Finding the electric field magnitude, and angle below the horizontal

AI Thread Summary
To find the electric field at a specified point due to multiple charges, it is essential to calculate the electric field contributions from each charge as vectors rather than just adding their magnitudes. The charges involved are 10 nC, 5 nC, and -5 nC, with distances affecting their respective electric field strengths. The correct resultant electric field magnitude is 7.9 × 10^4 N/C, directed at an angle of 5.2° below the horizontal. Proper vector addition must consider both the magnitude and direction of each electric field component. Understanding these principles is crucial for accurately solving electric field problems.
jheld
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Homework Statement


In the figure below d = 2.6 cm. What is the electric field at the position indicated by the dot in the figure?


Homework Equations



E = charge/4*pi*epsilon_0*distance^2
The figure is on the attachment.

The Attempt at a Solution


I decided to first find only the magnitude from each of the fields on the point.
charge 1 = 10 nC;
charge 2 = 5 nC;
charge 3 = -5nC
*I put all of the charges into Coulomb before any calculations
E1 = Kc1/5d^2;
E2 = Kc2/d^2;
E3 = kc3/4d^2;
After adding all of the Es together I get the wrong answer (as far as magnitude).
Answer is: 7.9 × 104 N/C, at 5.2° below the horizontal
 

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jheld said:
*I put all of the charges into Coulomb before any calculations
E1 = Kc1/5d^2;
E2 = Kc2/d^2;
E3 = kc3/4d^2;
After adding all of the Es together I get the wrong answer (as far as magnitude).
You can't just add up the magnitudes like ordinary numbers (if that's what you did). You must add them as vectors. What is the direction of each field?
 
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