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Need help with Continuous Charge Distributions

Blogs > HeavOnEarth
Post a Reply
HeavOnEarth
Profile Blog Joined March 2008
United States7087 Posts
Last Edited: 2009-02-11 09:33:13
February 11 2009 09:00 GMT
#1
In class we're preparing for gauss's Law with
continuous charge distributions like finding the electric field across from a ring and then a disk, using coulomb's law.

and i'm like wut is going on?!

Part 1 )For example, a thin ring shaped object of radius "a" holds a total charge "Q" distributed uniformly around it. Determine field at point P on it's axis a distance "x" from center Let "λ" be the charge per unit length (C/M)
and λ= Q/l

Part 2) Then find E on a uniformly charged disk, of radius R, and total charge Q.

However, im not even close to the disk yet, and thus the questions start!

+ Show Spoiler [answer for ring] +
Ex= k(Q)(x)/(x^2+a^2)^3/2

+ Show Spoiler [answer for disk] +
[image loading]


First, i don't understand what to integrate across, limits of integration , or even how to start it properly ;E= ?
[image loading]

or
[image loading]

??


[image loading]


Those questions aside, this is what i have done so far
(still skeptical actually) and i used the picture above-


+ Show Spoiler +

Actually to be honest this stuff i didn't come up with, i copied it from the book.
But i read it and followed the steps and it makes sense to me, although im not sure i could transcribe this for... another problem. im just not very confident in my physics capabilities.
λ= Q/l = Q/2πa ; l = circumference

dQ = λdl= Q(dl/2πa)

λ= dQ/dl

-and dE = k(dQ/r^2)


And then from here on i get this physics; vector stuff.

dE has a component on the x-axis + y-axis, the y components cancel
dEx = k(dQ/r^2)(cosΘ)

cos(Θ) = x/r

r^2 = x^2 +a^2

dEx= (k)(dQ)(x)/(x^2+a^2)^3/2

plug in -> ∫dEx= ∫k(dQ)(x)/(x^2+a^2)^3/2

and now i'm lost as to where to integrate from, and i dunno how to relate any terms Like Let u = something du= wee and it all works out like in calc but i dont see it o_o''

Thanks in advance for any help, but for now this will serve as my little note taking blog entry ^^'


*
"come korea next time... FXO house... 10 korean, 10 korean"
HeavOnEarth
Profile Blog Joined March 2008
United States7087 Posts
Last Edited: 2009-02-11 09:38:15
February 11 2009 09:07 GMT
#2

plug in -> ∫dEx= ∫k(dQ)(x)/(x^2+a^2)^3/2

and now i'm lost as to where to integrate from, and i dunno how to relate any terms Like Let u = something du= wee and it all works out like in calc but i dont see it o_o''

Oh that's weird. The book just pulls everything out except for dQ and integrates and gets
Ex= k(Q)(x)/(x^2+a^2)^3/2
O_O'''i guess all the terms were just constants? lol
oh boy

So let's assume i knew how to do that and move onto the disk!
[image loading]

Using that image ^
"By symmetry E, on the axis of the disk, is along the axis. A ring of radius "a" and width "da", The area of this ring is dA = 2πada and its charrge is dq = σdA = 2πσada, where σ= Q/πR^2 is the charge per unit area. The field produced by this ring is given by the equation for a ring, if we replace Q with dq = 2πσada.
(from the book)
Therefore,
dEx= k(x)(2πσada)/(x^2+a^2)^3/2
o.o'
blah i dunno how to write limits on the integral thingy D:
"come korea next time... FXO house... 10 korean, 10 korean"
Malongo
Profile Blog Joined November 2005
Chile3479 Posts
February 11 2009 10:30 GMT
#3
I dont understand what makes you lost.

∫dEx= ∫k(dQ)(x)/(x^2+a^2)^3/2 here what are you "adding": small parts of E due to small parts of ring no? so simply dQ= λdl= λdTheta*r (a small part of the ring and since your integrating to all the ring 2pi>theta>0 and the integral is k(x)/(x^2+a^2)^3/2*2pi*r* λ=k(x)/(x^2+a^2)^3/2*Q
Help me! im still improving my English. An eye for an eye makes the whole world blind. M. G.
Malongo
Profile Blog Joined November 2005
Chile3479 Posts
February 11 2009 10:41 GMT
#4
In the disk example you are adding rings.
dEx= k(x)(2πσada)/(x^2+a^2)^3/2 => the question is how do you add all this different rings? simple, the variable for "each" ring is a and you want all the rings from a=0 to a= R those are you integral limits.
Help me! im still improving my English. An eye for an eye makes the whole world blind. M. G.
HeavOnEarth
Profile Blog Joined March 2008
United States7087 Posts
Last Edited: 2009-02-11 11:06:15
February 11 2009 10:56 GMT
#5
I just don't get the whole , cut things into little parts, integrate all of them to get the total.
I mean i can do the calculation i just ... really skeptical, meaning i'll only know how to do a problem, after i've seen it, so basically, completely worthless


Oh that's weird. The book just pulls everything out except for dQ and integrates and gets
Ex= k(Q)(x)/(x^2+a^2)^3/2
O_O'''i guess all the terms were just constants? lol
oh boy

don't get why that happened either, and if they are just constants, i don't get why ( i think it ties into not seeing the "tiny chunks" and taking the integral.

simple, the variable for "each" ring is a and you want all the rings from a=0 to a= R those are you integral limits.

And this, it all makes sense when someone tells me, Oh u integrate from x->y because of xyz.. etc.
But i could never come up with that on my own =[ Oh hey, let's cut it up into rings... use what i got from the last problem... and integrate across the radius!

And, also the really basic part at the beginning throws me off; but i'll just commit coulombs law to memory as well as what Q/V= , Q/A= and Q/L=
and mess around with that.

Strangely enough, i have no problem doing the calculus itself, (finishing finding E for the disk) integrating is fun :D
"come korea next time... FXO house... 10 korean, 10 korean"
Malongo
Profile Blog Joined November 2005
Chile3479 Posts
February 11 2009 11:56 GMT
#6
I get it now. First thing try to change your approach "But i could never come up with that on my own" is not going to help you. Second note that it is always the same procedure to get to the calculus: you divide certain object in small parts to wich you know how do they interact (since you dont know how the whole body interacts). In the case of the ring the division is in puntual parts, and you find the properties of those smaller parts using the continuous distribution: dQ=lambda*dL (L is the circumference). My first post used another way to express the same division dQ=lambda*r*dtheta. As you can check using dL you have to ask what are your limits for L? obviously 0 and 2*pi*r that gives you the whole circumference. In the second case your limits are related to the angle theta limits for a full circumference in terms of the central angle? easy 0 and 2*pi. Either method gives you the same result, just try to avoid the mechanic and actually think about what are you "adding". The disk is also obvious once i asked what are you adding? rings. So how are the rings related to the formulae? by the radii, so you need to cross all the posible rings from the center to the limit of the disk -> R>a>0. Its not too hard if you take the time to check intuitively what are you doing because normally you can check at once what is constant. In the ring you can even say dQ ok you have to "integrate" dQ for all the small pieces. Does dQ change for each "piece" nop so the integral of dQ comes instantly to be Q. (here you forget about limits and think ok im "adding" all the small dQ and they dont change for each piece what is the total? Q. Damn i hope this helped, gl with these try to understand at maximum because the next stage is a little more trickier but basically the same. (The field for an void sphere is done by integrating rings, and a full sphere integrating void spheres ahaha).
Help me! im still improving my English. An eye for an eye makes the whole world blind. M. G.
HeavOnEarth
Profile Blog Joined March 2008
United States7087 Posts
Last Edited: 2009-02-15 20:06:40
February 11 2009 12:04 GMT
#7
OMG SEXY POST But
i have school i'll read you in 8 hours~~~
actually hmm i'll print it out bwahah
;edit;
l0l i got a 50 curved to an 80 on my test.
failure
"come korea next time... FXO house... 10 korean, 10 korean"
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