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

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Homework Help Overview

The problem involves determining the electric field at the origin due to two point charges located on the y-axis, one with charge +q and the other with charge +2q. The participants are exploring the implications of the charges' positions and magnitudes on the resulting electric field direction and magnitude.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the electric field using the formula E=kq/r^2 and consider the contributions from both charges. There are questions about the direction of the electric field and how to approach finding the electric potential at the origin.

Discussion Status

Some participants have confirmed the approach to calculating the electric field, while others are questioning the direction of the field based on the nature of the charges. There is an ongoing exploration of the relationship between electric fields and potentials, with some guidance provided regarding the distinction between vectors and scalars.

Contextual Notes

Participants are preparing for an AP exam, which may impose specific constraints on how concepts are applied. There is also a discussion about the assumptions related to the behavior of electric fields generated by positive 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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