Electric Fields, Finding the zero field point

[Oblivion77]

Homework Statement

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.

Homework Equations

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

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}

 

[gabbagabbahey]

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}

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.

 

[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?

 

[gabbagabbahey]

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?

Yes you can do that.

Have you not learned the quadratic equation yet?!

 

[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.

 

[Hurkyl]

Yea this is what it came out to, which is wrong unfortunately.

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?

 

[Oblivion77]

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?

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

 

[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.

 

[Oblivion77]

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

 

[Hurkyl]

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?

 

[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

 

[Hurkyl]

I assumed using the "-" would be somewhere left of q1

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?