Yahoo Answers: Answers and Comments for Calculate the Magnitude and Direction of the NET Electric Field? [Physics]
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From Kelsey
enUS
Thu, 05 Sep 2019 18:53:56 +0000
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Yahoo Answers: Answers and Comments for Calculate the Magnitude and Direction of the NET Electric Field? [Physics]
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https://answers.yahoo.com/question/index?qid=20190905185356AA2m6id
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From lunchtime_browser: The electric field (E) due to a point charge (...
https://answers.yahoo.com/question/index?qid=20190905185356AA2m6id
https://answers.yahoo.com/question/index?qid=20190905185356AA2m6id
Thu, 05 Sep 2019 20:15:19 +0000
The electric field (E) due to a point charge (Q) at a point at a distance (r) is given by:
E = kQ / r²
k is the coulomb constant = 8.99×10⁹ N·m²·C⁻²
The field is away from a positive point charge and towards a negative point charge.
{Remember the way to test a field is to put a unit positive test charge at the point in question and see which way the resulting force acts on it.}
So we need the diagonal distances (r) from the red to black and blue to black.
Uncle Pythagoras:
r² = 0.2² + 0.2²
r² = 0.08
E due to red dot = (8.99×10^9 * 2.0×10^6 / 0.08) = 2.25 ×10^5 V/m { to the north west :}
Similarly
E due to blue dot = (8.99×10^9 * 4.0×10^6 / 0.08) = 4.50 ×10^5 V/m { to the south west :}
The resultant field at the black dot is just the vector sum of the two field components {purple}.
magnitude from Pythagoras
R = 5.03 ×10^5 V/m
direction = arctan (2.25/4.5) = 26.6 degrees north of south west = 18.4 degrees south of west
{So invent your own coordinate system then! :) }