# Electric field lines between surfaces of hollow sphere

I was wondering what was the direction of the electric field between the two surfaces of a hollow sphere with constant charge density $\rho$. With the help of Gauss' Law I got the following absolute values for $\vec{E}$:

$r<r_1$: $E = 0$

$r_1<r<r_2$: $E = \frac{\rho}{3 \epsilon_0} (r - \frac{r_1^3}{r})$

$r_2 < r$: $E = \frac{\rho}{3 \epsilon_0} \frac{r_2^3 - r_1^3}{r^2}$.

I would think that the field lines are directed outwards for $r>r_2$ and inwards for $r_1<r<r_2$ because there is more positive charge on the outer border of the shell than on the inner one (the volume element is there greater). But when I graph $E(r)$ it then creates a surprising discontinuity at $r_2$ that makes me think it might also be directed radially outwards. Can someone please clear up my doubts?

Julien.

Due to the spherical symmetry, the electric field can never be directed inwards (assuming $\rho$ is positive.) To prove this, proceed by contradiction:

• Suppose there was some point in space where the electric field pointed inwards. Let $R$ denote the distance between this point and the center of the sphere.

• By spherical symmetry, all points at this same distance $R$ from the center of the sphere must have an inward-directed electric field.

• If we then take a Gaussian surface to be a sphere of radius $R$, the integral of the electric flux over this surface will be $\oint \vec{E} \cdot d\vec{a} < 0$, since $\vec{E}$ and $d\vec{a}$ point in opposite directions.

• By Gauss's Law, the total charge enclosed in this sphere must therefore be negative. This is a contradiction, because $\rho$ is positive or zero everywhere in space.

By the way, your electric field magnitude isn't "discontinuous"; the value of the electric field as you approach $r_2$ (or $r_1$) from the inside is the same as you approach that same point from the outside. It does, however, have a "kink" in the plot. A general rule of thumb is that if the charge distribution $\rho$ is defined at all points in space and has a discontinuous $n$th derivative, then $\vec{E}$ will have a discontinuous $(n+1)$th derivative. Here, since the function $\rho$ is itself discontinuous at $r = r_1$ and $r = r_2$ (we can think of this as the "0th" derivative being discontinuous), then the first derivative of $\vec{E}$ is also discontinuous.

• Alright thank you very much. – Jxx Nov 17 '16 at 15:57

EDITED INTERPRETATION: Uniform Volume Charge Density between at $r_1$ and $r_2$

If $\rho$ is a uniform volume charge density for $r_1\le r\le r_2$, then we have from Gauss's Law for $r\in [r_1,r_2]$

\begin{align} \oint_{|\vec r|=r}\vec E(\vec r)\cdot \hat n\,dS&=4\pi r^2E_r(r)\\\\ &=\frac{1}{\epsilon_0}4\pi (r^3-r_1^3)\rho \end{align}

Therefore, for $r_1<r<r_2$, the electric field $\vec E(\vec r)$ is given by

$$\bbox[5px,border:2px solid #C0A000]{\vec E(\vec r)=\hat r \frac{\rho}{\epsilon_0}\left(r-\frac{r_1^3}{r^2}\right)} \tag 1$$

For $r_2<r$, application of Gauss's Law reveals that

$$\oint_{|\vec r|=r}\vec E(\vec r)\cdot \hat n\,dS=4\pi r^2E_r(r)=\frac{4\pi \rho}{\epsilon_0}\left( r_2^3-r_1^3\right)$$

Therefore, for $r_2<r$, the electric field $\vec E(\vec r)$ is given by

$$\bbox[5px,border:2px solid #C0A000]{\vec E(\vec r)=\hat r \frac{\rho}{\epsilon_0}\left(\frac{r_2^3-r_1^3}{r^2}\right)} \tag 2$$

Note that for $r=r_1$, the electric field as given by expression in $(1)$ is $0$ and the electric field is continuous at $r_1$.

For $r=r_2$, the electric field as given by expression in $(1)$ is $\vec E(\vec r)=\hat r \frac{\rho}{\epsilon_0}\left(\frac{r_2^3-r_1^3}{r_2^2}\right)$, which is equal to the electric field as given by $(2)$ for $r=r_2$. Hence, the electric field is continuous at $r_2$.

ORIGINAL INTERPRETATION: Surface Charge Layers at $r_1$ and $r_2$

The result in the OP is not quite correct. The charge density $\rho$ is a surface charge density with units C/$\text{m}^2$. The total charge on the surface $r=r_1$ is $4\pi r_1^2 \rho$ while the total charge on the surface $r=r_2$ is $4\pi r_2^2 \rho$. The total charge enclosed is therefore is the sum $4\pi (r_1^2+r_2^2)\rho$.

Applying Gauss's Law to the region $r_1<r<r_2$ reveals

\begin{align} \oint_{|\vec r|=r}\vec E(\vec r)\cdot \hat n\,dS&=4\pi r^2E_r(r)\\\\ &=\frac{1}{\epsilon_0}4\pi r_1^2\rho \end{align}

Therefore, for $r_1<r<r_2$, the electric field $\vec E(\vec r)$ is given by

$$\bbox[5px,border:2px solid #C0A000]{\vec E(\vec r)=\hat r \frac{\rho}{\epsilon_0}\left(\frac{r_1}{r}\right)^2}$$

For $r_2<r$, application of Gauss's Law reveals that

$$\oint_{|\vec r|=r}\vec E(\vec r)\cdot \hat n\,dS=4\pi r^2E_r(r)=\frac{4\pi \rho}{\epsilon_0}\left( r_1^2+r_2^2\right)$$

Therefore, for $r_2<r$, the electric field $\vec E(\vec r)$ is given by

$$\bbox[5px,border:2px solid #C0A000]{\vec E(\vec r)=\hat r \frac{\rho}{\epsilon_0}\left(\frac{r_1^2+r_2^2}{r^2}\right)}$$

NOTE:

The electric field is discontinuous across the surface charge layers with

$$E_r(r_i^+)-E_r(r_i^-)=\frac{\rho}{\epsilon_0}$$

for $i=1,2$.

• I assumed that the setup was for a "thick shell" with inner radius $r_1$ and outer radius $r_2$, in which case the given answer is correct. I agree, though, that the description of the configuration wasn't clear. – Michael Seifert Nov 17 '16 at 20:31
• @MichaelSeifert You could be correct. It is unclear. – Mark Viola Nov 17 '16 at 20:43