Solving Electric Problem: Q/q & Net Force on Particles

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Homework Help Overview

The discussion revolves around a problem involving electrostatics, specifically the forces acting on charged particles arranged in a square configuration. The charges are defined as q1 = q4 = Q and q2 = q3 = q, with the objective of determining the ratio Q/q under conditions where the net electrostatic force on certain particles is zero.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss applying Coulomb's law to analyze the forces between the charged particles. There are attempts to set up equations based on the forces acting on particles 1 and 4, and questions arise regarding the implications of charge signs and equilibrium conditions. Some participants also explore the dimensions involved, questioning the units of the distance "a."

Discussion Status

The conversation is ongoing, with participants providing insights into the relationships between the charges and the forces involved. There is a mix of interpretations regarding the setup and calculations, with some participants clarifying concepts related to the geometry of the square and the forces acting on the particles.

Contextual Notes

There is mention of potential confusion regarding the units of "a" and its role in the problem setup. Participants are also considering the implications of charge symmetry and equilibrium conditions based on previous discussions from their professor.

KyoPhan
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My semester just started and I'm already struggling.

The diagram is a square with a dot on each corner representing a charged particle. 1 at top left, 2 at top right, 3 at bottom left, and 4 at bottom right. The distance on one side of the square is a
The problem reads In fig. 21-22, four particles form a square. The charges are q1 = q4 = Q and q2 = q3 = q.

(a) What is Q/q if the net electrostatic force on particles 1 and 4 is zero?

(b) Is there any value of q that makes the net electrostatic force on each of the four particles zero? Explain


For (a)
I think I have to apply Coulomb's law and calculate all the forces from each individual particle. Assuming that 1/4 are both negative and 2/3 are positive, I calculated the force upward (by adding the force caused between 1/4 and 1/3) on particle 1. Then I set them equal to 0. I solved for Q in relation with q and got Q = -2q/a . So I calculated Q/q and got -2/a.

For (b)
I want to say that q=Q but one of opposite sign because I remember my professor saying that if one charge is in equilibrium, then the rest are at well. (correct me if I'm using the terms or concept incorrectly). Or is this only when they are semetric because I remember him talking about it when there was a square with an electron on each corner, with a proton in the middle. Do I have the right idea or am I completely lost? Do I have to apply Coulomb's equation in some way?

Thanks you for taking your time, I really appreciate it.
 
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Well, since Q and q are both quantities of charge, in [Coulomb],
is it possible that "a" has no units? Is "a" a distance, in [meter] ?

The horizontal component of Force on 1 must be zero ... so that :
kQq/a^2 = {kqq/(1.414 a)^2}sin 45 => Q/a^2 = .707q/(2a^2) ...
 
A has no units, its just an unknown distance

I'm sorry I'm kinda slow, but where did 1.414a come from?
 
KyoPhan said:
A has no units, its just an unknown distance

I'm sorry I'm kinda slow, but where did 1.414a come from?
It's the square root of two. Since we're dealing with a square, the length of the diagonal is [tex]s\sqrt{2}[/tex], where [tex]s[/tex] is the side length.
 
lightgrav said:
Well, since Q and q are both quantities of charge, in [Coulomb],
is it possible that "a" has no units? Is "a" a distance, in [meter] ?

The horizontal component of Force on 1 must be zero ... so that :
kQq/a^2 = {kqq/(1.414 a)^2}sin 45 => Q/a^2 = .707q/(2a^2) ...

O okay I understand it now, thanks a lot

sorry for late reply btw
 

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