# Electric Field magnitude problem help

• Seraph404
In summary, if you are having trouble with calculating the magnitude of an electric field, there are a few key steps to follow. First, determine the distance between the source charge and the point where you want to calculate the field. Then, calculate the Coulomb's constant and the charge of the source. Finally, use the formula for electric field magnitude, which is equal to the Coulomb's constant multiplied by the source charge divided by the distance squared. By following these steps, you can easily solve for the magnitude of the electric field.
Seraph404

## Homework Statement

A very long uniform line of charge has charge per unit length 4.88E-6 C/m and lies along the x-axis. A second long uniform line of charge has charge per unit length -2.44E-6 C/m and is parallel to the x-axis at y1 = 0.408 m.

What is the magnitude of the net electric field at point y2 = 0.198 m on the y-axis?

E = kq/r^2

## The Attempt at a Solution

E1 = (9E9)(4.88E-6)/(.198)^2
E2 = (9E9)(-2.44E-6)/(.21)^2

net E = E1 + E2 (both point in the positive y direction), but I'm not getting the right answer. Apparently, my answer is close, but not correct. Can somebody help me to spot my mistake?

Also, I know the problem gave linear charge density, but it didn't say how long the wire was. Maybe that's where my error lies. If so, how do I correct it?

Seraph404 said:

## Homework Statement

A very long uniform line of charge has charge per unit length 4.88E-6 C/m and lies along the x-axis. A second long uniform line of charge has charge per unit length -2.44E-6 C/m and is parallel to the x-axis at y1 = 0.408 m.

What is the magnitude of the net electric field at point y2 = 0.198 m on the y-axis?

E = kq/r^2

## The Attempt at a Solution

E1 = (9E9)(4.88E-6)/(.198)^2
E2 = (9E9)(-2.44E-6)/(.21)^2

net E = E1 + E2 (both point in the positive y direction), but I'm not getting the right answer. Apparently, my answer is close, but not correct. Can somebody help me to spot my mistake?

Also, I know the problem gave linear charge density, but it didn't say how long the wire was. Maybe that's where my error lies. If so, how do I correct it?
One cannot simply take the expression for the electric field of a point charge and apply it to an infinite line charge. Conceptually, to find the electric field do to a line charge one must sum up the contributions from each individual point charge. Since the charge distribution is continuous one does this via an integral rather than a sum.

Firstly, it is important to note that the electric field from a line of charge is not constant, and therefore cannot be calculated using the formula E = kq/r^2. Instead, we must use the formula E = (kλ)/(r^2), where λ is the linear charge density.

Now, to find the net electric field at point y2, we need to consider the contributions from both lines of charge. The distance between the first line of charge and point y2 is 0.408 m, while the distance between the second line of charge and point y2 is 0.198 m. Therefore, our equations for E1 and E2 should be:

E1 = (9E9)(4.88E-6)/(0.408)^2 = 4.9 N/C
E2 = (9E9)(-2.44E-6)/(0.198)^2 = -29.3 N/C

Note that the negative sign in E2 indicates that the electric field is pointing in the opposite direction to E1. Therefore, the net electric field at point y2 is:

E = E1 + E2 = 4.9 N/C - 29.3 N/C = -24.4 N/C

This value is negative because the two electric fields are pointing in opposite directions. We can also check our answer by considering the symmetry of the problem. Since the two lines of charge are parallel and have equal but opposite charge densities, the net electric field at point y2 should be zero. Our answer of -24.4 N/C is close to zero, so our calculations are likely correct.

In terms of the length of the wire, it is not necessary to know the exact length of the wire in order to calculate the electric field. As long as we have the linear charge density, we can use the formula E = (kλ)/(r^2) to find the electric field at any point along the wire. However, if the length of the wire was given, we could use the formula E = (kq)/(Lr), where L is the length of the wire and r is the distance from the wire to the point where we want to find the electric field.

I hope this helps clarify any confusion and helps you to identify any mistakes in your calculations. Remember to always double check your units and use the correct formula for the situation. Good luck with your studies!

## 1. How do I calculate the magnitude of an electric field?

To calculate the magnitude of an electric field, you will need to know the charge of the particle creating the field, the distance from the particle to the point where you want to calculate the field, and the permittivity of the medium surrounding the particle. You can use the equation E = kQ/r^2, where E is the electric field, k is the Coulomb constant (8.99 x 10^9 Nm^2/C^2), Q is the charge of the particle, and r is the distance from the particle to the point where you want to calculate the field.

## 2. How does the distance from the particle affect the magnitude of the electric field?

The magnitude of the electric field is inversely proportional to the square of the distance from the particle. This means that as the distance increases, the electric field decreases. This relationship is described by the inverse square law, which states that the intensity of a field is inversely proportional to the square of the distance from the source of the field.

## 3. How do I determine the direction of the electric field?

The direction of the electric field is always in the direction that a positive test charge would move if placed in the field. This means that the direction of the electric field points away from positive charges and towards negative charges. You can use the concept of electric field lines to visualize the direction of the field, where the lines point away from positive charges and towards negative charges.

## 4. Does the medium surrounding the particle affect the magnitude of the electric field?

Yes, the permittivity of the medium surrounding the particle can affect the magnitude of the electric field. The permittivity is a measure of how easily a medium can be polarized by an electric field. In a medium with a higher permittivity, the electric field will be stronger compared to a medium with a lower permittivity. The permittivity is represented by the symbol ε and is measured in units of Farads per meter (F/m).

## 5. How can I use the electric field magnitude to determine the force on a charged particle?

The force on a charged particle is determined by multiplying the magnitude of the electric field by the charge of the particle. The direction of the force will depend on the direction of the electric field and the charge of the particle. For example, a positive charge will experience a force in the same direction as the electric field, while a negative charge will experience a force in the opposite direction. You can use the equation F = qE, where F is the force, q is the charge of the particle, and E is the electric field.

• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
16
Views
241
• Introductory Physics Homework Help
Replies
12
Views
1K
• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
525
• Introductory Physics Homework Help
Replies
5
Views
871
• Introductory Physics Homework Help
Replies
4
Views
2K