Electric field = 0 at what point on x axis?

[connor02]

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?

Homework Equations

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?

 

[Villyer]

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

 

[ehild]

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

 

[connor02]

ah, r=1.707?

 

[ehild]

What is r?

 

[connor02]

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

right?

 

[ehild]

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

 

[connor02]

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]

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]

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

 

[connor02]

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]

It is correct now. Well done!

ehild

 

[connor02]

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]

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

 

[connor02]

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]

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