Electrostatic field at the square center

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Homework Help Overview

The discussion revolves around calculating the electrostatic field at the center of a square configuration with four charges: +q, -2q, +2q, and -q. Participants are exploring the implications of charge placement and the necessary parameters for the calculation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to compute the total electric field by calculating contributions from each charge and using trigonometric projections. Some participants question the need for additional information, such as the side length of the square and the specific positions of the charges.

Discussion Status

The discussion is ongoing, with participants clarifying the parameters needed for the calculation. One participant has provided the side length and the arrangement of the charges, while another has asked for the formula related to the electric field from a point charge, indicating a search for foundational concepts.

Contextual Notes

There is a noted requirement for the side length of the square and the specific arrangement of the charges to proceed with the calculations. The original poster expresses uncertainty about their reasoning and projections in the calculations.

tom75
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I have just begun studying electrostatic and I'm trying to do this exercize:

We have a square with charges +q , -2q, +2q, -q1)Compute the electrostatic field [tex]\vec{E}[/tex]at the center of the square.

I did this way :

I find [tex]\vec{E_A}=\frac{q}{2 \pi \epsilon_0} \vec{u}[/tex]
[tex]{E_B}=\frac{-q}{ \pi \epsilon_0} \vec{u}[/tex]
[tex]{E_C}=\frac{q}{ \pi \epsilon_0} \vec{u}[/tex]
[tex]{E_D}=\frac{-q}{2 \pi \epsilon_0} \vec{u}[/tex]

Then with projection :

[tex]E_A=\frac{q}{2 \pi \epsilon_0}*cos(45)=\frac{\sqrt{2}q}{4\pi \epsilon_0}[/tex]

[tex]E_B=\frac{-q}{ \pi \epsilon_0}*cos(45)=\frac{-\sqrt{2}q}{2\pi \epsilon_0}[/tex]

[tex]E_C=\frac{q}{2 \pi \epsilon_0}*sin(-45)=\frac{-\sqrt{2}q}{2\pi \epsilon_0}[/tex]

[tex]E_D=\frac{-q}{2 \pi \epsilon_0}*sin(45)=\frac{-\sqrt{2}q}{4\pi \epsilon_0}[/tex]

Finally [tex]E_{total}=\frac{-\sqrt{2}q}{\pi \epsilon_0}[/tex]

Is-it correct ? I'm not sure of my way of reasoning and the projection.

Thank you
 
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tom75 said:
I have just begun studying electrostatic and I'm trying to do this exercize:

We have a square with charges +q , -2q, +2q, -q1)Compute the electrostatic field [tex]\vec{E}[/tex]at the center of the square.
You need to give the length of one of the sides; also the position of the four charges.
 
rude man said:
You need to give the length of one of the sides; also the position of the four charges.

Yes sorry the length of each side is 1 and this is a square with A (upper left), B(upper right) C(lower right) D(lower left) with respectively charges +q,-2q,+2q,-q
 
What's the formula for the E field a distance d from a point source q?
 

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