Electric field finding third charge

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Homework Help Overview

The problem involves determining the position of a third point charge in an electric field created by two other charges, with the goal of achieving a net electric field of zero at a specific point. The charges are +5c and -6c, located at different positions along a line.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the vector nature of electric fields and the need to consider the directions of forces exerted by the charges. There is confusion regarding the correct formulation of the equations governing the electric fields and the implications of signs in the calculations.

Discussion Status

Some participants have provided guidance on how to approach the problem, emphasizing the importance of vector addition of electric fields. There is ongoing clarification about the roles of the charges and how to set up the equations correctly. Multiple interpretations of the charge interactions are being explored.

Contextual Notes

Participants are grappling with the implications of charge signs and their effects on the electric field, as well as the mathematical treatment of these signs in the context of the problem. There is mention of a specific position where the electric field is to be zero, which may influence the setup of the equations.

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Homework Statement


a point charge of +5c is located at x=-2cm, and a second point charge of -6c is located at x=1cm
where should a thid charge of +4c be placed so that electric field at x=0cm is zero?


Homework Equations



E(total)=E1+E2+E3

E=k*q/(r^2)

The Attempt at a Solution



E(total)=0
E3=-E1-E2

what I did first is divide the k out so now the equation will look like this
q3/(r3^2)=-(q1/(r1^2))-(q2/r2^2))
and then solve for r3 and put a negative sign because the force is going to the right to the electric field at x=0
i got r3=-.917cm
It is saying I am getting this wrong can anyone tell me what i am doing rong
 
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You forgot that electric field adds vectorially, not arithmetically. Think about the configuration: if you put a positive test charge at x=0, both the +5c charge and the -6c charge would push it to the right. The third charge's field has to counter the combined effect of these two charges. Try writing an equation with only the magnitudes of the electric fields; forget about +/- signs.
 
Well the first charge is pushing to the right and the second charge is pushing to the left thus making first plus and second negative because of their displacement and the position of the eletric field at x=0
but if we do forget about the +/- then this cannot be solve
because the E3=-E1-E2
since we have negavtives
R^2 is in E3 we have to square root negatives number...if you see where i am getting at.
thus we need +5 and -6
 
No, both charges push to the right: the one on the left repels while the one of the right attracts, so their effects combine.

Also, when I said we should ignore all +/- signs, I meant all of them. So don't use E3=-E1-E2; use E3=E1+E2. Don't use q=-3; use q=3.
 
ok i got the answer correct but i am not understand when you mean by both charges are pushing to the right. This i what i see as.

q1=5c----->E=0<------q2=-6c

to me this diagram makes sense because q1 is positive and q2 is negative
and the other one that doesn't make sense to me is that
how does E3=E1+E2
where the standard equation was E1+E2+E3=0
and when you move E1 and E2 to the right they become negative.
E3=-E1-E2
thanks for your help though
 

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