What is the Electric Field at the Origin Due to Two Point Charges?

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Homework Help Overview

The discussion revolves around calculating the electric field at the origin due to two point charges: a +8 µC charge located at (3,5) m and a -2 µC charge at (3,-2) m. Participants are tasked with finding the electric field in x and y components.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the formula E=kq/r^2 for calculating electric fields and the importance of determining the correct direction of the electric field vectors based on the sign of the charges. There are questions about the correct treatment of the components and the potential impact of sign errors on the final answer.

Discussion Status

Some participants have identified potential errors in the calculations, particularly regarding the direction of the electric field vectors and the treatment of the negative charge. There is acknowledgment of differing values for the electric field calculations, and one participant mentions resolving their issue through careful rounding.

Contextual Notes

Participants are working under the constraints of homework rules that require them to show understanding without providing complete solutions. There is an emphasis on clarifying assumptions about vector directions and the mathematical approach used.

negatifzeo
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Homework Statement


A +8·µC charge is located at coordinates (3,5)·m, and a -2·µC charge is located at coordinates (3,-2)·m. Find the electric field at the origin.
E= ? V/m x + ? V/m y

(It asks for the solution in x and y components)

Homework Equations


I was using E=kq/r^2, where E is the electric field vector (Away if positive, towards if negative) K was 9x10^9, q is the value of the charge.


The Attempt at a Solution


First I used pythagorean theorem to determine distance to origin. Then I used this value for r.
Using the formula for the +8 charge I get E=2117.61 and for the -2 charge I got E=1376.63. Then I used the trig functions from the triangles to solve for the components, which I thern added together for the wrong answer. Where have I gone wrong here?
 
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-2 charge, E=1376.63, it should be negative, or pointing into the right so when you add the 2 E vectors you will be subtracting them.

I get the same values for E's, just that negative sign, Hope that's the problem!
 
The angles from the original triangles are used to calculate the components, correct? I was treating the +8 charge as my negative vector as electric fields point outwards from positive charges, right?
 
Yes,

I think if you use the 8 charge as the negative, you will get the wrong answer, maybe just -1*answer wil fix it but I'm not 100% sure on that.

You should find the 2 vector pointing away, then inverse it so it points inward and then find the x-y components and subtract or add the components.

Also for your -2 charge, I got E=1384.62 not 1376.63
 
Hi negatifzeo,

negatifzeo said:

Homework Statement


A +8·µC charge is located at coordinates (3,5)·m, and a -2·µC charge is located at coordinates (3,-2)·m. Find the electric field at the origin.
E= ? V/m x + ? V/m y

(It asks for the solution in x and y components)

Homework Equations


I was using E=kq/r^2, where E is the electric field vector (Away if positive, towards if negative) K was 9x10^9, q is the value of the charge.


The Attempt at a Solution


First I used pythagorean theorem to determine distance to origin. Then I used this value for r.
Using the formula for the +8 charge I get E=2117.61 and for the -2 charge I got E=1376.63. Then I used the trig functions from the triangles to solve for the components, which I thern added together for the wrong answer. Where have I gone wrong here?

What components and wrong answer did you get?
 
I finally got the correct answer, my problem was rounding a bit too much.
 

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