# Homework Help: Point Charges and electric field

1. Jan 26, 2008

### marissag

1. The problem statement, all variables and given/known data
A positive point charge Q1 = 1.9 x 10^-5C is fixed at the origin of coordinates, and aa negative charge Q2 = -6.5 x 10^-6C is fixed to the axis at x = +2.0m. Find the location of the place(s) along the x axis where the electric field due to these two charges is zero. IF there is only one location then enter zero for the second location.
x = ______m
x = ______m

3. The attempt at a solution

for the distance I used Q1 = x and Q2 = abs val of 2-x and then used the equation of Q1/r1^2 = Q2/r2^2

so I started out with ::::
first (1.9 x 10^-5C)/x^2 = (-6.5 x 10^-6C)/((2-x)^2)
then (1.9 x 10^-5C)/x^2 = (-6.5 x 10^-6C)/(x^2-4x+4)
then multiplied both denominators to get rid of them and got:::
(1.9 x 10^-5C)(x^2-4x+4) = (-6.5 x 10^-6C)(x^2)
then multiplied them all out and got:::
(2.55 x 10^-5C)(x2) - (7.6 x 10^-5)(x) + ((7.6 x 10^-5) = 0

at this point I used the quadratic formula and used:::
a = 2.55 x 10^-5
b = 7.5 x 10^-5
c = 7.5 x 10^-5

Now...here is the problem. When I plug these all in to the quadratic formula I end up with a negative number in the square root which you cant take the square root of...

any help is appreciated...anybody to show me where I went wrong is appreciated as well!

2. Jan 26, 2008

### PiratePhysicist

First let's look at Coulomb's law, just because LaTeX is pretty if nothing else:
$$\vec{E}=\frac{1}{4\pi\epsilon_0}\sum{\frac{q_i}{r'_i^2}\hat{r}}$$
Now, I would like to point out that r' here is the seperation between the point in space we are looking at and the charged particle.

Okay, right away because everything is on the x axis we can ignore the vector aspects of this and lets get rid of the summation and look at it that way:

$$E_x=\frac{1}{4\pi\epsilon_0}\left( \frac{q_1}{x_1^2}+\frac{q_2}{x_2^2}}\right)$$
Where $$x_1$$ is the distance between some point and the first charge, and $$x_2$$ is the distance between the point and the second charge. Since that's two variables dependent on the point in space (let's call it $$x_0$$ since E is zero there), lets substitute the point in:
$$E_x=\frac{1}{4\pi\epsilon_0}\left( \frac{q_1}{x_0^2}+\frac{q_2}{(x_0-2)^2}\right)$$
Now set that equal to zero, and solve for $$x_0$$.
Comes out as:
$$x=\frac{2\sqrt{q_1}*(\sqrt{q_1}\pm\sqrt{-q_2})}{q_1+q_2}$$

Which does not have any imaginary numbers.

Last edited: Jan 26, 2008
3. Jan 27, 2008

### marissag

Okay...so I set it up as this says...

(2sqrt 1.9x10^-5)(sqrt 1.9 x 10^-5 * sqrt -(-6.5 x 10^-6)

answer I received is 6.02 x 10^-5 however the answer is showing incorrect in my homework program. I am just totally confused on this one. I guess the way I was doing it at first was completely incorrect...

am I still missing something?

4. Jan 27, 2008

### PiratePhysicist

Well, if that one doesn't work, what does the other possibility say? (remember that's a plus or minus sign in there).

5. Jan 27, 2008

### marissag

okay...totally forgot about the plus and minus signs.

So now I have a for teh first equation (plus) 6.02 x 10^-5
and for the second equation (minus) 1.58 x 10^-5

The distance given for Q2 is on the x axis at 2m. So I am needing to find where the electrc field due to the charges is zero.

6. Jan 27, 2008

### Staff: Mentor

Hint: First figure out which region the answer must be in. (x < 0, 0 < x < 2, x > 2) Then set up the expression for total field.

7. Jan 27, 2008

### marissag

The hint is kinda sorta helping a little...

considering the charges I am thinking that the point is going to have to be 0 < x < 2

as far as the equation I am not really sure what equation I should be using to find the point on this problem once having both charges...

8. Jan 27, 2008

### Biest

If as you stated the charges are of opposite sign (one negative and one positive) the potential may be zero at a point between them, but not the electric field. Remember what the direction of the electric field originating from the charges, one is going radial outward and one is radial inward (in 2D).

Look at this thread:

https://www.physicsforums.com/showthread.php?t=153273

9. Jan 27, 2008

### marissag

That post was full of drama is really all that i noticed...as far as this problem I will just go to the physics help lab tomorrow...I dont understand it and at this point all I am doing is getting more and more frustrated. I dont know the next step...I have beeen working on this for two days and finally posted to ask for help last night...I dont know what else to do.

10. Jan 27, 2008

### Biest

11. Jan 27, 2008

### Staff: Mentor

The way to see that this could not be the correct region is to figure out which way the field from each charge points. Q1, a positive charge, creates a field pointing outwards--thus for x > 0 it points to the right. Q2, a negative charge, creates a field pointing inwards--thus for x < 2 it also points to the right. So there's no way that the two fields can cancel in the region 0 < x < 2.

So it's got to be in one of the other two regions. You can try to reason it out--by considering the relative size of the charges. Or you can just set up the field equation for each region (one at a time, of course) and see if you get a sensible answer.

The field from each charge has a magnitude of kq/r^2, where q is the magnitude of the charge and r is the distance from the charge. You will assign the direction depending on the region you are examining. For example, for x > 0, the field from Q1 is positive (since it points to the right); but for x < 0, the field from Q1 is negative.

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