Electric field of a line of charge with symmetry

[Regtic]

Homework Statement

http://imgur.com/W4Ntkfb

Homework Equations

E=KQ/R²
e= electric field
Q = charge
R = radius from point to charge
K is a constant, 9x10⁹

The Attempt at a Solution

http://imgur.com/4CTEwDw

If my handwriting sucks, I basically did the standard integral of k\intdq /r² but kept the r hat vector since there is symmetry and I know that the magnitude will be be in the y direction. I used dq=\lambdadx and r²=x²+a² which gave me an integral that was an arctan integral. By substituting a for x, I got E= \frac{k \lambda \pi}{2a} in the radial direction.

The answer should be √2k\lambda/a upwards and perpendicular to the line of charge

 

[SammyS]

Homework Statement

http://imgur.com/W4Ntkfb

Homework Equations

E=KQ/R²
e= electric field
Q = charge
R = radius from point to charge
K is a constant, 9x10⁹

The Attempt at a Solution

http://imgur.com/4CTEwDw

If my handwriting sucks, I basically did the standard integral of k\intdq /r² but kept the r hat vector since there is symmetry and I know that the magnitude will be be in the y direction. I used dq=\lambdadx and r²=x²+a² which gave me an integral that was an arctan integral. By substituting a for x, I got E= \frac{k \lambda \pi}{2a} in the radial direction.

The answer should be 2k\lambda/R r hat.

$\hat{r}\$ depends on x, it's not constant.

You failed to take that into account.


Here's the image for the link to your problem.

$\hat{r}\$ depends on x, it's not constant.

You failed to take that into account.

 

[Regtic]

$\hat{r}\$ depends on x, it's not constant.

You failed to take that into account.Here's the image for the link to your problem.

$\hat{r}\$ depends on x, it's not constant.

You failed to take that into account.

Oh right it's not constant in direction... was just thinking of that as a constant. Derp, shouldn't be studying this late. Thank you. Redid it with Ey and got it.

also it seems whenever I post photos directly into the forum they become ridiculously large, or maybe that's just photos from Iphone...