# Electric Field from two lines of charge on the y-axis.

• ArtemRose
In summary, the conversation discusses a homework problem involving the calculation of the electric field at a point on the x-axis due to two uniformly distributed charges along the y-axis. The conversation includes an attempted solution using calculus and a suggestion to consider the problem in two segments. The final advice is to find the x and y components of the electric field at the given point and add them with proper signs.
ArtemRose

## Homework Statement

I was given this assignment for homework, and got it wrong, but now I'm studying for the first exam, and I still can't find out where I went awry. The problem is:

A charge per unit length λ = +8.00 μC/m is uniformly distributed along the positive y-axis from y = 0 to y = +a = +0.500 m. A charge per unit length λ = -8.00 μC/m, is uniformly distributed along the negative y-axis from y = 0 to y = –a = -0.500 m. What is the magnitude of the electric field at a point on the x-axis a distance x = 0.371 m from the origin?

1.567×105 N/C This being the answer I am supposed to get.

## Homework Equations

Now, I know that I can find this using the calculus with dE=k*(dq/r^2) dy

## The Attempt at a Solution

Now, set the equation to be dE=k*lambda*y/(y^2 + (.371)^2)^(3/2) dy

Then when I took the derivative, it would be then E=k*lambda/(sqrt(y^2 + (.371)^2) + Constant

And then I plug in for y, but it's nowhere near the right answer. Where am I going wrong? I've tried another method to solve for line of charge
[ ((k*8e-6)/.371)*(.5/sqrt(.5^2 + .371^2)) ] (that is, [ ((k*lambda)/x)*(y/sqrt(y^2 + x^2)) ] equaling the E Field.
without using calculus, and that gave me an answer within the bounds, but I didn't have to double it when it should be giving me only the answer for one of the lines of charge, so that makes me doubt that method.

Any ideas?

hi rose

you can consider two different segments...one from y=0 to y=0.5 which has positive charge and one from y=0 to y=-0.5 which has negative charge. then the problem reduces to find the magnitude of the E due to these two configurations at point P (say) on the x-axis and then add them with proper signs...for each of the segment, the point P is at some distance from one end of the segment... you need to find the x and y components of the total electric field
at point P due to both the segments... for example the x component would be given by

$$E_x=\frac{1}{4\pi\epsilon_o}\int_0^L \frac{\lambda\,dy}{(y^2+x^2)}\cos{\theta}$$

where $$\cos{\theta}=\frac{x}{(y^2+x^2)^\frac{1}{2}}$$
a
where $\lambda$ would be line charge density and x is the distance of the point P from the origin and y is the distance of the differential element dy from the origin. L in each case would be 0.5...

that may help a bit

## 1. How do you calculate the electric field from two lines of charge on the y-axis?

The electric field from two lines of charge on the y-axis can be calculated by using the formula E = k(q/r²), where k is the Coulomb's constant, q is the charge of each line, and r is the distance between the lines.

## 2. Does the direction of the electric field depend on the type of charges on the lines?

Yes, the direction of the electric field depends on the type of charges on the lines. If the charges are of the same sign, the electric field will be repulsive and point away from the lines. If the charges are of opposite signs, the electric field will be attractive and point towards the lines.

## 3. Can the electric field from two lines of charge on the y-axis cancel out?

Yes, the electric field from two lines of charge on the y-axis can cancel out if the charges on the lines are equal in magnitude and opposite in sign. In this case, the electric field at the midpoint between the lines will be zero.

## 4. How does the distance between the lines affect the electric field?

The distance between the lines has an inverse square relationship with the electric field. This means that as the distance between the lines increases, the electric field decreases, and vice versa.

## 5. Is the magnitude of the electric field the same at all points along the y-axis?

No, the magnitude of the electric field varies at different points along the y-axis. It is strongest at the ends of the lines and decreases as you move away from the lines. The magnitude is also affected by the distance between the lines and the charges on the lines.

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