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Electromagnetism Qby Monster007
Tags: electromagnetism 
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#1
Nov612, 10:53 PM

P: 3

Hi, studying for exam and checking if i am doing this question correctly ,
Q: Cylinder radius R=1m length L=2m exists in free space with no other charges nearby. Cylinder volume is charged with uniform charge Q= 7C. (a)Find the charge density in Coulombs per cubic metre. (b) charge distribution now changes to p = a.s (s=radial distance from central axis of cylinder) find the total charge. Answer: (a) I think p=Q/V as uniform charge so; p=7/(volume cylinder) = 7/2(pi) (b) total charge =Q = integral(charge dens . volume) hence Q= ∫∫∫(a.r).(2∏r^{2}).r.dr.dθ.dz (where 0<θ<2∏, 0< z < 2, 0<r<1) and doing the math i get : Q = (8∏^{2}.a)/5 There are no answers to the questions in this course for some reason so hard to know if im right or on completely wrong track so any help would be appreciated cheers 


#2
Nov712, 03:40 AM

PF Gold
P: 941

Your integrand, I think, is wrong. You are, I think, intending to integrate ρ dV, in which dV is the volume element. But instead of using the volume element, which is rdθ dr dz, you are using 2πr^{2} times this. You'd have spotted this if you'd considered the UNITS of your integrand.
The extra 2πr^{2} probably arises because you're mixing integration going right back to the basic volume element with integration by cylindrical shell  perfectly o.k in itself, but you can't mix the two methods  you're integrating too many times! 


#3
Nov712, 08:19 AM

P: 1,020

volume element is simply 2πrdrdz,what you have really written there(I integrated over angle because of symmetry)



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