After 8 years, I have forgotten a thing or two

Printable View

August 31st 2008, 05:56 PM
ktagore

After 8 years, I have forgotten a thing or two

Hey everyone, first post. I found your website when I was looking for a forum regarding mathematics help. I am getting close to graduating with a non science major, but I have to complete my lab hours that I started 8 years ago.

My background is Physics I in 1998 B, and Cal I and Cal II in 1998 and 1999.. B and a C I think...

Anyway I am slowly relearning everything but I have come to a factoring issue that I am needing someone to step through so I can understand the right way.

Since it is early algebra, I think, I am not finding any examples in my Physics or Calculus Texts.

I believe I have the correct beginning formulas, and the book has provided the answer, I am looking for the meat in between.

Any help would be much appreciated.

http://www.jtheb.com/phys.png
August 31st 2008, 06:20 PM
skeeter

if you put a small (+) test charge, +q, at point P, it will be affected by both charges +Q and -Q using the principle of superposition.

charge +Q exerts a force ...

$F_1 = \frac{kQq}{(x+a)^2}$

charge -Q exerts a force in the opposite direction ...

$F_2 = -\frac{kQq}{(x-a)^2}$

the net force is the sum ...

$F_1 + F_2 = \frac{kQq}{(x+a)^2} - \frac{kQq}{(x-a)^2}$

$F_1 + F_2 = kQq \left(\frac{1}{(x+a)^2} - \frac{1}{(x-a)^2}\right)$

$F_1 + F_2 = kQq \left(\frac{(x-a)^2}{(x-a)^2(x+a)^2} - \frac{(x+a)^2}{(x-a)^2(x+a)^2}\right)$

$F_1 + F_2 = kQq \left(\frac{(x-a)^2 - (x+a)^2}{(x-a)^2(x+a)^2} \right)$

$F_1 + F_2 = kQq \left(\frac{(x^2 -2xa + a^2) - (x^2 + 2xa +a^2)}{(x-a)^2(x+a)^2} \right)$

$F_1 + F_2 = kQq \left(\frac{-4xa}{(x-a)^2(x+a)^2} \right)$

$F_1 + F_2 = -kQq \left(\frac{4xa}{(x-a)^2(x+a)^2} \right)$

the electric field at point P is $E_P = \frac{F_1 + F_2}{q}<br />$ ...

$E_P = -kQ \left(\frac{4xa}{(x-a)^2(x+a)^2} \right) = \frac{-4kQxa}{(x^2 - a^2)^2}$

the (-) sign indicates the direction of the electric field (left) as you would expect since the charge -Q is nearer point P than the +Q charge.

hope you can follow the algebra ... I tried hard not to leave out any step.
August 31st 2008, 06:24 PM
ktagore

Perfect....

That is exactly what I was looking for Skeeter... Thanks SO MUCH!
August 31st 2008, 06:40 PM
ktagore

I worked through this step by step, and the only part that I am missing is how $(x-a)^2 (x+a)^2 = (x^2 - a^2)^2$ .

I broke it down to $(x^2 - 2ax +a^2)(x^2 + 2ax +a^2)$ but I am unsure of the next step, or I could be completely wrong at that.
August 31st 2008, 07:00 PM
skeeter

$(x-a)^2(x+a)^2 =$

$(x-a)(x-a)(x+a)(x+a) =$

$[(x-a)(x+a)][(x-a)(x+a)] =$

$(x^2-a^2)(x^2-a^2) = (x^2-a^2)^2$
August 31st 2008, 07:01 PM
ktagore

Guess I owe you a beverage...

Thanks man.

Re-learning is fun!