Sphere-with-non-uniform-charge-density ρ= k/r

[Taxi1337]

I am working on the same problem as a previous post, but he already marked it as answered and did not post a solution.

https://www.physicsforums.com/threads/sphere-with-non-uniform-charge-density.938117/

1. Homework Statement
A sphere of radius R carries charge Q. The distribution of the charge inside the sphere, however, is not homogeneous, but decreasing with the distance r from the center, so that ρ(r) = k/r.

1. Find k for given R and Q.

2. Using Gauss’s Law (differential or integral form), find the electric field E inside the sphere, i.e., for r < R.

Homework Equations

$$\int_V \rho \, dV = Q$$
$$\oint \vec E \cdot d \vec A = \frac {Q_{enclosed}} {\epsilon_0}$$
$$\vec {\nabla} \cdot \vec E = \frac {\rho} {{\epsilon}_0}$$

The Attempt at a Solution

1. $$\int_V \rho \, dV = \int_{0}^{2 \pi} \int_{0}^{\pi} \int_{0}^R \frac k r r^2 \sin \theta \, dr \, d \theta \, d \phi = 2 k \pi R^2 = Q$$
The units check out here.

2.
Here's where I ran into a problem. I tried using both the differential and integral forms of Gauss's Law, and in both cases the r canceled out, leaving me with an expression for the electric field I know is wrong. Oddly, the units work out here as well.

$$\oint \vec E \cdot d \vec A = 4 \pi r^2 E = \frac {Q_{enclosed}} {\epsilon_0} = \frac {2 k \pi r^2} {{\epsilon}_0}$$
$$\vec {\nabla} \cdot \vec E = \frac 1 {r^2} \frac {\partial} {\partial r}(r^2 E_r)= \frac {\rho} {{\epsilon}_0} = \frac k {r {{\epsilon}_0}}$$
$$E = \frac k {2{{\epsilon}_0}}$$

Thank you so much for any help! Please let me know if you need any further information or edits for clarification.

I am curious as to a method of finding the $k$ and substituting into the electric field. I tried this way.

$$ρ=\frac {k} {r}\,\,\,\,\,\,(1) $$
$$ρ=q \cdot (Volume)\,\,\,\,\,\,(2)$$
$$ρ=ρ\,\,\,\,\,\,\,(3)$$
$$\frac {k} {r}=q \cdot (Volume)\,\,\,\,\,\,(4)$$
$$k=q \cdot {(Volume)} \cdot r\,\,\,\,\,\,(5)$$
$$k=q \cdot {4 \pi r^2} \cdot r\,\,\,\,\,\,(6)$$
$$k=q \cdot {4 \pi r^3}\,\,\,\,\,\,(7)$$
then substitute $k$ from into $E=\frac {k} {2 ε} \,\,\,\,\,\,(8)$
$$E_{inside}=\frac {q \cdot {4 \pi r^3}} {2 ε}\,\,\,\,\,\,(9)$$
Finally
$$E_{inside}=\frac {2 { \pi q}} {ε} r^3\,\,\,\,\,\,(10) $$
which goes like
$$E_{inside}= r^3\,\,\,\,\,\,(11) $$

When I plot the electric field for the inside and outside I get ( i took all the constants as =1 for the graph)
for $figure A$ I plotted equation $(8)$ ( i took all the constants as =1 for the graph) so $E=1$
for $figure B$ i plotted equation $(11)$.

What is interesting to me is if I solve for $k$ and substitute into $(8)$ It makes a large difference in $E_{inside}$. BUT when I solve for $k$ for the outside i still get an equation that goes like $E_{outside}=\frac {1} {r^2}\,\,\,(12)$.

I found $$E_{outside}=\frac {k \cdot {4 R^2}} {2 ε r^2}\,\,(13)$$ and when I plug in $k=q \cdot {4 \pi R^3}\,\,(14)$ into $(13)$ i get $$E_{outside}=\frac {q \cdot {4 \pi R^3} \cdot {4 R^2}} {2 ε r^2}=\frac {q \cdot {\pi } \cdot {R^5}} {2 ε r^2}\,\,(15)$$

where in $(15)$ it still goes like $\frac {1}{r^2}$ no matter if i put in $k$ or not.

Thank you for your help, any input big or small is appreciated. It doesn't have to be elaborate.
Jared

 

[kuruman]

Your equation (2) is incorrect and so is are the equations that follow because they are based on it. If nothing else $k$ is a constant therefore it cannot depend $r$ (a variable) as you show in equation (7). To do find $k$, start from $Q=\int \rho~dV$, do the integral with $\rho=k/r$ and solve for $k$. See "Attempt at a solution, part 1" in the thread that you referenced.

On another note, why are you surprised that the electric field goes as $1/r^2$ outside the distribution? What does Gauss's law say about the field outside a spherical distribution of total charge $Q$?

 

[Taxi1337]

1.

Your equation (2) is incorrect and so is what it results in, equation (7). If nothing else $k$ is a constant therefore it cannot depend $r$ which is a variable. To do find $k$, start from $Q=\int \rho~dV$, do the integral with $\rho=k/r$ and solve for $k$. See "Attempt at a solution, part 1" in the thread that you referenced.

$$\int_V \rho \, dV = \int_{0}^{2 \pi} \int_{0}^{\pi} \int_{0}^r \frac k r r^2 \sin \theta \, dr \, d \theta \, d \phi = 2 k \pi r^2 = q\,\,\,\,\,\, (16) $$
$$ q=2 k \pi r^2 \,\,\,\,\,\, (17) $$

solving for $k$ from $(17)$

$k=\frac {q} {2 \pi r^2} \,\,\,\,(18)$
Plugin in $(18)$ into $(8)$

$$E_{inside}=\frac {q} {2 \pi r^2} \cdot \frac {1} {2ε}=\frac {q} {4\pi r^2} \cdot \frac {1} {ε}\,\,\,\,(19)$$

and for $E_{outside}=\frac {k \cdot {4 R^2}} {2 ε r^2}\,\,(13)$ and when I substitute

$k=q \cdot {4 \pi R^3}\,\,(14)$ into $(13)$ i get $$E_{outside}=\frac {q \cdot {4 \pi R^3} \cdot {4 R^2}} {2 ε r^2}=\frac {q \cdot {\pi } \cdot {R^5}} {2 ε r^2}\,\,(15)$$

2.

On another note, why are you surprised that the electric field goes as $1/r^2$ outside the distribution? What does Gauss's law say about the field outside a spherical distribution of total charge $Q$?

Sorry that i was not clear on my concern, its not that I am surprised that that out side the sphere of radius $R$ has a $E$ that goes like the inverse square law. I am surprised that when I solve for $k$ for both $E_{outside}$ and $E_{outside}$ only $E_{inside}$ changes relation of $r$ and $E_{outside}$ has the same relation of $\frac {1} {r^2}$

tldr; $E_{outside}$ is fine, i expect the field to look like a point charge.
$E_{inside}$ is what bothers me here.

3. Lastly, which of the figures is correct in my first post?

thank you Kuruman for your help

 

[kuruman]

1. You still don't get it. Equation (18) is incorrect. Parameter $k$ is constant and cannot depend on $r$. This is how you do it step by step.
a. Find constant $k$ using $\int_V \rho \, dV =Q$, where $Q$ is the total charge as given by the problem.
b. Find the enclosed charge $q_{enc}$ enclosed by a Gaussian sphere of radius $r$. That would be equation (16), $q_{enc}=2k\pi r^2$ in which $k$ is replaced by the value you found for it in the previous step.
c. Use Gauss's law $\int E_{inside}dA=q_{enc}/\epsilon_0$ to find the electric field inside.

Sorry that i was not clear on my concern, its not that I am surprised that that out side the sphere of radius $R$ has a $E$ that goes like the inverse square law. I am surprised that when I solve for kk for both $E_{outside}$ and $E_{outside}$ only $E_{inside}$ changes relation of $r$ and $E_{outside}$ has the same relation of $\frac {1} {r^2}$

I see. That will go away when you proceed as I outlined above and you find a constant value for $k$.

3. Lastly, which of the figures is correct in my first post?

That will be obvious when you do it right as I already explained.

 

[Taxi1337]

So the plot I'm getting is

, which is what I got all along (just not in terms of the constants $Q_{total}$ and $R^2$ ) . This question has bothered me a long time because I cannot think of anything where the electric field is constant and not a function of radius inside of a solid sphere. are there any examples that are observed in nature or experimentally?

I followed your step by step directions to get to where I am. Being more careful of my notation of $Q_{total}$ and $q_{enclosed}$ , the domain when I am integrating over gaussian surface $r$, vs. the entire radius of the sphere, $R$ also the order of my steps.

a. Find constant $k$ using $\int_V \rho \, dV =Q$, where $Q$ is the total charge as given by the problem.

$$\int_V \rho \, dV =Q_{total}$$
$$2 k \pi R^2 =Q_{total}$$
$$ k = \frac {Q_{total}} {2 \pi R^2}$$

b. Find the enclosed charge $q_{enc}$ enclosed by a Gaussian sphere of radius $r$. That would be equation (16), $q_{enc}=2k\pi r^2$

$$\int_V \rho \, dV =q_{enclosed}$$
$$2k\pi r^2=q_{enclosed}$$

in which $k$ is replaced by the value you found for it in the previous step.

$$2 \frac {Q_{total}} {2 \pi R^2} \pi r^2 =q_{enclosed}$$
$$ \frac {Q_{total}} {R^2} r^2 =q_{enclosed}$$

c. Use Gauss's law $\int E_{inside}dA=q_{enc}/\epsilon_0$ to find the electric field inside.

$$\int E_{inside}dA=q_{enclosed}/\epsilon_0$$
$$ E_{inside} \cdot (4\pi r^2) =q_{enclosed}/\epsilon_0$$
$$E_{inside} \cdot (4\pi r^2) =\frac {Q_{total}} {\epsilon_0 R^2} r^2$$

where I find that the electric field inside the solid sphere of non uniform volume charge density as
$$E_{inside}=\frac {Q_{total}} {4\pi \epsilon_0 R^2} $$

which is a constant related to inverse square (so the larger my sphere gets, the smaller the electric field will be inside).

I am going to redo my solution for the outside (its not required but i want to make sure I have a firm grasp on the concept of electric fields and Gauss' law.)

Thank you for your time kuruman, i have suffered over this problem for some time and It will stick to me now.

 

[Taxi1337]

I am going to redo my solution for the outside (its not required but i want to make sure I have a firm grasp on the concept of electric fields and Gauss' law.

For the outside:

Step 1. find find k in terms of the total charge of the system (since non uniform volume charge density), if it was uniform, skip to step 2.
$$\int_V \rho \, dV =Q_{total}$$
$$2 k \pi R^2 =Q_{total}$$
$$ k = \frac {Q_{total}} {2 \pi R^2}$$step 2. Find the charge enclosed by a Gaussian sphere of radius $r$ where $r>R$ so we would integrate over $(0,R)$.
$$\int_V \rho \, dV =q_{enclosed}$$
$$2k\pi R^2=q_{enclosed}$$
$$2 \frac {Q_{total}} {2 \pi R^2} \pi R^2 =q_{enclosed}$$
$$ {Q_{total}}=q_{enclosed}$$

Step 3. Apply Gauss' Law:
$$\int E_{outside}dA=q_{enclosed}/\epsilon_0$$
$$ E_{outside} \cdot (4\pi r^2) =q_{enclosed}/\epsilon_0$$
$$E_{outside} =\frac {Q_{total}} {4 \pi \epsilon_0 r^2} $$

I would expect the sphere to have the electric field outside $E_{outside} =\frac {Q_{total}} {4 \pi \epsilon_0 r^2}$ being the same as a point charge, so my answer looks correct for the outside.

 

[kuruman]

I think you got it now. It might be worth your while also to get the electric field inside from Poisson's equation $\vec{\nabla}\cdot \vec E_{inside}=\rho/\epsilon_0$. Sorry, I don't know of any "real" cases where the electric field is constant inside a spherical distribution.

 

[Taxi1337]

I think you got it now. It might be worth your while also to get the electric field inside from Poisson's equation $\vec{\nabla}\cdot \vec E_{inside}=\rho/\epsilon_0$. Sorry, I don't know of any "real" cases where the electric field is constant inside a spherical distribution.

Thank you so much! I will try using Poisson's next to check if my answer comes up the same for $E_{inside}.$

I guess what was bugging me so much is i graph everything from our lecture and the problems i do in the book and check my units. everything looked okay for me except when i attempted to solve $k$ and got a different answer. also the graph for this problem (figure a) looked very strange to me as all the other graphs of solid spheres looked like figure b.

 

[haruspex]

all the other graphs of solid spheres looked like figure b.

Really? For a conductor the internal field would be zero, and for a uniformly charged sphere it would be proportional to radius, no?