Find Min Force on Particle 3 with q1,q2,q3 Charges

[derek181]

Homework Statement

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.

Homework Equations

F=(1/4piε)qq/r^2

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!

Please help.

 

[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)...

 

[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

 

[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 that's 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).

 

[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.

 

[BvU]

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