What Is the Net Force on the -2 mC Charge in a Triangular Charge Configuration?

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SUMMARY

The net force on the -2 mC charge in a triangular charge configuration involving charges of +1 mC at (0, 0) and +3 mC at (-5 mm, 15 mm) has been analyzed. The calculations for the forces between the charges were initiated using Coulomb's Law, specifically F = k * q1 * q2 / r^2. The angle at point 3 was incorrectly calculated as 32.5 degrees, while the correct angle should be determined using atan(-2/15). The discussion emphasizes the importance of accurately determining angles based on the specific charge positions.

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yjk91
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Homework Statement



In a region of two-dimensional space, there are three fixed charges: +1 mC at (0, 0), −2 mC at (15 mm, -2 mm), and +3 mC at (-5 mm, 15 mm). What is the net force on the −2-mC charge?


? magnitude
? direction ° counter-clockwise from the +x-axis



The Attempt at a Solution


3mc = point 1
1 mc = point 2
-2 mc = point 3

length (1-3) = 26.25 mm
length (2-3) = 15.133 mm
length(1-2) = 15.811 mm

angle at point 3 is 32.5 degree using the sss triangle formula

F(1to3) = k *q1* q3 / l^2
Fx13 = F(1to3) Cos(32.5)
Fy13 = F(1to3) sin(32.5)

F(2 to 3) = k *q2* q3 / l^2
Fx23 = F(2to3) Cos(32.5)
Fy23 = F(2to3) sin(32.5)

Fx13 + Fx23 = Fxnet
Fy13 + Fy23 = Fynet

rad (Fxnet ^2 + Fynet^2 ) = Fnet which is the magnitude but it seems like I'm off by few points the answer has to be pretty close since i have to plug it into the system. Does anyone know where i made the mistake?
 
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yjk91 said:
angle at point 3 is 32.5 degree using the sss triangle formula

It's not 32.5 degrees. Point 3 is at (15,-2), which means it's at an angle of atan(-2/15). Try drawing out the three points to get a better feel for the configuration.

F(2 to 3) = k *q2* q3 / l^2
Fx23 = F(2to3) Cos(32.5)
Fy23 = F(2to3) sin(32.5)

You seem to be assuming that the angle from 2 to 3 is the same as the angle from 1 to 3, but there's no reason for this to be the case.
 

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