# Electric Fields, Finding the zero field point

1. Jan 24, 2009

### Oblivion77

1. The problem statement, all variables and given/known data

Two particles with positive charges q1 and q2 are separated by a distance s. Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges zero.

2. Relevant equations

$$E= \frac{kq}{r^2}$$

3. The attempt at a solution

Assuming x1 is the distance from q1 to the point where the field is 0.

This is what I get, I believe it is correct but I can't seem to solve for x1.

$$\frac{kq1}{x1^2}$$ = $$\frac{kq2}{(s-x1)^2}$$

2. Jan 24, 2009

### gabbagabbahey

Just "cross multiply" and expand:

$$\frac{kq_1}{x_1^2} =\frac{kq_2}{(s-x_1)^2}\implies kq_1(s-x_1)^2=kq_2 x_1^2$$

$$\implies q_1(s-x_1)^2=q_2 x_1^2 \implies q_1 x_1^2-2q_1 s x_1 +q_1 s^2=q_2 x_1^2$$

Then collect all terms on one side of the equations and solve the resulting quadratic.

3. Jan 24, 2009

### Oblivion77

Thanks, but I already got to that part above. How can I start to solve for x1 from that?
can I do this? (q2 - q1)x^2 + 2q1sx - q1s^2 = 0. What would I do from here?

Last edited: Jan 24, 2009
4. Jan 24, 2009

### gabbagabbahey

Yes you can do that.

Have you not learned the quadratic equation yet?!

5. Jan 24, 2009

### Oblivion77

Yea this is what it came out to, which is wrong unfortunately. Is there something I can simplify? I also tried the "+" instead of the "-", no luck.

6. Jan 24, 2009

### Hurkyl

Staff Emeritus
Check your reference: you got the quadratic formula wrong. (Were you using a computer algebra system? It looks like you forgot a pair of parentheses)

Incidentally, in order to try simplifying things, did you think about expanding things first?

7. Jan 24, 2009

### Oblivion77

I was doing it by hand =), I will try and expand and see if anything simplifies. Thanks.

8. Jan 24, 2009

### rl.bhat

Easier method will be to take the square root of the equation you have written and solve for x1. If the charge are same take + sign. If they are opposite take - sign.

9. Jan 24, 2009

### Oblivion77

Ok, after doing it again I got this? Does it look better? Stuff indeed canceled under the square root sign.

10. Jan 24, 2009

### Hurkyl

Staff Emeritus
That looks plausible. (Note things still simplify more!) You can always try substituting back into the original equation.... (which is usually a very good idea: there is rarely a better way to check if you've found a solution to an equation than actually plugging it in)

Incidentally, how did you decide that was the solution that lies on "the line connecting the two charges" (which I assume meant to say 'line segment'), rather than the other solution?

11. Jan 24, 2009

### Oblivion77

It turns out this answer is acceptable but it does simplify more. I chose the + instead of the - in the formula because the distance is somewhere between the 2 charges. I assumed using the "-" would be somewhere left of q1

12. Jan 24, 2009

### Hurkyl

Staff Emeritus
Don't assume, prove! The argument isn't difficult, once you know what the right idea is. But you have to make note of your assumptions, and cover all of the cases... what if q1 > q2?

By the way, I note that your derivation (and your solution) isn't complete; it doesn't cover all possible cases... in particular, what if q1 = q2?

Last edited: Jan 24, 2009