Finding the value of the element dq

  • Thread starter Thread starter warrior_1
  • Start date Start date
  • Tags Tags
    Element Value
AI Thread Summary
The discussion focuses on deriving the expression for the element of charge, dq, within an infinitesimally thin spherical shell of radius r. It is established that dq is equal to the product of charge density, ρ(r), and the volume of the shell, leading to the equation dq = ρ(r) * 4π * r² * dr. The reasoning involves recognizing that the volume of the shell can be approximated as 4π * r² * dr when considering an additional infinitesimal layer. Participants clarify that ρ(r) is treated as a constant for this derivation, and the provided equations sufficiently demonstrate the required proof. The explanation emphasizes the geometric interpretation of the spherical shell's volume and charge distribution.
warrior_1
Messages
10
Reaction score
0

Homework Statement


Hi guys, i need a bit of help with the following question. Ok the question states the following, explain why the element of charge, dq, located within an infinitesimally thin spherical shell or radius r is equal to rho(r)*4*pi*(r)^2dr, where dr is the thickness of the shell.

Homework Equations


The Attempt at a Solution


ok i know that total charge is equal to charge density multiplied by volume, which is equal to rho*4/3pi*r^2. Hence if i were to find dq, i would have to differentiate with respect to r, dq/dr and solve for dq... thus dq/dr=rho(r)*dv/dr
where dv/dr=4*pi*r^2, therefore if i solve for dq i should get dq=rho(r)*4*pi*r^2*dr...

ok i have no idea if that was right or not... any help here would be greatly appreciated and also why don't we differentiate rho(r)
 
Last edited:
Physics news on Phys.org
*bump*
 
yeah i kinda need help with a question very similar to this... i have the same basic idea as warror_1 but I am still unsure as to how to explain it
 
Yes, that approach is right. And we do don't differentiate rho because it's a constant, not a function.

As for explaining it, you don't need to; the OP's equations prove what the question asked to prove, so they're perfectly sufficient. For a more intuitive explanation, think of adding an extra layer of thickness dr on top of a sphere of radius r. If you flatten out the sphere onto a map, it would have an area of 4pi*r^2; with the extra layer, it would be 4pi*(r+dr)^2. If dr is small, there's no difference in their size, so now you have two layers of equal shape and area separated by distance dr. The volume of contained in that is 4pi*r2dr, so the contained charge must be rho*4pi*r2dr
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top