Electric Fields and source charge

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SUMMARY

The discussion focuses on calculating the electric field and force on a negative charge placed in the vicinity of a positive source charge. The electric field at location D is given as < -8000, 8000, 0 > N/C. To find the unit vector in the direction of the electric field, the calculation involves normalizing the electric field vector. The electric force on the -6e-09 C charge can be determined using the formula F = qE, where q is the charge and E is the electric field. The relationship between the direction of the electric field and the force on the charge is clarified, indicating that they are antiparallel due to the nature of the charges involved.

PREREQUISITES
  • Understanding of electric fields and forces, specifically Coulomb's Law.
  • Familiarity with vector normalization and unit vectors.
  • Knowledge of basic physics equations involving electric charge and field strength.
  • Ability to perform trigonometric calculations, particularly with angles and vector components.
NEXT STEPS
  • Study vector normalization techniques in physics.
  • Learn about Coulomb's Law and its applications in electric field calculations.
  • Explore the relationship between electric fields and forces on different types of charges.
  • Investigate the concept of antiparallel vectors in the context of electric forces.
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Students studying electromagnetism, physics educators, and anyone looking to deepen their understanding of electric fields and forces in electrostatics.

DeadFishFactory
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Homework Statement


14P022_PosPattern.jpg



The source charge (the magenta circle at the origin) is a positive charge. A particle whose charge is -6e-09 C is placed at location D.

1. The electric field at location D has the value < -8000, 8000, 0 > N/C. What is the unit vector in the direction of E at this location?

2. What is the electric force on the -6e-09 C charge?

3. What is the unit vector in the direction of this electric force?

Homework Equations


E = kq/r^2


The Attempt at a Solution



1. E = sqrt ( -8000^2 + 8000^2 + 0^2) = 11300
tan^-1(-8000/8000) = 135 degrees

11300=(8.99E9)(6E-9) / r^2
r=0.07 m
r(x)=0.07cos(135) = -0.05 m
r(y)=0.07sin(135) = 0.05 m

This is wrong. Can anyone show me how to do it? I assume you can't do #2 without knowing the value of #1, and you can't do #3 without knowing the value of #2? What exactly is question #2 and #3 asking?
 
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DeadFishFactory said:
11300=(8.99E9)(6E-9) / r^2

Here's the problem. You were trying to use the definition of the electric field to calculate the distance D must be from the origin, but you used the value of the test charge instead of the value of the source charge.

Hint: The value of the source charge is not given because you don't need it. You can figure out what the unit vector is without knowing how far D is from the origin. All you have to do is calculate E/|E|.

Hint 2: For part three, there is a very simple relationship between the direction of the electric field and the direction of the force on the test charge. Their directions are either the same (in which case the unit vectors are the same), or they are exactly opposite (in which case the unit vectors are antiparallel.) Which one is it in this situation?
 

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