Finding the electric field magnitude, and angle below the horizontal

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SUMMARY

The discussion focuses on calculating the electric field magnitude and direction at a specific point due to three point charges: 10 nC, 5 nC, and -5 nC, positioned 2.6 cm apart. The correct formula for the electric field is E = charge/(4 * π * ε₀ * distance²). The final calculated electric field magnitude is 7.9 × 104 N/C, directed at an angle of 5.2° below the horizontal. Participants emphasize the importance of vector addition for electric fields rather than simple arithmetic summation.

PREREQUISITES
  • Understanding of Coulomb's Law and electric fields
  • Familiarity with vector addition in physics
  • Knowledge of the constant ε₀ (permittivity of free space)
  • Basic skills in unit conversion (nC to C)
NEXT STEPS
  • Study vector addition of electric fields in multiple charge systems
  • Learn about the concept of electric field lines and their implications
  • Explore the effects of charge distribution on electric fields
  • Investigate the role of ε₀ in different mediums
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in understanding electric field calculations and vector analysis in electrostatics.

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
 

Attachments

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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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