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

1. Feb 4, 2007

### imranq

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

2. Feb 4, 2007

### mjsd

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
$$\displaymath{\frac{2}{5^2}+\frac{-8}{(5+4)^2}= \frac{2}{25}-\frac{8}{81} \neq 0}$$

3. Feb 4, 2007

### imranq

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

4. Feb 4, 2007

### mjsd

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)

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