Find the magnitude and the direction of the electric field at the thir

AI Thread Summary
The discussion focuses on calculating the electric field at the third corner of a triangle formed by two point charges, q1 = -11.3 μC and q2 = 15.4 μC. The correct answer for the electric field's magnitude is 1.07e+07 N/C at an angle of 71.2 degrees above the -x-axis. Participants highlight the importance of unit conversion, noting that errors in converting microcoulombs and centimeters can lead to incorrect results. The sign of the electric field components also plays a crucial role in determining the direction of the resultant field. Proper unit conversion and understanding the impact of charge signs are essential for accurate calculations.
Richard Ros
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Homework Statement


Two point charges are placed at two of the corners of a triangle as shown in the figure. (Take q1 = -11.3 μC, and q2 = 15.4 μC.) Find the magnitude and the direction of the electric field at the third corner of the triangle. {Correct Answer: 1.07e+07 N/C at 71.2 degrees above the -x-axis}


Homework Equations


F = ((1)/(4πε0))(q1*q2/r^2)

The Attempt at a Solution


Attached files. What am i doing wrong? I can't seem to get the correct answer.
 

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Did you convert units correctly? Your answers appear to be off by some factors of 10.

Is Ey from q1 positive or negative? [Edited]
 
Hmm I don't know how I got a negative. But either way when I solve for the magnitude it'll end up being positive. Which unit should I convert? The μc or cm?
 
Richard Ros said:
Hmm I don't know how I got a negative. But either way when I solve for the magnitude it'll end up being positive.
Yes, but the sign of Ey will affect the answer for the direction angle.
Which unit should I convert? The μc or cm?
Neither μC nor cm is an SI unit.
 
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Thanks, turns out it was just my conversion.
 
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