Finding the net field strength due to a line of charge

[Ocasta]

I want to apologize ahead of time for my apparent inability to get LaTeX to display properly.

Homework Statement

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.

$dE = λk \int_0^a \frac{dx}{r^2}$

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

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

$dE_x = λk \int_0^a \frac{xdx}{r^3}$

$u = R^2 + x^2 → \frac{du}{2} = xdx$

$dE_x = λk \int_0^a (u)^{-3/2} du$

$dE_x = λk [ -2 u^{-1/2} ]_0^a$

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

$dE_x = -2λk [ \frac{1}{\sqrt{R^2 + x^2}} ]_0^a$

$dE_x = -2λk [ \frac{1}{\sqrt{R^2 + \inf^2}} - \frac{1}{\sqrt{R^2}} ]$

$dE_x = -2λk [ 0 - \frac{1}{\sqrt{R^2}} ]$

$dE_x = -2λk [ \frac{1}{R} ]$$dE_x ≈ -2.13465 x 10^3$

 

[Mindscrape]

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.

 

[Ocasta]

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?

 

[Mindscrape]

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.

 

[Ocasta]

So I talked to my Physics professor, and he took a look at it. Evidently I forgot to carry over an x.

$dE = λk \int_0^a \frac{dx}{r^2}$

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

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

$dE_x = λk \int_0^a \frac{x dx}{r^3}$

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

$dE_x = λk [ \frac{-1}{\sqrt{R^2 + x^2}} ]_0^a$

$dE_x = λk [ \frac{-1}{\sqrt{R^2 + ∞^2}} - \frac{-1}{\sqrt{R^2 + 0^2}} ]$

$dE_x = λk [ 0 + \frac{+1}{\sqrt{32^2}} ]$

$dE_x = λk \frac{1}{\sqrt{32^2}}$

$dE_x ≈ 1.0673 * 10^3 \frac{N}{C}$

Why the solution is positive still confounds me.

 

[Mindscrape]

Ah, whoops, that is easy to do. Yeah, I didn't really check all your steps, just that the integration was correct.

 

[Ocasta]

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!