imagine that the point of interest is a positive charge. Which should be the default. A positive charge to the righthand side of it pushes it left. and a positive charge to the lefthand side of it will push right. So you will have to make appropriate changes to your answer due to that.
 
  • #3
point A is between the charges. The direction of the electric field is the direction of the force on a positive test-charge.
What is the magnitude and direction of the field due to the positive charge 0.25 m to the left of A?
What is the magnitude and direction of the field due to the positive charge 0.75 m to the right of A?
Now you can add those fields like vectors.
 

1. What is an electric field?

An electric field is a physical field that surrounds electrically charged particles and exerts a force on other electrically charged particles within its range. It can be described as the force per unit charge that a charged particle would experience when placed in the field.

2. How is the electric field between two charges in 1D calculated?

The electric field between two charges in 1D is calculated using Coulomb's law, which states that the magnitude of the electric field is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them. It can be represented by the equation E = k(q1q2)/r^2, where k is the Coulomb's constant, q1 and q2 are the charges, and r is the distance between them.

3. What factors affect the strength of the electric field between two charges in 1D?

The strength of the electric field between two charges in 1D is affected by the magnitude of the charges, the distance between them, and the medium in which the charges are located. In a vacuum, the electric field is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them. However, in a medium other than vacuum, the electric field can be affected by the material's dielectric constant.

4. How does the direction of the electric field between two charges in 1D affect the motion of charged particles?

The direction of the electric field between two charges in 1D determines the direction of the force that a charged particle will experience when placed in the field. If the two charges have opposite signs, the electric field will push the charged particle towards one of the charges, while if the two charges have the same sign, the electric field will push the charged particle away from both charges.

5. Can the electric field between two charges in 1D be negative?

Yes, the electric field between two charges in 1D can be negative. This occurs when the two charges have opposite signs, and the electric field points towards the negative charge. In this case, the electric field is considered to be negative in relation to the positive charge, but it is still a valid and measurable value.

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