Electric Field for a line of charge

[VitaX]

Homework Statement

Homework Equations

E = λ/2π*r*ε0 (Line of charge)

The Attempt at a Solution

E1 = E2 = 0

I can set each equation equal to each other and have 2π*ε0 cancel out on both sides. Leaving me with: λ1/r1 = λ2/r2. My problem now is that I'm not sure what to use for the r values. What expression and how to incorporate L into it. I tried .17 - r on one side leaving the other side as simply r and my value for r was .56 which is way too large since it exceeds L. What do you suggest I use for my expression for r on each side?

 

[SammyS]

To find the electric field at position x₀ along the x-axis:

How far is x₀ from Line 1, which is at x = -L/2 ?
At position x = x₀, what is the direction of the E field produced bu Line 1?

How far is x₀ from Line 2, which is at x = +L/2 ?
At position x = x₀, what is the direction of the E field produced bu Line 2?

 

[VitaX]

To find the electric field at position x₀ along the x-axis:

How far is x₀ from Line 1, which is at x = -L/2 ?
At position x = x₀, what is the direction of the E field produced bu Line 1?

How far is x₀ from Line 2, which is at x = +L/2 ?
At position x = x₀, what is the direction of the E field produced bu Line 2?

-.085 - x = x0 for Line 1
.085 - x = x0 for Line 2

Should it be .17 instead of .085?

So ur saying to plug these values in for r and solve for x and that is my answer?

 

[SammyS]

x₀ is the x-coordinate at which you want to evaluate the E field.

Suppose x_{0 = 5 cm. How far is x0} from Line 1? In other words, how far is x = 5 cm from x = -8.5 cm?

For Line 2: How far is x = 5 cm from x = +8.5 cm?

Your formulas don't seem to give the right answers for these questions.

 

[VitaX]

x₀ is the x-coordinate at which you want to evaluate the E field.

Suppose x_{0 = 5 cm. How far is x0} from Line 1? In other words, how far is x = 5 cm from x = -8.5 cm?

For Line 2: How far is x = 5 cm from x = +8.5 cm?

Your formulas don't seem to give the right answers for these questions.

I don't understand why they would be wrong. It says in the hint use the derived equation for an electric field produced by a long line charge, which I noted in the original post. Are you saying I can't set these equations equal to each other?

To answer your question it would be: 5-8.5 = 13.5 so I guess I could say x - L/2 = R1. And L/2 - x = R2 right?

But are you saying that E1 = E2 = 0 can't be done?

 

[SammyS]

I don't understand why they would be wrong. It says in the hint use the derived equation for an electric field produced by a long line charge, which I noted in the original post. Are you saying I can't set these equations equal to each other?

To answer your question it would be: 5-8.5 = 13.5 so I guess I could say x - L/2 = R1. And L/2 - x = R2 right?

But are you saying that E1 = E2 = 0 can't be done?

How is 5 minus 8.5 equal to 13.5 ? 5 - (-8.5) = 13.5.

So, x - (-L/2) = R1 is the distance that x is from Line 1; (This should be |x - (-L/2)| = R1)

and L/2 - x = R2 is the distance that x is from Line 2. (This should be |L/2 - x| = R2)

Notice: R1 + R2 = L, if x is between Line 1 and Line 2.

I'm NOT saying "E1 = E2 can't be done"?. It can be done. However what you really need is that their magnitudes are equal but their signs are opposite.

Can that happen any place between Line 1 & Line 2? Why can't it happen there? -- and it definitely can't.

 

[VitaX]

How is 5 minus 8.5 equal to 13.5 ? 5 - (-8.5) = 13.5.

So, x - (-L/2) = R1 is the distance that x is from Line 1; (This should be |x - (-L/2)| = R1)

and L/2 - x = R2 is the distance that x is from Line 2. (This should be |L/2 - x| = R2)

Notice: R1 + R2 = L, if x is between Line 1 and Line 2.

I'm NOT saying "E1 = E2 can't be done"?. It can be done. However what you really need is that their magnitudes are equal but their signs are opposite.

Can that happen any place between Line 1 & Line 2? Why can't it happen there? -- and it definitely can't.

My bad on the typo. And yeah I understand that the magnitudes have to equal one another. So I guess I could write E1 = -E2 or vice versa. I thought that for the net electric field to be zero, it has to be between Line 1 and 2. So now you are saying that it can't happen there and has to be outside of those boundaries?

How is this:

λ1/r1 = - λ2/r2 where r1 = x - (-L/2) and r2 = L/2 - x and solve for x.

 

[SammyS]

My bad on the typo. And yeah I understand that the magnitudes have to equal one another. So I guess I could write E1 = -E2 or vice versa. I thought that for the net electric field to be zero, it has to be between Line 1 and 2. So now you are saying that it can't happen there and has to be outside of those boundaries?

How is this:

λ1/r1 = - λ2/r2 where r1 = x - (-L/2) and r2 = L/2 - x and solve for x.

If x > L/2, then L/2 - x < 0, therefore, r₂ = x - L/2, to the right of Line 2.

The reason that the electric field can't be zero between the line charges is that E₁ points away from Line 1, and E₂ points towards from Line 2. In between, they both point to the riht.

 

[VitaX]

If x > L/2, then L/2 - x < 0, therefore, r₂ = x - L/2, to the right of Line 2.

The reason that the electric field can't be zero between the line charges is that E₁ points away from Line 1, and E₂ points towards from Line 2. In between, they both point to the riht.

I'm a bit confused at how the you figure out the direction of the E fields based on the placement of the charge. How exactly do you know that if the point were in between, both lines would point to the right?