# Why is the electric field intensity zero at only one point?

Suppose you have a Cartesian coordinate system with:

a 2uC charge at (0,0) and a -8uC charge at (4,0), why is there only one point in which the EFI is zero (-4,0)? Shouldn't that be the first point where the EFI is zero (so (-5,0),(-6,0)... all have Electric Fields of zero)

The reason I think this is because if you drew the Electric field lines from one charge to the other, they wouldn't go either way infinitely.

Thanks

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mjsd
Homework Helper
Electric field strength is zero only at (-4,0) because field strength are given by (up to a multiplicative constant)
$$\displaymath{\frac{q}{r^2}}$$
so at (-4,0)
you have
$$\displaymath{\frac{2}{4^2}+\frac{-8}{(4+4)^2}= \frac{2}{16}-\frac{8}{64}=0}$$
whereas at (-5,0) you get
$$\displaymath{\frac{2}{5^2}+\frac{-8}{(5+4)^2}= \frac{2}{25}-\frac{8}{81} \neq 0}$$

I know that, but can you explain the field itself, since drawing the electric field lines doesn't seem to explain this

mjsd
Homework Helper
it is very hard to draw field lines to explain this, you will need a computer to help to do this properly. mind you an E-field is a vector field and at each spacetime point it has a magnitude and direction, so strictly speaking, to draw this properly you need to insert at each point a little vector, do this for each charge separately and you will find that only at (-4,0) will the magnitude be the same AND direction is opposite (ie cancel after doing the vector sum)