Finding the value of the element dq

  • Thread starter Thread starter warrior_1
  • Start date Start date
  • Tags Tags
    Element Value
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 4K views
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
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