# Electric force

1. Sep 6, 2014

### derek181

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

Particles 1 and 2 are fixed in place on an x-axis at a separation of L=8cm. Their charges are q1=e q2=-27e. Particle 3 with charge q3=4e is to be placed on the line between particles 1 and 2, so that they produce a net electrostatic force F3net on it. a) at what coordinate should particle 3 be placed to minimize the magnitude of that force.

2. Relevant equations

F=(1/4piε)qq/r^2

3. The attempt at a solution

Both forces are acting rightward.
So Fnet=(1/4piε)(q1*q3)/x^2+(1/4piε)(q2*q3)/(8-x)^2

This simplifies down to Fnet=e^2/piε(1/x^2+27/(8-x)^2)

Now take the derivative to minimize function

d(Fnet)/dx=(54e^2x^3-2e^2(8-x)^3)/piε(8-x)^3x^3

Set it to zero and my polynomial in the numerator is 8e^2(7x^3-6x^2+48x+128) and there is no min value in between 0 and 8!!!!!!!!!!!!

2. Sep 6, 2014

### BvU

Can you show how you go from 54x^3-2(8-x)^3 to 8(7x^3-6x^2+48x+128) ? they are not the same.

(sorry for the notation, you too should use "Go Advanced" or read the guidelines point 6 for something more legible)....

3. Sep 6, 2014

### derek181

54e2x3-2e2(8-x)3
first step
(8-x)3=512-192x+24x2-x3
Multiply that by -2e2 to get -1024e2+384e2x-48e2x2+2e2x3

Then ad the 54e2x3 to get 56e2x3-482x2+384e2x-1024e2

Then factor out 8e2 so 8e2(7x3-6x2+48x+128)

set that to zero and consequently get rid of the 8e2 leaving you with the polynomial 7x3-6x2+48x+128 in the numerator

4. Sep 7, 2014

### BvU

Check the signs factoring out 8e^2 : why do only 3 out of four signs remain the same ?

Tip: factor out 4e^2/piε (a.k.a. $4e^2\over 4\pi\epsilon_0^2$) right at the beginning. You are doing real work and you'll have less work and less chance of errors. The $e^2$ really blurs the picture, especially in the notation you use (but I suppose thats on PF only )

Then: If you are really stuck (and I'm with you there: it took me a long time to sort things out just as well), you can always do several more things:
1. Make a graph - qualitatively at first. F runs away at x=0 and at x=8, and it definitely isn't infinite all over, so |F| MUST have a minimum.
2. Do some numerical tests, x=1 F≈1.55, x=7 F≈27 so you'll have to end up somewhere near the smaller charge (of course).

5. Sep 7, 2014

### derek181

Ahhhhh, geez. Thanks. It's always the simplest mistakes that get past me. Should have been -128 and I get a root of 2.

6. Sep 7, 2014

### BvU

Happens to everybody. Hope the tips come in useful someday...