# Circular cylinder integral question

the question is here:
http://i42.tinypic.com/9j2alv.gif

the charge on one small ring is
$$dq=\frac{Qdx}{h}\\$$
the problem is with the distance squared part
its constantly changes
i dont know how to write in the distance square part
??
$$E=\frac{kQ }{h} \int_{h+d}^{d}d x$$

and it a ring not a small partical
so the normal formula of a field kq/r^2
is not logical here

Related Calculus and Beyond Homework Help News on Phys.org
tiny-tim
Homework Helper
the problem is with the distance squared part
its constantly changes
i dont know how to write in the distance square part
??
Just use Pythagoras

(and the result of Example 3.8)

this is the solution:
http://i40.tinypic.com/2d14t2.gif

regarding part a:
it seems that they use one ring at a time.
but the Pythagorean distance suggests that they use only a small part from that ring.

if they use only a small part of a ring then the whole this should be a double integral
one for the sum of every small part of a ring
the other is for all rings.

but if they use one ring at a time then they should take the distance from the center of a ring to the point,not the Pythagorean distance of a small part from a ring
??

tiny-tim
Homework Helper
… but if they use one ring at a time then they should take the distance from the center of a ring to the point,not the Pythagorean distance of a small part from a ring
??
No … they're using √(x2 + R2),

which is the distance from every part of the ring (in other words, for every value of theta) …

I agree if the ring wasn't circular, then those distances would be different, and you'd need a second integral (over theta) …

but they're the same, so the field is exactly the same as if all the mass of the ring were concentrated in one part

i dont know how to choose a small part of a ring:
first way:
dtheta is a small angle from from the center of a ring.
i say that the whole ring is 2pi so a small part of that ring is dtheta/2pi
$$dq=\frac{Qdx}{ h}\frac{d\theta}{2\pi}$$

the other way is saying that dx is the width and r*dtheta is the length of that small
square.the density is $$\frac{Q}{2\pi r h}$$
so i get
$$dq=\frac{Q}{ h2\pi r}dx Rd\theta$$
what way of choosing a small part is wrong?

Last edited:
tiny-tim
Homework Helper
They're both right!

(I'm assuming you meant r not R in the second one )

If you cancel the r/r, they're the same.

So you can either do it that way, which gives you a ∫∫dr dθ, and the integral doesn't depend on θ, so you can immediately integrate wrt θ and get 2π times the ∫dr, which cancels out the 2π on the bottom

or you can do what the book does, which is to take the whole ring together, which had no 2π on the bottom, so it comes out the same.

in the second one:
i meant R
because the radius is R do its dR
is it ok?

i cant understand the whole ring method:
why they do sqrt(R^2+X^2) if its a whole ring and not a small part of it

tiny-tim
Homework Helper
(hmm … just noticed I kept typing ∫dr when I meant ∫dx in my last post )
in the second one:
i meant R
because the radius is R do its dR
is it ok?
Yup … I got confused between R and r

but then you should have had 2πR on the bottom instead of 2πr, and the R/R still cancels.
i cant understand the whole ring method:
why they do sqrt(R^2+X^2) if its a whole ring and not a small part of it
Perhaps I'd better write it out in full …

∫∫ (Qdx/2πRh)√(x2 + R2) dx Rdθ

= ∫ [∫ (Qdx/2πh)√(x2 + R2) dx] dθ

= 2π[∫ (Qdx/2πh)√(x2 + R2) dx]

= ∫ (Qdx/h)√(x2 + R2) dx,

which as you can see is what the book does …

and it woks because the √(x2 + R2) is the same at all points on the ring

this is my calculations.
i got got a different double integral then you
why??
$$dq=\frac{Q}{ h2\pi R}dx Rd\theta\\$$
$$\cos \alpha=\frac{x}{\sqrt{x^2+R^2}}\\$$
$$k\int_{d}^{h+d}\int_{0}^{2\pi}\frac{dq}{(R^2+x^2)}\cos \alpha= k\int_{d}^{h+d}\int_{0}^{2\pi}\frac{x\frac{Q}{ h2\pi R}dx Rd\theta}{(R^2+x^2)^{\frac{3}{2}}}\\$$

tiny-tim
Homework Helper
oh yes … I forgot about the cosine

i solved the integral but i get an expression which differs alot from the books answer
where is the mistake
?
$$k\int_{d}^{h+d}\int_{0}^{2\pi}\frac{x\frac{Q}{ h2\pi R}dx Rd\theta}{(R^2+x^2)^{\frac{3}{2}}}=k\int_{d}^{h+d}\frac{x\frac{Q}{ h2\pi R}dx R(2\pi-0)}{(R^2+x^2)^{\frac{3}{2}}}=\frac{kQ2\pi}{2\pi h}\int_{d}^{h+d}\frac{xdx}{(R^2+x^2)^{\frac{3}{2}}}\\$$
$$R^2+x^2=t\\$$
$$2xdx=dt\\$$
$$\frac{kQ}{ 2h}\int_{R^2+d^2}^{R^2+h^2+2hd+d^2}\frac{dt}{t^{\frac{3}{2}}}= \frac{kQ}{ 2h}\int_{R^2+d^2}^{R^2+h^2+2hd+d^2}{-2t^{\frac{-1}{2}}}\\$$
the last part is not an integral i just dont know how to write this straight line with intervals on it in LATEX code.
so its like that straight line on the right side.

Last edited:
tiny-tim
Homework Helper
Well, that looks ok so far.

Maybe the ends of the cylinder are supposed to be solid also (it isn't clear from the question)?

What answer does the book give?

tiny-tim
Homework Helper
But that is the same as your solution

(assuming, in your last line, the ∫… is really a […] )

wow i think i solved it i got identical expression.
HIP HIP HOORAY.

ok now in part b:
its a solid cylinder .
i need to take a 3d small part
i divide the cylinder into small plates
the density is $$\frac{Q}{\pi r^2 h}$$
$$dq=\frac{Q}{\pi r^2 h}rd\theta dr$$
but i am not sure regarding the dr part
i know that our radius is temporaty
why i cant write dR
i cant see how the dr part changes from one radius to the other
we can have a constant length that stays the same for every radius and get the same result

correct
?

tiny-tim
Homework Helper
wow i think i solved it i got identical expression.
HIP HIP HOORAY.
ok now in part b:
its a solid cylinder .
i need to take a 3d small part
i divide the cylinder into small plates
the density is $$\frac{Q}{\pi r^2 h}$$
$$dq=\frac{Q}{\pi r^2 h}rd\theta dr$$
but i am not sure regarding the dr part
i know that our radius is temporaty
why i cant write dR
i cant see how the dr part changes from one radius to the other
we can have a constant length that stays the same for every radius and get the same result

correct
?
I'm not really following this

The answer in the book uses the formula for the field at distance x from a solid disc …

why don't you go back to Example 23.9, and see how they got it?

tiny-tim
Homework Helper
23.8 is not about disks at all
http://i44.tinypic.com/2lihmjn.gif
That's strange … part b clearly ends "Use the result of Example 23.9 to find the field it creates at the same point"
regarding part b:
is it ok to write dR instead of dr(we take a small part of radius R,not r)
Yes … in fact, you must, because R is fixed …

you'll get the whole of one disk by integrating over dr from r = 0 to r = R.