Charge and Electric Field Problem

Tags:
1. Oct 8, 2015

Callix

1. The problem statement, all variables and given/known data
You have been hired by Brockovich Research and Consulting (BRC) to research a new water purification device that uses seeds from the Moringa Oleifera trees.1 A protein in the seed binds to impurities causing them to aggregate so that the clusters can be separated from the water. For this research, you are asked to build an electron microscope to investigate the structure of the Moringa oleifera seed. Your new device consists of a charged, conducting circle which is divided into two half circles separated by a thin insulator so that half of the circle can be charged positively (+q) and half can be charged negatively (–q). To complete the design of the electron microscope, you calculate the electric field in the center of the circle as a function of: the amount of positive charge on the half circle, the amount of negative charge on the half circle, and the radius of the circle (+q, –q, R).

2. Relevant equations
E=kQ/r^2
E=QV?

3. The attempt at a solution
I know that since the circle is constructed out of a conductive material, then the electric field would be 0. However, there is an insulating strip that runs down the middle of the circle. This is where I got stuck. I interpreted this as the circle is now a dipole, but I am unsure how to calculate the field inside the insulating strip. Isn't it just E=kQ/r^2?

Any help or further direction is appreciated! :)

2. Oct 8, 2015

Staff: Mentor

You're thinking of the electric field within the conducting material, but that's not the issue here.

You have two charged half circles. You need to find the field due to those charges at the center of the circle. (Treat each half circle as if it were uniformly charged with a total charge of +/- q.)

3. Oct 8, 2015

Callix

Ohhh, right!

So both emit a field, one kQ/r^2 and the other -kq/r^2

4. Oct 8, 2015

Callix

Or will this require integration?

5. Oct 8, 2015

Staff: Mentor

I'm afraid it will, since the field from each element of charge on the half circle will have a different direction.

6. Oct 8, 2015

Callix

Sure, that makes sense. Would you be able to explain the integration to me for this scenario?

7. Oct 8, 2015

Staff: Mentor

Why don't you give it a shot yourself? (If you're totally stuck, a little Googling should help get you moving.)

Hint: Use symmetry. Break the charge into charge elements, and their resulting electric field into components.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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}