# Electric field = 0 at what point on x axis?

## Homework Statement

you have a +3C and a -1C charge on the x axis. the +3C charge is at the origin and the -1C is at x=1. At what point on the positive x axis does E=0?

E=kQ/r^2

## The Attempt at a Solution

So i have

E=(-1k/r^2)+(3k/(1+r)^2)

Setting E=0

1k/r^2=3k/(1+r)^2

r= sqrt(0.5) = 0.707

is this correct?

## Answers and Replies

Will your answer be a point in between the two charges, or to the right of each of them?

ehild
Homework Helper

## The Attempt at a Solution

So i have

E=(-1k/r^2)+(3k/(1+r)^2)

What does "r" mean?

Setting E=0

1k/r^2=3k/(1+r)^2

r= sqrt(0.5) = 0.707

is this correct?

No, r= sqrt(0.5) is not correct. How did you get it?

ehild

ah, r=1.707?

ehild
Homework Helper
What is r?

r=the point on the x axis where the E firld is 0

right?

ehild
Homework Helper
You used the expression of the electric field at distance "r" from a point charge. "r" is distance, not a point. What do you substitute for "r" for the 3C and -1C charges?

ehild

my bad. r=distance to the right of the -1C charge. x=1.707 is the point on the x axis where E=0.

right?

HallsofIvy
Homework Helper
r=the point on the x axis where the E firld is 0

right?
No, it isn't. Your r is clearly a number, not a point. I suspect you mean it is the x-coordinate of such a point. In that case, the distance to 0 is r but the distance to 1 is |1- r| not 1+ r.

ehild
Homework Helper
my bad. r=distance to the right of the -1C charge. x=1.707 is the point on the x axis where E=0.

right?

If so, your equation 1k/r^2=3k/(1+r)^2 is correct, but the solution is wrong. Check it.

ehild

WOW i feel STUPID.

r=-0.366 or 1.366

Since x is to the right of the -1C charge, x = 2.366 when E=0.

I hope i am right now.

ehild
Homework Helper
It is correct now. Well done!

ehild

Thanks man!

Now, I'm just curious, but if I wanted to find the spot where V=0, I would have

0=3Q/(1+r) + (-Q/r)

so r=1/2

Does that mean V=0 at x=1.5 or V=0 at both x=1.5 and x=0.5?
I think it is the second one as all potential lines have to form a closed surface.
Am I right?

thank you.

ehild
Homework Helper
It is less confusing if you calculate with the position of the point on the x axis, that is with its x coordinate.
If a point charge Q is at position x=a, its contribution to the potential at x is kQ/|x-a|. Note the absolute value in the denominator!

In this problem, U(x)=3k/|x|-k/|x-1|. U(x)=0 at x=1.5 and x=3/4.

ehild

Yes you are right, it is a lot easier to make sense of the answer if i use the point on the x axis.

Why did you use absolute value?

I used

0 = 3k/x + [(-Q)/(x-1)]

and i get the solution x = 1.5 only

thanks.

ehild
Homework Helper
The term "r" in the denominator of the potential is distance. The distance is positive (or zero). The coordinate can be both positive and negative.
At distance r from a point charge, the potential is the same either at the right hand side of the charge or at the left hand side.
Find the potential of the -1 C charge at x=3/4. What do you get without absolute value and what is the right value of the potential?

ehild