Were is the electric field zero?

In summary, the net electric field intensity at the origin due to two charges with magnitudes Q1= 2\muC and Q2 = -10\muC, located at (3,4) and (6,-8) respectively, is E= 3236\widehat{i}-10069\widehat{j}. To find the coordinates for a third charge of magnitude 5\muC that would result in a net electric field of zero at the origin, the third charge would need to supply a field equal to the negative of E, and would be located on the line in the direction of E at a distance appropriate to the given strength.
  • #1
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Homework Statement



What is the net electric field intensity at the origin due to two charges Q1= 2[tex]\mu[/tex]C at (x=3,y=4) and Q2 = -10[tex]\mu[/tex]C at (x=6,y=-8). And where should a third charge of magnitude 5 [tex]\mu[/tex]C be placed so the net electric field at the origin is zero?

Homework Equations


E=KQ/R[tex]^{2}[/tex] [tex]\widehat{r}[/tex]

The Attempt at a Solution



I have found the first part of the question to be E= 3236[tex]\widehat{i}[/tex]-10069[tex]\widehat{j}[/tex] but I am not sure how you would find out were to place the third charge. Please help
 
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  • #2
Hi blackblanx! :smile:

(have a mu: µ :wink:)
blackblanx said:
… And where should a third charge of magnitude 5 [tex]\mu[/tex]C be placed so the net electric field at the origin is zero?

I have found the first part of the question to be E= 3236[tex]\widehat{i}[/tex]-10069[tex]\widehat{j}[/tex] but I am not sure how you would find out were to place the third charge. Please help

The third charge will have to supply a field of minus E. :wink:
 
  • #3
tiny-tim said:
Hi blackblanx! :smile:

(have a mu: µ :wink:)


The third charge will have to supply a field of minus E. :wink:

but what are the coordinates of where that charge should be placed?
 
  • #4
oh wait never mind i think i got it know thanks it would just be the negative of the electric field vector divided by k(5[tex]\mu[/tex]C)
 
  • #5
Forget I realized that does not work any ideas of how to find the coordinates?
 
  • #6
It'll be on the line in the direction of E, at the distance appropriate to the given strength. :smile:
 
  • #7
ok thanks
 

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