Total energy of a shell charge

[Buffu]

Suppose a spherical shell is kept at the orgin. Its surface charge density is $\sigma$ and radius is $R$.

I think I remember the formula for its electric potential energy is $U = \displaystyle {Q^2 \over 2R}$.

Now I want to derive it. I used $\displaystyle U =\displaystyle\dfrac{1}{8\pi}\int_{\text{Region}} \vec E\cdot \vec E\ \ dv$.

I know that electric field by an sperical shell is $Q\over R^2$.

So I got,

$\displaystyle U =\displaystyle\dfrac{1}{8\pi}\int_{\text{Region}} {Q^2\over R^4}\ \ dv = \displaystyle\dfrac{Q^2 }{8\pi}\int_{\text{Region}} {1 \over R^4 }\ \ dv$

Then I used $v = {4\pi \over 3}R^3$ to get $\displaystyle \left({3v \over 4 \pi}\right)^{1/3} = R$

Back in integral, $U = \displaystyle\dfrac{Q^2 }{8\pi}\int_{\text{Region}} \left({4 \pi \over {3v} }\right)^{4/3}\ \ dv$

For limits $v \to \infty$ I got the desired answer.

Is this correct ? I have some confusions because the proofs I saw on internet involve triple integral and spherical coordinates.

I think I am doing something wrong as why would anybody use spherical coordinates for such a simple proof ?

 

[Arman777]

I didnt quite understand your solution but you can look this
.http://www.feynmanlectures.caltech.edu/II_08.html

 

[Buffu]

I didnt quite understand your solution but you can look this
.http://www.feynmanlectures.caltech.edu/II_08.html

Where ? I am willing to clarify.

 

[Charles Link]

Suppose a spherical shell is kept at the orgin. Its surface charge density is $\sigma$ and radius is $R$.

I think I remember the formula for its electric potential energy is $U = \displaystyle {Q^2 \over 2R}$.

Now I want to derive it. I used $\displaystyle U =\displaystyle\dfrac{1}{8\pi}\int_{\text{Region}} \vec E\cdot \vec E\ \ dv$.

I know that electric field by an sperical shell is $Q\over R^2$.

So I got,

$\displaystyle U =\displaystyle\dfrac{1}{8\pi}\int_{\text{Region}} {Q^2\over R^4}\ \ dv = \displaystyle\dfrac{Q^2 }{8\pi}\int_{\text{Region}} {1 \over R^4 }\ \ dv$

Then I used $v = {4\pi \over 3}R^3$ to get $\displaystyle \left({3v \over 4 \pi}\right)^{1/3} = R$

Back in integral, $U = \displaystyle\dfrac{Q^2 }{8\pi}\int_{\text{Region}} \left({4 \pi \over {3v} }\right)^{4/3}\ \ dv$

For limits $v \to \infty$ I got the desired answer.

Is this correct ? I have some confusions because the proofs I saw on internet involve triple integral and spherical coordinates.

I think I am doing something wrong as why would anybody use spherical coordinates for such a simple proof ?

You could write the integral as (a much simplified version of spherical coordinates) $\int \limits_{R_o}^{+\infty} \frac{1}{r^4} (4 \pi r^2) dr$ and you get the same result. Meanwhile, a (editing: scratch "alternative") proof of the original formula is to add $dQ$ from $Q= 0$ to $Q_o$ with $dU=(\frac{Q}{R_o}) dQ$ so that $U=Q_o^2/(2 R_o)$.

 

[Buffu]

You could write the integral as (a much simplified version of spherical coordinates) $\int \limits_{R_o}^{+\infty} \frac{1}{r^4} (4 \pi r^2) dr$ and you get the same result. Meanwhile, a (editing: scratch "alternative") proof of the original formula is to add $dQ$ from $Q= 0$ to $Q_o$ with $dU=(\frac{Q}{R_o}) dQ$ so that $U=Q_o^2/(2 R_o)$.

I did not understand the integral.

Isnt $\displaystyle \int E\cdot E dv = \iiint E\cdot E dx dy dz$.
How do you get $R$ in here ?

 

[Charles Link]

I did not understand the integral.

Isnt $\displaystyle \int E\cdot E dv = \iiint E\cdot E dx dy dz$.
How do you get $R$ in here ?

I did two different integrals. The first one is one that you did, over $dv$ where you replaced $r^4$ with an expression for $v$. The simpler way to do that is to use the surface area of a sphere is $4 \pi r^2$ so that with spherical symmetry $dv=4 \pi r^2 \, dr$. (Otherwise, in polar coordinates $dv=r^2 sin(\theta) d \theta \, d \phi \, dr$ where $\theta$ integrates from $0$ to $\pi$ and $\phi$ integrates from $0$ to $2 \pi$.) $\\$ The second integral was one where you assemble the charge on a sphere, and the potential energy $dU$ for each $dQ$ that you add is $dU=(Q/R_o)dQ$. The energy required depends on the charge that is already there. The charge $Q$ that is already there determines the potential $V$ at that instant which the charge $dQ$ encounters when it gets added to the shell. (You gave the formula for this one at the very beginning, but without a derivation.)