Spherical Shell-Potential Energy, Energy density

[Arman777]

Homework Statement

Homework Equations

The Attempt at a Solution

I fount these

Part(a)its $E=\frac {ρa^3} {3ε_0}$ and
$υ=\frac 1 2ε_0E^2$
Part (b)
$dU=4πr^2drυ$
Part (c)
$U=\int_0^a 4πr^2udr$ but it gives me $U=\frac {-Q^2} {8πε_0a}$

This"-" bothers me.

 

[TSny]

Part(a)its $E=\frac {ρa^3} {3ε_0}$

How is the charge Q distributed for a metallic sphere?
What is $E$ for points inside the sphere?
What is $E$ for points outside the sphere?

[EDIT: Note that your expression for $E$ does not have the correct dimensions for an electric field.]

$U=\int_0^a 4πr^2udr$ but it gives me $U=\frac {-Q^2} {8πε_0a}$

This"-" bothers me.

The "-" sign should bother you. Your integrand is positive (including the dr). So, the integral must be positive. But are you sure you want to integrate from 0 to $a$? If you can answer my questions above, it should help you see what you should use for the range of integration.

 

[Monci]

Someone was faster :P

 

[Arman777]

How is the charge Q distributed for a metallic sphere?

Just in surface

What is EEE for points inside the sphere?

E=0

What is EEE for points outside the sphere?

$E=\frac {ρa^3} {3ε_0r^2}$ where $r≥a$ I know in the upside I forget $r^2$

But are you sure you want to integrate from 0 to aaa?

What else it could be.. ?

The idea of the problem is that you bring the charge from infinity, since at infinity there is potential energy zero, while using a spherical distribution of charge. With this in mind, what should be the integration limits?

I thought about that but it still give me nothing...
I am thinking there should be minus cause I am thinking like bringing charges from $r=0$ to $r=a$ but I should bring charges $r=∞$ to $r=a$ but still there appears "-" sign also in $r=∞$ case what should I use the volume...? Or is it make sense ?
The result of integral will be $U=Constant\int_∞^a\frac {1} {r^2}dr=constant\frac {-1} {r}$ I can't get rid of "-".I think its cause of potential energy case.

 

[Arman777]

Is it $U=Constant\int_a^r\frac {1} {r^2}dr$ where r goes to ∞ ?

 

[TSny]

Just in surface

So, all the charge Q is on the surface. So, there is no volume charge density $\rho$.

E=0

Yes, $E = 0$ inside the sphere. So, what is the energy density inside the sphere? What would you get if you integrated the energy density over the volume of the inside of the sphere ($0<r<a$)?

$E=\frac {ρa^3} {3ε_0r^2}$

There is no volume charge density. For points outside the sphere, try to express E in term of Q and r.

I thought about that but it still give me nothing...
I am thinking there should be minus cause I am thinking like bringing charges from $r=0$ to $r=a$ but I should bring charges $r=∞$ to $r=a$ but still there appears "-" sign also in $r=∞$ case what should I use the volume...? Or is it make sense ?
The result of integral will be $U=Constant\int_∞^a\frac {1} {r^2}dr=constant\frac {-1} {r}$ I can't get rid of "-".I think its cause of potential energy case.

I'm not sure what you are doing here. Follow the outline given in the problem. How would you express the energy contained in a spherical shell of radius r and thickness dr?

 

[Arman777]

Yes, E=0E=0E = 0 inside the sphere. So, what is the energy density inside the sphere? What would you get if you integrated the energy density over the volume of the inside of the sphere (0<r<a0<r<a0

Zero ?
$E=\frac {Q} {4πε_0r^2}$
$dU=4πr^2drυ$

 

[TSny]

Zero ?
$E=\frac {Q} {4πε_0r^2}$
$dU=4πr^2drυ$

Yes Yes Yes

 

[Arman777]

Yes Yes Yes

soo?

 

[TSny]

soo?

What is preventing you from finishing the problem?

 

[Arman777]

What is preventing you from finishing the problem?

Limits of integral

 

[TSny]

Limits of integral

You need to add up the energy in every spherical shell for which $E \neq 0$.

 

[Arman777]

from a to ∞ ?

 

[Arman777]

You need to add up the energy in every spherical shell for which $E \neq 0$.

a to ∞ ?

 

[TSny]

a to ∞ ?

Yes

 

[Arman777]

Yes

İnteresting... ok thanks a lot

 

[TSny]

İnteresting... ok thanks a lot

Yes, it is very interesting. The potential energy stored in the system can be thought of as stored in the field that extends from the surface of the sphere all the way out to infinity!

 

[Arman777]

Yes, it is very interesting. The potential energy stored in the system can be thought of as stored in the field that extends from the surface of the sphere all the way out to infinity!

Thats just amazing...

 

[Arman777]

I used gaussian law and surface charge density etc to find E

 

[TSny]

I used gaussian law and surface charge density etc to find E

OK