Question

Work done in assembling a solid charged sphere of uniform charge density

Answer 1

can you draw the picture??

Answer 2

I meant bringing a solid ball of charge together what is the amount of work need too be done.You know after bringing a charge to bring the next one you need to do work right?So like that finally we get a ball of radius r.What is the work need to be done

Answer 3

you mean bringing charges from infinity to from a charged sphere?

Answer 4

Yep work need to be done to make a charged sphere

Answer 5

I haven't worked out this yet ... well let's try together.

Answer 6

Ill put it more clearly,"What is the work needed to be done in assembling a solid ball of charge"

Answer 7

Yeah

Answer 8

I was thinking to use layers of length dr and charge dq and then integrate from 0 to r but havent been able to put it on paper

Answer 9

http://en.wikipedia.org/wiki/Gravitational_binding_energy
Ah ... similar here

Answer 10

Oh Ill see.

Answer 11

Yeah he used charge density of a ball of radius r and added layers such that dq/dr is the same charge density and integrated from 0 to R if said

Answer 12

I think this would work out!! we will just have to replace Density With Surface charge density and Charge Density!!!

Answer 13

Yeah!!

Answer 14

Well first there is a ball of charge q with radius r.Now to make the complete sphere where the charge would be Q and radius R then charge density would be Q/R where now we initially take a small ball of charge with charge density as q/r.Not we add a layer of thickness dr with charge dq and dq/dr=charge density.And it will be potential at the sphere surface multiplied by dq

Answer 15

\[\int\limits_{0}^{R}[1/(4pi \epsilon)q/r].dq\] where dq=dr.charge density.yeah got it thanx

Answer 16

Charge Dentisty \( \rho = Q/V = Q/(4/3 \pi r^3)\) and Surface Charge Density = \( \sigma = Q/A = Q/(4\pi r^2)\)

Answer 17

Why would we need surface charge density?I mean would it be a conducting ball of charge?

Answer 18

Isn't it a non conducting sphere??

Answer 19

Nothing as such is told.We just need to get a ball of charge together.So wouldnt it be the same all around?i mean \[\rho=Q/V=q/v=dq/dr\]

Answer 20

I meant dq/2(pi)r.dr

Answer 21

No ... we cannot do that. Instead we assume that charge density is constant and find Charge interms of Charge Dentisty
\( Q_1 = \rho V\) <--- initial charge in sphere
\( Q_1 = \rho A*dr \) <--- added charge on sphere

Answer 22

But we add charge too right?

Answer 23

Oh yeah it's dq/(4.pi.r^2)dr

Answer 24

yeah ... that's how charged sphere is formed,