Determine the electric field on the y axis

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SUMMARY

The discussion focuses on calculating the electric field at a point on the y-axis due to two 2.04 µC point charges located symmetrically on the x-axis at x = 1.02 m and x = -1.02 m. The electric field is determined using the formula E = k*q/r², where k is Coulomb's constant (8.99e9 N·m²/C²). The solution involves calculating the distance from each charge to the point on the y-axis and resolving the electric field vectors into their x and y components. The key conclusion is that the y-components of the electric fields from the two charges cancel each other out while the x-components add, leading to a net electric field along the x-axis.

PREREQUISITES
  • Understanding of Coulomb's Law and electric field calculations
  • Knowledge of vector resolution techniques (SOH CAH TOA)
  • Familiarity with trigonometric functions and their applications in physics
  • Basic algebra for combining vector components
NEXT STEPS
  • Review vector resolution techniques in physics
  • Study the effects of charge sign on electric field direction
  • Learn about superposition principle in electric fields
  • Explore advanced topics in electrostatics, such as electric field lines and equipotential surfaces
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This discussion is beneficial for physics students, educators, and anyone studying electrostatics, particularly those working on problems involving electric fields from multiple point charges.

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Homework Statement


Two 2.04 µC point charges are located on the x axis. One is at x = 1.02 m, and the other is at x = -1.02 m. Determine the electric field on the y-axis at y = 0.470 m.


Homework Equations


E = k*q/r^2


The Attempt at a Solution



First I find the distance from one of the charges to the point on the y-axis.
[tex]\sqrt(1.02^2 + 0.47^2) = 1.123m[/tex]

I also find the angle between this line and the y-axis.
[tex]tan^1 \frac{1.02}{0.47} = 65.26[/tex]

Then I found the component of the electric field at the point on the axis which lies along the y-axis.

[tex]E = k_e*\frac{q}{r^2}[/tex]
[tex]E = 8.99e9 * \frac{2.04\mu C}{1.123m}*cos(65.26)[/tex]

As this only accounts for one charge, and the other charge is equal in magnitude and position, I double this number to get the answer.

Howerver, this gives me the wrong answer. Any help would be appreciated.
 
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The E field due to one of the charges (the one at x=1.02) on the y-axis is given by your equation E=k*q/r^2.
if you draw a line from your charge to the point on the y-axis, this E vector points in this direction. If your other charge (at x=-1.02) has opposite sign, then the y components of the E vectors cancel each other out and the x-components add. Else, the x-components cancel and the y-components add.

In either case break the E vector that you found into its x and y components using SOH CAH TOA. Then do the simple algebra depending on the signs of your charges.
 

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