# Help approximating an electric field between points.

## Homework Statement

A strip of invisible tape 0.13 m long by 0.019 m wide is charged uniformly with a total net charge of 5 nC
(nano = 1 ✕ 10−9)
and is suspended horizontally, so it lies along thex axis, with its center at the origin, as shown in the figure below. Calculate the approximate electric field at location 0, 0.02, 0 m (location A) due to the strip of tape. Do this by dividing the strip into three equal sections, as shown in the figure below, and approximating each section as a point charge. (Assume each point charge is located at the center of the section it approximates. Express your answer in vector form.)
http://prntscr.com/9xvusu picture of the tape.

(a) What is the approximate electric field at A due to piece 1?

E = kq/r^2 rhat

## The Attempt at a Solution

I did K * Q which is (9 * 10^9)* ((5/3) * 10 ^ -9)) which comes out to 15. Then I attempted to find r^2, which I believe would be (.02^2 + .13^2) and do 15 / .0173. Then I tried to multiply by the unit vector r <.98837, .15205, 0> to get <856.9697, 131.6415, 0> as my answer. I'm guessing there's some problem with how I'm calculating the r values since when I did part b (b) What is the approximate electric field at A due to piece 2?
it worked, but I was only managing the y component (.02m) since it was right below point A.

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Simon Bridge
Homework Helper
There would be something wrong with how you are handling the x component then?
First, what did you get for the location and size of each of each point charge used in the approximation.

I see you have put the end charges a distance 0.13m from the center ... considering that the strip is only 0.13m long, that puts the end charges off the ends of the strip.
You should sketch the layout - the strip is centered on the origin so it cannot extend more than 0.13/2 m to either side.
Mark off the boundaries of the 3 sub-strips. Where is the logical place to put the point charges?
Hint: not at the ends of the strip.

• Sheolfire
I tried using (15/2 * (1/3) ) as my distance (.02166) and ended up with something that seems closer, but it still not right. I did 15 / (.02^2 + .0266^2) = 17252.396 and then found rhat again with .0266/.02949 and .02/.02949. I ended up getting <12677.12, 11701.96, 0>.

Nevermind, I figured it out, it should be .13/2 * 2/3 not 1/3. Thanks for the help.

Simon Bridge