Were is the electric field zero?

AI Thread Summary
The net electric field intensity at the origin due to charges Q1 and Q2 is calculated to be E = 3236i - 10069j. To achieve a net electric field of zero at the origin, a third charge of 5 µC must be placed such that it generates an electric field equal in magnitude but opposite in direction to the calculated E. The placement of this charge should be along the line of the electric field vector, at a distance determined by the required field strength. The discussion highlights the need for further calculations to find the exact coordinates for the third charge. Overall, the focus is on balancing the electric fields to achieve equilibrium at the origin.
blackblanx
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Homework Statement



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

Homework Equations


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

The Attempt at a Solution



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

(have a mu: µ :wink:)
blackblanx said:
… And where should a third charge of magnitude 5 \muC 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\widehat{i}-10069\widehat{j} 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:
 
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?
 
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\muC)
 
Forget I realized that does not work any ideas of how to find the coordinates?
 
It'll be on the line in the direction of E, at the distance appropriate to the given strength. :smile:
 
ok thanks
 
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