This preview shows pages 1–2. Sign up to view the full content.
CAPA #3 solutions.
1)
This problem is basically solved for you as an Example of the text. I HIGHLY recommend
that you work it out for yourself  just taking an example from the book doesn't teach you much.
But the solution is nicely written up, so I won't repeat it here.
So, E(z) = k z Q / (z^2 + R^2)^(3/2).
They gave you ALL these quantities, just plug in.
Remember to convert all distances to meters, and charges to Coulombs, before plugging in!!
2) We want to maximize E(z) , that means set dE/dz = 0.
Taking the derivative of that function
is a little nasty (Remember, k, Q, and R are constants.
I use the
"(bottom times deriv of top  top times deriv of bottom) / bottom^2" rule,
maybe you have your own! You need to use the chain rule, the derivative of
(z^2 + R^2)^(3/2) all by itself is (3/2) (z^2+R^2)^(1/2)*(2z).
This is about the worst derivative I think we'll have to take all semester!? I won't give you ones
quite this nasty on exams (!)
but I *do* expect you to be able to use the chain rule for simple
cases, so this is good practice.
I get k Q
[
(z^2 + R^2)^(3/2)  z*(3/2)*(z^2+R^2)^(1/2)*(2z) ] /
(z^2+R^2)^3]
= 0
Looks like a mess. But it's not so bad, you just have to believe you can clear the dust! There
really is some pleasure to be found in "simplifying" nasty expressions like this, it gives you some
faith in the orderliness and simplicity of the universe :)
Multiply both sides by the denominator (z^2_R^2)^3, it just goes away!!!
Divide by kQ, gone.!
Lastly:
divide through by the common (z^2+R^2)^(1/2).
Here's what's left:
(z^2+R^2)^(1)  z*(3/2)*(2z) = 0,
which means
z^2 + R^2  3 z^2 =0,
or R^2 = 2 z^2.
z = R/Sqrt[2].
That's it. (Don't you love it when nasty algebra simplifies away to almost
nothing?)
If you have trouble with this sort of exercise, my best suggestions are to take a deep
breath, use a nice big clean piece of paper, write neatly, keep yourself organized.
.. It's a great
skill to develop.
3) Recall, from #6, E(z) = k z Q / (z^2 + R^2)^(3/2).
Near the origin, where z very small compared to R, R^2 + z^2 becomes R^2, so E becomes
E = k z Q / (R^2)^(3/2).
You can neglect z in the denominator because when you ADD something big to something
small, the small thing is neglected. You don't neglect it in the numerator, because it's
MULTIPLYING things, it isn't small *compared* to anything.
..
(Think about this! It's an idea
that crops up everywhere in science and engineering, and is worth wrapping your head around)
So now we have E(z) = k z Q / R^3.
Then, the force on an electron is F = q*E = eE
F = ( e *k
Q / R^3) * z
Just as advertised, that's of the form c*z, where c = e*k*Q/R^3.
Plug in your numbers.
(e = 1.6E19 C,
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 10/08/2008 for the course PHYS 1120 taught by Professor Rogers during the Fall '08 term at Colorado.
 Fall '08
 ROGERS
 Physics, Work

Click to edit the document details