# Charge density in sphere that makes constant radial E-field inside

• prosteve037
prosteve037
Homework Statement
What should the charge density ρ(x, y, z) in a sphere be so that there's a constant radial (it's always parallel to the radius) electric field E_0 at every point inside the sphere?
Relevant Equations
Maxwell's equations, specifically Gauss' law for electric fields, maybe some boundary conditions?
I'm having trouble understanding how a charge distribution in a sphere can make this happen.

My instinct is that the fact that it's radially directed is a big hint of something, but I don't know what that hint might be alluding to. If the net E-field is constant inside the sphere and is always directed radially outward, as one moves in any direction from the center, wouldn't the charges have to move/redistribute (as in the case for a conductor) in order for the E-field to remain the same value?

I'm not sure if any of the other Maxwell's equations gives any hints, but I don't think so.

Newton's shell theorem is a good place to start.

MatinSAR and topsquark
No movement or redistribution necessary. Assume such a fixed charge distribution exists but you don’t know the functional form. What can you write down about the electric field as a function of radius?

topsquark
jbriggs444 said:
Newton's shell theorem is a good place to start.
So from what I've read, this theorem states that the total charge can be considered a point charge from outside the shell and that the field inside a uniform shell of charge is 0. Does this also apply to solid spheres?

Cutter Ketch said:
No movement or redistribution necessary. Assume such a fixed charge distribution exists but you don’t know the functional form. What can you write down about the electric field as a function of radius?
Well I suppose I can write ##\varepsilon_{0} \vec{E} (4 \pi r^2) = \iiint_{V}{\rho(r)\ dV} = \iiint_{V}{\rho(r)\ r^{2}\ sin(\theta)\ dr\ d\theta\ d\phi}##, but I'm not sure where to go from there, since we still need ##\rho(r)##.

Yes, that integral is going to be very useful. Now, can you put the two ideas together?

prosteve037 said:
So from what I've read, this theorem states that the total charge can be considered a point charge from outside the shell and that the field inside a uniform shell of charge is 0. Does this also apply to solid spheres?
Yes. It applies to any object with spherical symmetry. That is to say, any object that consists of a collection of uniform spherical shells. A sphere is an example.

This is why we can correctly treat a gravitating sphere as if it were a point mass. This applies, even if the mass density of the sphere varies with depth. Or, as in this case, if the charge density varies with radius.

It also tells us that in the interior of a gravitating sphere, we can igore the gravitational (or Coulomb in this case) force from all layers above the point we are considering.

Gavran and prosteve037
I would solve Poisson’s equation for ##\rho##. We know what the field looks like, so all one has to do is write the divergence in spherical coordinates and go.

On edit
Actually I meant to suggest using Maxwell's equation for Gauss's law $$\vec{\nabla}\cdot \vec E=\rho/\epsilon_0$$ which is more straightforward than Poisson's equation.

Last edited:
prosteve037
prosteve037 said:
Well I suppose I can write ##\varepsilon_{0} \vec{E} (4 \pi r^2) = \iiint_{V}{\rho(r)\ dV} = \iiint_{V}{\rho(r)\ r^{2}\ sin(\theta)\ dr\ d\theta\ d\phi}##, but I'm not sure where to go from there, since we still need ##\rho(r)##.
It might help to write the mathematics more carefully. You have a vector equal to a scalar, and you have dummy variable ##r## on the righthand side appearing as an independent variable on the lefthand side.

Hint: think fundamental theorem of calculus.

prosteve037
vela said:
It might help to write the mathematics more carefully. You have a vector equal to a scalar, and you have dummy variable ##r## on the righthand side appearing as an independent variable on the lefthand side.

Hint: think fundamental theorem of calculus.
Yes, sorry about that. I'll try to keep these things sorted better.

Cutter Ketch said:
Yes, that integral is going to be very useful. Now, can you put the two ideas together?

jbriggs444 said:
Yes. It applies to any object with spherical symmetry. That is to say, any object that consists of a collection of uniform spherical shells. A sphere is an example.

This is why we can correctly treat a gravitating sphere as if it were a point mass. This applies, even if the mass density of the sphere varies with depth. Or, as in this case, if the charge density varies with radius.

It also tells us that in the interior of a gravitating sphere, we can igore the gravitational (or Coulomb in this case) force from all layers above the point we are considering.

kuruman said:
I would solve Poisson’s equation for ##\rho##. We know what the field looks like, so all one has to do is write the divergence in spherical coordinates and go.

I think these replies have helped me to get the same answer in two different ways:

Method 1:​
\begin{align} \nonumber E_{0} \ 4 \pi \varepsilon_{0} \ r^{2} & = \int_{0}^{r}{\rho(r) \ r^{2} \ dr} \ \int_{0}^{\pi}{\sin(\theta)\ d\theta} \ \int_{0}^{2 \pi}{d\phi} \\ \nonumber E_{0} \ 4 \pi \varepsilon_{0} \ r^{2} & = 4 \pi \int_{0}^{r}{\rho(r) \ r^{2} \ dr} \\ \nonumber E_{0} \ \varepsilon_{0} \ r^{2} & = \int_{0}^{r}{\rho(r) \ r^{2} \ dr} \\ \nonumber \frac{\partial}{\partial r}(E_{0} \ \varepsilon_{0} \ r^{2}) & = \rho(r) \ r^{2} \\ \nonumber \varepsilon_{0} \ E_{0} \ 2r & = \rho(r) \ r^{2} \\ \nonumber \frac{2 \ \varepsilon_{0} \ E_{0}}{r} & = \rho(r) \end{align}

Method 2:​
\begin{align} \nonumber \nabla^{2} \phi & = -\frac{\rho}{\varepsilon_{0}} \\ \nonumber \nabla \cdot \nabla \phi & = -\frac{\rho}{\varepsilon_{0}} \\ \nonumber \nabla \cdot \vec{E} & = \frac{\rho}{\varepsilon_{0}} \\ \nonumber \end{align}

Using the divergence in spherical coordinates (from Wikipedia):​
$$\nabla \cdot \vec{A} = \frac{1}{r^2} \frac{\partial (r^{2} A_{r})}{\partial r} + \frac{1}{r\sin(\theta)}\frac{\partial (A_{\theta}\sin(\theta))}{\partial \theta} + \frac{1}{r\sin(\theta)}\frac{\partial (A_{\phi})}{\partial \phi}$$

\begin{align} \nonumber \nabla \cdot \vec{E} & = \frac{1}{r^{2}}\frac{\partial (r^{2} E_{0})}{\partial r} \\ \nonumber & = \frac{E_{0}}{r^{2}} \ 2r \\ \nonumber & = \frac{2E_{0}}{r} \end{align}

Therefore:​
\begin{align} \nonumber \frac{\rho}{\varepsilon_{0}} & = \frac{2E_{0}}{r} \\ \nonumber \rho & = \frac{2 \varepsilon_{0} E_{0}}{r} \nonumber \end{align}

prosteve037 said:
Therefore:
Please don't leave us with a cliffhanger. Therefore what?

MatinSAR
kuruman said:
Please don't leave us with a cliffhanger. Therefore what?
Not sure if my browser is at fault, but the LaTex should be there in the post. I did just edit my response; maybe it was a syncing issue In any case, the "Therefore" is:
\begin{align} \nonumber \frac{\rho}{\varepsilon_{0}} & = \frac{2E_{0}}{r} \\ \nonumber \rho & = \frac{2 \varepsilon_{0} E_{0}}{r} \nonumber \end{align}

Cutter Ketch and kuruman
Whatever it was, it's fixed now in post #9. Good job!

prosteve037
prosteve037 said:
Not sure if my browser is at fault, but the LaTex should be there in the post. I did just edit my response; maybe it was a syncing issue In any case, the "Therefore" is:
\begin{align} \nonumber \frac{\rho}{\varepsilon_{0}} & = \frac{2E_{0}}{r} \\ \nonumber \rho & = \frac{2 \varepsilon_{0} E_{0}}{r} \nonumber \end{align}
Actually, the cliffhanger is there when I first load the page by entering the thread (see screenshot below) and disappears when I reload. Oh, well.

prosteve037
Weird. Same thing happened here.

vela said:
It might help to write the mathematics more carefully. You have a vector equal to a scalar, and you have dummy variable ##r## on the righthand side appearing as an independent variable on the lefthand side.

Hint: think fundamental theorem of calculus.
As a side note (and I'm not sure if this is appropriate to chat about here, so please feel free to move/tag this as needed), this class on electromagnetic wave theory is a bit of a challenge for me, in that I find myself having difficulty choosing the appropriate "strategies" of knowing when to use certain mathematical identities, theorems, shortcuts, tricks, etc.

As an example, you can see in my work above that this hint of using the fundamental theorem of calculus was the theorem I needed to employ in getting the answer using Method 1; it wouldn't have crossed my mind had @vela not mentioned it. Similarly, it wasn't obvious to me that I could use the Poisson equation to solve this problem as well (plus, that the formula for the divergence of a vector in spherical coordinates is needed to get the right divergence!).

I'm concerned about this, as it appears that derivations from some form/manipulation of Maxwell's equations (or even more fundamental definitions) are needed to complete a large majority of the homework problems for this class. I recognize that this won't just come naturally without a lot of practice, so I'd like to throw this out there and ask if anyone had recommendations to any (preferably free/accessible) materials out there that are specifically dedicated to "sharpening" one's skills on this front.

Thank you.

Tom.G

## What is charge density?

Charge density is a measure of electric charge per unit volume in a given space. It is often denoted by the symbol ρ (rho) and is measured in units of coulombs per cubic meter (C/m³).

## How can a sphere create a constant radial electric field inside it?

A sphere can create a constant radial electric field inside it if the charge density varies in such a way that the resulting electric field at any point inside the sphere is uniform. This typically requires a non-uniform charge distribution that increases with the radius.

## What is the mathematical expression for the charge density in a sphere with a constant radial electric field?

The charge density ρ(r) in a sphere of radius R with a constant radial electric field E inside can be found using Gauss's law. The charge density is given by ρ(r) = 3ε₀E/R, where ε₀ is the permittivity of free space.

## Why is the charge density not uniform in this case?

The charge density is not uniform because a uniform charge density would result in an electric field that varies with the distance from the center of the sphere. To achieve a constant radial electric field, the charge density must increase with the radius.

## What are the practical applications of a sphere with a constant radial electric field?

Spheres with constant radial electric fields can be used in various applications, including particle accelerators, electrostatic shielding, and scientific experiments that require uniform electric fields within a specific volume. They are also useful in theoretical studies of electrostatics and charge distributions.

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