Finding the net field strength due to a line of charge

AI Thread Summary
The discussion focuses on calculating the net electric field strength due to a line of charge with a given linear charge density and distance parameters. The integral for the electric field is derived, and the calculations involve determining the contributions from different segments of the line charge. There is confusion regarding the sign of the final answer, as the expected result is positive despite the vector direction indicating otherwise. The importance of correctly defining the coordinate system is emphasized, as it affects the interpretation of the results. Ultimately, the calculations yield a positive value for the electric field strength, which raises questions about the conventions used in the problem.
Ocasta
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I want to apologize ahead of time for my apparent inability to get LaTeX to display properly.

Homework Statement


prob2-33.png


λ = 3.8 x 10^(-6) C/m

R = 32m

r = magnitude of the hypotenuse (the book's convention, a poor choice IMHO)
r = \sqrt{R^2 + x^2}

k = 8.988 x 10^9

P:(0,-32)

cosθ = x/r

a→∞

Homework Equations


dE = λk \int_0^a \frac{dx}{r^2}

The Attempt at a Solution



We're supposed to figure out the overall field strength, which is finite.

dE = λk \int_0^a \frac{dx}{r^2}

dE_x = λk \int_0^a \frac{dx}{r^2} cosθ

dE_x = λk \int_0^a \frac{dx}{r^2} \frac{x}{r}

<br /> dE_x = λk \int_0^a \frac{xdx}{r^3}<br />

<br /> u = R^2 + x^2 → \frac{du}{2} = xdx<br />

<br /> dE_x = λk \int_0^a (u)^{-3/2} du<br />

<br /> dE_x = λk [ -2 u^{-1/2} ]_0^a<br />

<br /> dE_x = -2λk [ (R^2 + x^2)^{-1/2} ]_0^a<br />

<br /> dE_x = -2λk [ \frac{1}{\sqrt{R^2 + x^2}} ]_0^a<br />

<br /> dE_x = -2λk [ \frac{1}{\sqrt{R^2 + \inf^2}} - \frac{1}{\sqrt{R^2}} ]<br />

<br /> dE_x = -2λk [ 0 - \frac{1}{\sqrt{R^2}} ]<br />

<br /> dE_x = -2λk [ \frac{1}{R} ]<br /><br /> dE_x ≈ -2.13465 x 10^3<br />
 
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Looks good, though the sign is important to how you're defining your coordinate system. Make sure your coordinate system is the one the program is looking for.
 
I literally pasted all the information I was given. >.< Do you have any guidance as to how to figure out what coordinate system I should use?
 
I mean, typically we would define y is being up and x as being right. But you could do any permutation of those. I guess I don't understand what your questions about the problem are.
 
So I talked to my Physics professor, and he took a look at it. Evidently I forgot to carry over an x.

Ocasta said:
dE = λk \int_0^a \frac{dx}{r^2}

dE_x = λk \int_0^a \frac{dx}{r^2} cosθ

dE_x = λk \int_0^a \frac{dx}{r^2} \frac{x}{r}

<br /> dE_x = λk \int_0^a \frac{x dx}{r^3}<br />

<br /> dE_x = λk \int_0^a \frac{x dx}{(R^2 + x^2)^{3/2}}<br />

<br /> dE_x = λk [ \frac{-1}{\sqrt{R^2 + x^2}} ]_0^a<br />

<br /> dE_x = λk [ \frac{-1}{\sqrt{R^2 + ∞^2}} - \frac{-1}{\sqrt{R^2 + 0^2}} ]<br />

<br /> dE_x = λk [ 0 + \frac{+1}{\sqrt{32^2}} ]<br />

<br /> dE_x = λk \frac{1}{\sqrt{32^2}}<br />

<br /> dE_x ≈ 1.0673 * 10^3 \frac{N}{C}<br />

Why the solution is positive still confounds me.
 
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Ah, whoops, that is easy to do. Yeah, I didn't really check all your steps, just that the integration was correct.
 
Mindscrape said:
Ah, whoops, that is easy to do. Yeah, I didn't really check all your steps, just that the integration was correct.

The most frustrating part is that I apparently was correct the first time, but the answer they wanted was positive rather than negative. But looking at the problem, the vector is clearly negative in the X and Y directions!
 
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