Electric Field of Three Point Charges

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SUMMARY

The discussion focuses on calculating the resultant electric field at the center of a circle formed by three point charges positioned at angles of 30°, 150°, and 270°. The formula used is \(\vec{E}=\frac{k_{e}q}{r^{2}}\), where \(k_{e}\) is Coulomb's constant. The two positive charges' horizontal components cancel, while their vertical components combine, resulting in the expression \(2\frac{k_{e}q}{r^{2}}[sin(30)+1]\). The negative charge contributes to the downward electric field, confirming that all contributions point downward.

PREREQUISITES
  • Understanding of Coulomb's Law and electric fields
  • Familiarity with trigonometric functions, particularly sine
  • Knowledge of vector addition in physics
  • Basic understanding of point charge configurations
NEXT STEPS
  • Study vector addition of electric fields in multiple charge systems
  • Learn about the superposition principle in electrostatics
  • Explore the implications of charge placement on electric field strength
  • Investigate the effects of varying charge magnitudes on resultant fields
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Students studying electrostatics, physics educators, and anyone interested in understanding electric field interactions among multiple point charges.

crybllrd
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Homework Statement



Three point charges lie along a circle of radius r at angles of 30°, 150°, and 270° as shown in the figure below. Find a symbolic expression for the resultant electric field at the center of the circle.

23-p-026.gif


Homework Equations



\vec{E}=\frac{k_{e}q}{r^{2}}

The Attempt at a Solution



The two positive charges' horizontal components will cancel out, and their vertical components will combine. This is for the two positive charges:

=2\frac{k_{e}q}{r^{2}}sin(30)

The negative charge is all in the vertical direction.

=-2\frac{k_{e}q}{r^{2}}

I pulled the -2 from -2q out front.

Combining them for the total field:

=2\frac{k_{e}q}{r^{2}}sin(30)+-2\frac{k_{e}q}{r^{2}}

=2\frac{k_{e}q}{r^{2}}[sin(30)+1]

How does that look?
 
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crybllrd said:
Combining them for the total field:

=2\frac{k_{e}q}{r^{2}}sin(30)+-2\frac{k_{e}q}{r^{2}}

=2\frac{k_{e}q}{r^{2}}[sin(30)+1)

How does that look?

Well, "+-" identical "+" does not look nice :-p Indicate the direction of the contribution of the separate charges to the electric field. So the two positive charges contribute with a downward electric field and that of the negative charge is also downward.

ehild
 
Last edited:

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