# Were is the electric field zero?

• blackblanx
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.

## Homework Statement

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

Hi blackblanx! (have a mu: µ )
blackblanx said:
… And where should a third charge of magnitude 5 $$\mu$$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$$\widehat{i}$$-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. tiny-tim said:
Hi blackblanx! (have a mu: µ )

The third charge will have to supply a field of minus E. 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$$\mu$$C)

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. ok thanks