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October 30, 2014

October 30, 2014

Posted by **peter** on Monday, September 13, 2010 at 9:49pm.

- physics 2 -
**bobpursley**, Monday, September 13, 2010 at 10:23pmSo I assume you have calculus based physics.

integrate the charge thru and angle. let theta begin at the bottom rotate 90 deg for all the neg charge, then from 90 to 180 for the positive charge.

E=INT kdq/r^2 where dq=Q/(pi*r/2)dTheta

integrate from 0 to PI/2, that is E for the lower charge, and it points to the center point of the quadrant (45 deg, by symettry).

Do the same for the + charge, it points away from the center of the upper quadrant.

Now add the two E components. Neat, it point downward, and has value of E*sqrt2

work that out.

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