Electrostatic potential energy of a non-uniformly charged sphere

[Anonymous243]

Homework Statement

Consider a cloud of electros in a three-dimensional space. The cloud has a spherical form of radius R₀ and its particle density distribution is given by ρ(r)=ρ₀(1-r/R₀) for 0<r<R₀, where r is the radial distance from the center of the cloud. Calculate the electrostatic potential energy of the cloud.

Relevant Equations

Gauss's law and electrostatic potential energy equation

Hi, I'm new here, so I don't know how to write mathematical equations, and I may not be fully aware of the rules here, so I'm sorry if I made a mistake.
I know how to calculate the electrostatic potential energy of a countable number of charged particles, but I don't know how to calculate the electrostatic potential energy of a continuous charge distribution. Can anybody help me with this?

 

[PeroK]

For typing mathematics, try the Latex guide:

https://www.physicsforums.com/help/latexhelp/

For your problem, you'll need to integrate the charge density function.

 

[haruspex]

For your problem, you'll need to integrate the charge density function.

Well, that will give the total charge.
@Anonymous243 , can you find the potential at any point?

 

[Anonymous243]

Well, that will give the total charge.
@Anonymous243 , can you find the potential at any point?

Yes!
First, we have to get the function of the electric field.
Let's say that -e is the charge of an electron. Then,
$$dQ=-4e\pi r^2 \rho_0 \left(1-\frac r R_0 \right) dr.$$
If we think of a spherical gaussian surface with radius r (0<r<R₀),
$$\varepsilon_0 E(r) 4\pi r^2 = \int_0^r -4e \pi r^2 \rho_0 \left(1- \frac r R_0\right)dr.$$
Then you get
$$E(r)=-\frac {e\rho_0 r \left(4R_0-3r \right)} {12R_0 \varepsilon_0} (0<r<R_0)$$
if r≥R₀, then
$$\varepsilon_0 E(r) 4\pi r^2 = \int_0^{R_0} -4e \pi r^2 \rho_0 \left(1- \frac r R_0\right)dr.$$
Then you also get
$$E(r)= -\frac {e\rho_0 R_0^3} {12\varepsilon_0r^2} (r \geq R_0) $$
Now, if we integrate the electric field, we can also calculate the electric potential.
$$V(r)=-\frac {e \rho_0 \left( 2R_0^3-2R_0r^2+r^3 \right)} {12 \varepsilon_0 R_0}$$ when 0<r<R₀, and
$$V(r)=-\frac {e\rho_0 R_0^3} {12\varepsilon_0 r}$$ when r≥R₀.

But I have no idea how to calculate the electrostatic potential energy with this V(r)..

 

[haruspex]

But I have no idea how to calculate the electrostatic potential energy with this V(r)..

Consider the outermost shell. If the charge there were dispersed to infinity, what would be its change in potential energy?

 

[Anonymous243]

Consider the outermost shell. If the charge there were dispersed to infinity, what would be its change in potential energy?

Well, I'm not so sure..

 

[PeroK]

Well, I'm not so sure..

Why not consider the cloud when partially formed, with some radius $r$, and calculate the energy needed to bring the next infinitesimal shell of charge from infinity?

You've already done the integration needed to calculate the potential for a partially formed cloud.

Then you could integrate the energy needed for shells from $r = 0$ to $r = R_0$.

 

[ergospherical]

Integrating $\dfrac{1}{2} e\rho(r) V(r)$ over all space (e.g. with respect to the measure $r^2 dr d\Omega$) would also work.

 

[Anonymous243]

Why not consider the cloud when partially formed, with some radius $r$, and calculate the energy needed to bring the next infinitesimal shell of charge from infinity?

You've already done the integration needed to calculate the potential for a partially formed cloud.

Then you could integrate the energy needed for shells from $r = 0$ to $r = R_0$.

So do I have to calculate the charge $$Q(r)=-e \int_0^r 4\pi r^2 \rho(r)dr,$$ which is the the charge of the cloud when its radius is $r$ and then calculate the electric field $E(s) (s>r)$ using Gauss's law like this: $$E(s)= \frac {Q(r)} {4\pi \varepsilon_0 s^2}?$$
Then I can derive $$dU=-\int_\infty^r E(s)ds \cdot dQ$$ and $$dQ=-e4\pi r^2 \rho(r)dr$$...
Whoa it's very complex..! Am I doing it correctly so far?

 

[Anonymous243]

Integrating $\dfrac{1}{2} e\rho(r) V(r)$ over all space (e.g. with respect to the measure $r^2 dr d\Omega$) would also work.

How does it work? I mean, I don't understand how that equation derives the electrostatic potential energy.. Could you please explain it more in detail?

 

[ergospherical]

The energy density of the electric field is $\dfrac{1}{2} \epsilon_0 E^2$, so the energy of the charge distribution is\begin{align*}
U = \dfrac{1}{2} \epsilon_0 \int_{\mathbf{R}^3} E^2 dv &= -\dfrac{1}{2} \epsilon_0 \int_{\mathbf{R}^3} \mathbf{E} \cdot \nabla V dv \\ \\

&= -\dfrac{1}{2} \epsilon_0 \int_{\mathbf{R}^3} (\nabla \cdot (\mathbf{E}V) - V \nabla \cdot \mathbf{E}) dv \\ \\

&= - \int_{\mathbf{R}^3} \nabla \cdot (\mathbf{E}V) dv + \dfrac{1}{2} \epsilon_0 \int_{\mathbf{R^3}} V \nabla \cdot \mathbf{E} dv
\end{align*}The first term vanishes by the divergence theorem ($\mathbf{E}$ goes to zero at infinity), whilst in the second term you put $\nabla \cdot \mathbf{E} = \dfrac{e\rho}{\epsilon_0}$ from Gauss' law. Here the integration can be done most easily in spherical coordinates with $dv = r^2 dr d\Omega$!

 

[PeroK]

So do I have to calculate the charge $$Q(r)=-e \int_0^r 4\pi r^2 \rho(r)dr,$$ which is the the charge of the cloud when its radius is $r$ and then calculate the electric field $E(s) (s>r)$ using Gauss's law like this: $$E(s)= \frac {Q(r)} {4\pi \varepsilon_0 s^2}?$$
Then I can derive $$dU=-\int_\infty^r E(s)ds \cdot dQ$$ and $$dQ=-e4\pi r^2 \rho(r)dr$$...
Whoa it's very complex..! Am I doing it correctly so far?

Yes, it is going to be complicated. If you have not previously done so, I would work the problem to get the potential energy of a uniformly charged sphere. This is a more complicated problem than that.

Both methods I can see require you to calculate the potential for a sphere of radius $R < R_0$. And perhaps the simplest way to do this is to calculate the electric field first.

Then, you either use the idea of @ergospherical and use that formula. Or, imagine building up the sphere shell by shell using the potential for $R < R_0$.

 

[bob012345]

Yes, it is going to be complicated. If you have not previously done so, I would work the problem to get the potential energy of a uniformly charged sphere. This is a more complicated problem than that.

Both methods I can see require you to calculate the potential for a sphere of radius $R < R_0$. And perhaps the simplest way to do this is to calculate the electric field first.

Then, you either use the idea of @ergospherical and use that formula. Or, imagine building up the sphere shell by shell using the potential for $R < R_0$.

But the integration is zero for $r>R_0$ isn't it because the charge density is zero?

 

[PeroK]

But the integration is zero for $r>R_0$ isn't it because the charge density is zero?

Yes, of course.

 

[PeroK]

@Anonymous243 I get the same answer both ways. Just to clarify:

Method 1: Calculate the potential for a partially built sphere of radius $R$ with the given charge density. Then, imagine bringing in an infinitesimal shell from infinity and calculate the differential energy to do that. Integrate this differential energy from $0$ to $R_0$.

Method 2: Calculate the electric field for the sphere (inside and outside the sphere). Using this, calculate the potential inside the sphere. Then integrate $\frac 1 2 \rho(r)V(r)$ over the sphere.

The first method is simpler, but I did it both ways to check I got the same answer.

 

[kuruman]

@Anonymous243 I get the same answer both ways. Just to clarify:

Method 1: Calculate the potential for a partially built sphere of radius $R$ with the given charge density. Then, imagine bringing in an infinitesimal shell from infinity and calculate the differential energy to do that. Integrate this differential energy from $0$ to $R_0$.

Method 2: Calculate the electric field for the sphere (inside and outside the sphere). Using this, calculate the potential inside the sphere. Then integrate $\frac 1 2 \rho(r)V(r)$ over the sphere.

The first method is simpler, but I did it both ways to check I got the same answer.

I used a different (maybe) method from these two straight out of my old E&M textbook (Reitz, Milford and Christy.)
$$U=\frac{1}{2}\int_V \rho(r) ~\varphi(r)~dV~~\text{where } \varphi(r)=\frac{1}{4\pi \epsilon_0} \int_0^r\frac{\rho(r')}{r}dV'+\frac{1}{4\pi \epsilon_0} \int_r^{R_0}\frac{\rho(r')}{r'}dV'.$$It's similar to method 2 without the intermediate step of finding the electric field and gives the same answer.

 

[bob012345]

I used a different (maybe) method from these two straight out of my old E&M textbook (Reitz, Milford and Christy.)
$$U=\frac{1}{2}\int_V \rho(r) ~\varphi(r)~dV~~\text{where } \varphi(r)=\frac{1}{4\pi \epsilon_0} \int_0^r\frac{\rho(r')}{r}dV'+\frac{1}{4\pi \epsilon_0} \int_r^{R_0}\frac{\rho(r')}{r'}dV'.$$It's similar to method 2 without the intermediate step of finding the electric field and gives the same answer.

Why is the integral split up and what happened to the potential terms?

 

[kuruman]

Why is the integral split up and what happened to the potential terms?

Second question first.

Think of assembling $N$ discrete charges, in which case the electrostatic potential energy is $$U=\frac{1}{2}\sum_{i\neq j}^N\frac{kq_iq_j}{r_{ij}}$$The potential energy of the $i$th charge is $U_i=q_i \varphi(r_i)$ where $\varphi(r_i)$ is the electric potential at that charge. When the discrete distribution goes to the continuous case $q_i$ is replaced with $\rho(r')dV'$ where primed coordinates denote points inside the distribution and the summation sign is replaced with an integral sign.

To find the potential $\varphi(r)$ inside the distribution, one uses $$\varphi(r)=\int_V\frac{k \rho(r')}{|\vec r-\vec r'|}dV'.$$Now for the first question.
One needs to be careful here because the point of interest is inside the distribution ($r<R_0$). As $r'$ varies from 0 to $R_0$, there are two distinct cases:
Case I, $r'<r$
In this case, the potential at $r$ is the sum of shells that contribute a $d\varphi = kdq/r$ potential which is the first of the two integrals. The limits of this integral are from $0$ to $r$ because that is the range in which the point of interest is outside the shells that are being added.
Case II, $r'>r$
In this case, a shell of radius $r'$ contributes a potential that goes as the inverse of the shell radius, $d\varphi=kdq/r'$. The limits of this integral are from $r$ to $R_0$ because that is the range in which the point of interest is inside the shells that are being added.

 

[TSny]

Why is the integral split up and what happened to the potential terms?

In the expression $$U = \frac{1}{2}\int_V \rho(r) ~\varphi(r)~dV$$ the integral is not being split up.

But, the potential function $\varphi(r)$ that occurs in the integral for $U$ can be written as the sum of two integrals: $$\varphi(r)=\frac{1}{4\pi \epsilon_0} \int_0^r\frac{\rho(r')}{r}dV'+\frac{1}{4\pi \epsilon_0} \int_r^{R_0}\frac{\rho(r')}{r'}dV'$$ The first integral here gives the contribution to $\varphi(r)$ from the charge that lies closer to the origin than $r$ while the second integral gives the contribution to $\varphi(r)$ from the charge that lies farther from the origin than $r$.

 

[ergospherical]

I used a different (maybe) method from these two straight out of my old E&M textbook (Reitz, Milford and Christy.)
$$U=\frac{1}{2}\int_V \rho(r) ~\varphi(r)~dV~~\text{where } \varphi(r)=\frac{1}{4\pi \epsilon_0} \int_0^r\frac{\rho(r')}{r}dV'+\frac{1}{4\pi \epsilon_0} \int_r^{R_0}\frac{\rho(r')}{r'}dV'.$$It's similar to method 2 without the intermediate step of finding the electric field and gives the same answer.

The way I thought about this formula was as follows. Consider the charge distribution as the superposition of two separate charge distributions: a sphere of those charges at radii smaller than $r$ [system 1: S1], and a shell of those charges at radii larger than $r$ [system 2: S2].

The potential at radius $r$ is the sum of those due to system 1 and system 2 independently.

For system 1, the potential at $r$ is simply that as if all the charge were a point at the centre, i.e. $\phi_1 = \dfrac{Q_1}{4\pi \epsilon_0 r}$. Note that $Q_1 = \int_{\mathrm{S1}} \rho(r') dV'$.

For system 2, the potential at $r$ is the same as the potential at the centre (there is no field inside a hollow shell). The potential at the centre due to a thin shell element $dV' = 4\pi^2 r' dr'$ is nothing but $\dfrac{\rho(r')}{4\pi \epsilon_0 r'}dV'$, and integrating this up from $r$ to $R_0$ gives the total potential at the centre (and thus at $r$) due to S2.

 

[PeroK]

And, of course, another option is to calculate the electric field everywhere and use:

The energy density of the electric field is $\dfrac{1}{2} \epsilon_0 E^2$, so the energy of the charge distribution is\begin{align*}
U = \dfrac{1}{2} \epsilon_0 \int_{\mathbf{R}^3} E^2 dv
\end{align*}

 

[bob012345]

In the expression $$U = \frac{1}{2}\int_V \rho(r) ~\varphi(r)~dV$$ the integral is not being split up.

But, the potential function $\varphi(r)$ that occurs in the integral for $U$ can be written as the sum of two integrals: $$\varphi(r)=\frac{1}{4\pi \epsilon_0} \int_0^r\frac{\rho(r')}{r}dV'+\frac{1}{4\pi \epsilon_0} \int_r^{R_0}\frac{\rho(r')}{r'}dV'$$ The first integral here gives the contribution to $\varphi(r)$ from the charge that lies closer to the origin than $r$ while the second integral gives the contribution to $\varphi(r)$ from the charge that lies farther from the origin than $r$.

It seems to me this will involve a double integral. In post #4 the OP calculated the potential everwhere so that would involve a single integral. This method assumed the potential is not yet known, just the charge distribution.

 

[ergospherical]

It seems to me this will involve a double integral. In post #4 the OP calculated the potential everwhere so that would involve a single integral. This method assumed the potential is not yet known, just the charge distribution.

It's a triple integral over a volume; by the notation $\displaystyle{\int_{r_1}^{r_2} dV'}$ @kuruman means over the shell with inner and outer radii $r_1$ and $r_2$.

 

[bob012345]

It's a triple integral over a volume; by the notation $\displaystyle{\int_{r_1}^{r_2} dV'}$ @kuruman means over the shell with inner and outer radii $r_1$ and $r_2$.

Then that makes it as messy as some quantum overlap integrals I did earlier this year. Very messy. Seems there is no need anyway since the OP already computed the potential.

 

[ergospherical]

Then that makes it as messy as some quantum overlap integrals I did earlier this year. Very messy. Seems there is no need anyway since the OP already computed the potential.

Well not particularly because you have spherical symmetry. In particular you can choose a volume element $dv = r^2 dr d\Omega$, and because all quantities depend only on $r$ the angular part $\int d\Omega = 4\pi$ separates out and you're left with integrals over $r$ only. You can equivalently think about it in terms of shells $dv = 4\pi r^2 dr$.

 

[bob012345]

Well not particularly because you have spherical symmetry. In particular you can choose a volume element $dv = r^2 dr d\Omega$, and because all quantities depend only on $r$ the angular part $\int d\Omega = 4\pi$ separates out and you're left with integrals over $r$ only. You can equivalently think about it in terms of shells $dV' = 4\pi r^2 dr'$.

Same for low order overlap integrals. But I don't see the attraction of doing this the hard way?

 

[kuruman]

It seems to me this will involve a double integral. In post #4 the OP calculated the potential everwhere so that would involve a single integral. This method assumed the potential is not yet known, just the charge distribution.

$$\frac{1}{4\pi \epsilon_0} \int_0^r\frac{\rho(r')}{r}dV'=\frac{e}{4\pi \epsilon_0} \int_0^r\frac{\rho_0\left(1-\frac{r'}{R_0}\right)}{r}4\pi r'^2dr'.$$
$$\frac{1}{4\pi \epsilon_0} \int_r^{R_0}\frac{\rho(r')}{r'}dV'=\frac{e}{4\pi \epsilon_0} \int_0^r\frac{\rho_0\left(1-\frac{r'}{R_0}\right)}{r'}4\pi r'^2dr'.$$As has already been stated, the nominally triple volume integral reduces to a single integral. Because of the spherical symmetry, the volume element becomes the volume of a shell of radius $r'$ and thickness $dr'$, hence $dV'=4\pi r'^2dr'.$

 

[bob012345]

$$\frac{1}{4\pi \epsilon_0} \int_0^r\frac{\rho(r')}{r}dV'=\frac{e}{4\pi \epsilon_0} \int_0^r\frac{\rho_0\left(1-\frac{r'}{R_0}\right)}{r}4\pi r'^2dr'.$$
$$\frac{1}{4\pi \epsilon_0} \int_r^{R_0}\frac{\rho(r')}{r'}dV'=\frac{e}{4\pi \epsilon_0} \int_0^r\frac{\rho_0\left(1-\frac{r'}{R_0}\right)}{r'}4\pi r'^2dr'.$$As has already been stated, the nominally triple volume integral reduces to a single integral. Because of the spherical symmetry, the volume element becomes the volume of a shell of radius $r'$ and thickness $dr'$, hence $dV'=4\pi r'^2dr'.$

I wasn't referring to the dimensions of the volume but the fact that you integrate over both $r,'r$

 

[kuruman]

I wasn't referring to the dimensions of the volume but the fact that you integrate over both $r,'r$

Sorry, I misunderstood. Yes, in that sense it is a double integral.