Homework Help: The electric field at the center of a square

1. Sep 5, 2011

Jay9313

http://helios.augustana.edu/~dr/203/probs/Set 1.pdf
Scroll down to #11 for a picture.

So I kind of understand the math. I am a little confuse though. So why is a divided by sqrt(2)?
If this is a vector, then the center is really (a(sqrt(2)))/2 and the x component would be divided by sqrt(2) which makes the equation equal a/2. Can someone explain this? Is my math wrong?

2. Sep 5, 2011

Spinnor

The diagonal distance from corner to corner is a*2^.5 and the distance from corner to the center is half of that,

a*2^.5/2 = a/2^.5

Yes?

3. Sep 5, 2011

Redbelly98

Staff Emeritus
The distance from the center to any corner is $a/\sqrt{2}$, that's why a is divided by sqrt(2).

Taking the x-component is accounted for by the cos(45) in the expression.

4. Sep 5, 2011

Jay9313

I'm not sure why it's a/sqrt(2). I do know that (a(sqrt(2))/2 is the exact same number. I don't know why.

Basically, I found the length of the diagonal, a(sqrt(2)) and divided it by 2. It's the same as a/sqrt(2)
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}