What is the Electric Field at the Origin Due to Two Point Charges on the Y-Axis?

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SUMMARY

The discussion focuses on calculating the electric field at the origin due to two point charges located on the y-axis: one charge of +q at position a and another charge of +2q at position -a. The electric field is determined using the formula E=kq/r², leading to the conclusion that the net electric field at the origin is E=kq/a², directed in the positive y-direction. Participants confirm that the direction of the electric field is influenced by the magnitudes of the charges, with the stronger charge (+2q) dominating the field direction.

PREREQUISITES
  • Understanding of electric fields and point charges
  • Familiarity with Coulomb's Law and the formula E=kq/r²
  • Knowledge of vector addition in physics
  • Basic concepts of electric potential and its relationship to electric fields
NEXT STEPS
  • Study vector addition of electric fields from multiple point charges
  • Learn about electric potential and its calculation using V=kq/r
  • Explore the concept of electric field lines and their representation
  • Review the principles of electrostatics in preparation for AP Physics exams
USEFUL FOR

Students preparing for AP Physics exams, educators teaching electrostatics, and anyone interested in understanding electric fields generated by point charges.

jwl322
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Homework Statement



Two point charges are on the y-axis. One, with charge a of +q, is located at point a. The other, with a charge of +2q is located at point -a. Determine the magnitude and direction of the electric field at the origin. Express answers in terms of q,a, and constants.

Homework Equations



E=kq/r^2

The Attempt at a Solution



Do i just do k(2q)/a^2 - k(q)/a^2 and that's the answer?

K(2q)/a^2 - k(q)/a^2= kq/a^2

this in preparation of the ap exam and i learned this last semester so i just need a little refreshing
 
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Yes, and the direction is the direction a positive test charge would go
 
Yup, that's how I would do it.

Since both are a distance "a" from the origin and we only care about what happens at the origin, you equations makes sense to me.

EDIT: oops, turdferguson beat me to it.
 
the direction is in the -y direction right?

also, if i were to find the electric potential at the origin i would use E=V/d? if i use that, I am not sure what i would use for d

thanks
 
Think about the direction again, all charges are positive

d is just the distance a. A good thing to remember for the AP is that fields are vectors and potentials are scalars. Voltage is also the scalar sum of each kq/r
 
ok so its in the positive y direction because the the electric field points outward from positive charges and since the 2q charge is greater the field at the origin will point in the positive y direction
 

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