Electrostatics - charges on spheres and work done in moving them

1. Jan 6, 2006

hi,
this is a great forum here and i was wondering if anyone could lend a helping hand , it would be great.

im currently studying at univercity for electronics engineering and have been given some physics questions to work onover the holidays. I ahve spend quite some time on this question. and got most of the way there (i believe).

question exactly as ive been presented it:

1.Consider a sphere of radius R containing a total charge Q, in which the charge is evenly distributed throughout the sphere.

i) what is the charge density p?

ii) calculate the total charge, Q1 within a sphere of radius r, and uniform charge density p, and express this in terms of Q.

iii) calculate the total charge Q2, within a spherical shell of radius r , thichkness dr and uniform charge density p, and express this in terms of Q

Now ,show that the work done in bringing Q2 from infinity to Q1 can be written:-

dW= ((3Q^2r^4)dr)/4 pi epsilon0 R^6

well

ive worked on this question anmd so far ive come up with the following

i)p = Q/(4/3)pi.R^3

ii)Q1=(4/3)pi.r^3.p (substituting the common p for both Q1 and Q)
---> Q1=(Q.r^3)/R^3

iii)
i gather the total charge Q2 on the shell is = p.vshell
where p is common to both Q1 and Q so
Q2=p.(4/3).pi(r^3-(r-dr)^3) i got this from Q2 = charge density x (voloume of sphere radius r - volume of sphere radius (r-dr))

but i just cant work ou the last part of the answer what Dw =???

i also tried using vshell= 4pir^2dr which is surface areax shell depth

so my question is , hve i gone wrong or if not how to i get the answer to part iii in their form?

thanks

2. Jan 6, 2006

StatusX

Can you simplify the expression for Q2? Do you know the formula for the voltage outside Q1? Do you know the formula for the work done?

3. Jan 6, 2006

yes i meant to write:
as Q2=p.(4/3pi(r^3-(r-dr)^3)
--->and p= Q/(4/3)pi.R^3
therefore Q2= (Q(r^3-(r-dr)^3))/R^3
thats the simplified as it was meant to be written in terms of Q.

Im not sure of the formula but work done is = qQ/4pi epsilon0 a
where a would be the distance one charge has moved towards the other.
But the answer on the sheet does not contain a distance as it is not work done it has not yet been integrated, thats where im lost!

the voltage outside Q1, would that be to work out the two potentials?
or is iot because change in potential is equal to the work done?

i.e if the potential has increased by 5j/c (5v) then does that mean 5 joules of work have been done?

im confuised here!

thats for the reply by the way!

4. Jan 6, 2006

StatusX

You need to simplify Q2 (or, maybe more appropriately, dQ) by expanding (r-dr)^3 and discarding all terms containing dr^2 or higher. Then the work done is the change in potential (where V=Q1/4πε0r outside the sphere) times the charge (Q2).

5. Jan 6, 2006

thanks, but one last question, i can discard all terms of dr^2 or higher as they are significantly small, is that correct?

also im confused in the potwential, arnt i moving one charge from infinity towards the other charge?

also how do i get this in the dw form?

6. Jan 6, 2006

StatusX

You can get rid of all the terms dr^2 and higher because, at the end, you'll be taking the limit as dr goes to zero, and so the only important term is dr. You are moving one charge (the infinitessimal one) in from infinity. And just think of dW as the change in W, so it will be related to dr, the change in the radius of the sphere.

7. Jan 6, 2006

thants so far or the excellent help status, however im still a little confused, why will dr go to zero? isnt that the thickness of the shell? why does that alter, am i looking at this in the wrong way, i though it was the shell moving from infinity ( infinitely small charge as it has thickness of dr) towards charge q1 a sphere of radius r.

im just confused why dr will decrease basically.

8. Jan 6, 2006

StatusX

dr is an infinitessimal. It is a shorthand way of expressing relations like this one. An equation like dW= ((3Q^2r^4)dr)/4 pi epsilon0 R^6 isn't really mathematically precise, but is a way of expressing dW/dr=((3Q^2r^4))/4 pi epsilon0 R^6. This is found by:

$$\frac{dW}{dr} = \mbox{lim}_{dr->0} \frac{W(r+dr)-W(r)}{dr}$$

That's where dr is sent to zero.

9. Jan 8, 2006