- #1
jb007
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Homework Statement
Here is the problem I am stuck on. I have checked my process multiple times, but have come up with the same wrong result. I would like to find out where by error of thinking lies.
"An infinitely long line charge of uniform linear charge density λ = -1.30 µC/m lies parallel to the y axis at x = -1.00 m. A point charge of 1.20 µC is located at x = 2.00 m, y = 3.00 m. Find the electric field at x = 3.00 m, y = 2.50 m."
The answer form is
Side note: since it's in vector form, does that mean they want the answer to be written in i hats and j hats?
Homework Equations
For a infinite line of charge a distance R away, the electric field produced is E = 2kλ / R.
For just a point charge some distance r away, the electric field is E = kq / r2.
The Attempt at a Solution
I drew a picture in the x and y coordinate plane with the line of charge at x = -1, the point charge at (2, 3), and the analyze point at (3, 2.5).
First, I calculated the electric field due to the infinite line of charge. Since the y-component cancels, the net electric field at (3, 2.5) due to the line of charge is just the horizontal x-component. So R = 3 - (-1) = 4.
So E = 2kλ / R = 2(9*109)(-1.3*10-6) / 4 = -5850 N/C.
Second I calculated the electric field due to the point charge. I used the distance formula to find the distance r between the point charge and (3, 2.5).
r = sqrt[(3-2)2+(2.5-3)2] = sqrt[1.25]
r2 = 1.25 meters
So E = kq / r2 = (9*109)(1.2*10-6) / 1.25 = 8640 N/C.
But this is the total. To get the x-direction of this electric field, I drew a vector triangle formed by the line between (2, 3) and (3, 2.5). The hypotenuse is sqrt(1.25) meters, the short side is |2.5-3| = 0.5 meters, and the long side is |3-2| = 1 meter. Using inverse trig, I got that the angle between the angle between the two points was 26.565 degrees.
Using the angle, I then calculated the x and y components of the E due to the point charge.
For the x-component I did Ecos(26.565) = 8640cos(26.565) = 7727.854 N/C.
For the y-component I did Esin(26.565) = 8640sin(26.565) = 3863.92 N/C. It's actually -3863.92, since the force would be in the negative y direction.
So the total E at the point (3, 2.5) is the total x and y-components.
The total x-component is -5850 + 7727.854 = 1877.854 N/C
The total y-component is -3863.92 N/C.
This is where I am unsure about how to answer. Do I do the square root? Like E = sqrt[(1877.854)2 + (-3863.92)^2]? This result gives 4296.07 N/C. But wouldn't this be just the magnitude of the electric field at (3, 2.5), thus excluding direction? Or is the question wanting me to answer in some other form? Is some part of my process wrong?