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

[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?

 

[issacnewton]

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