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A solid insulating sphere of radius R = 10 cm carries charge spread through its volume…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A solid insulating sphere of radius $R = 10\ \text{cm}$ carries charge spread through its volume. The charge is not uniform. Its density is greatest at the centre and falls off steadily to zero at the surface:

\displaystyle \rho(r) = \rho_0\left(1 - \frac{r}{R}\right), \qquad \rho_0 = 9.0\ \mu\text{C/m}^3

where $r$ is the distance from the centre, as in the figure. Such a sphere is a simple model of a charged cloud that is densest in its middle. The material of the sphere does not affect the field, so treat the charge as if it sat in empty space. There is no other charge nearby. Take $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$ .

A sphere of radius R equal to 10 cm, shaded dark blue at its centre and fading to white at its surface, to show a charge density rho equal to rho nought times one minus r over R, with rho nought equal to 9.0 microcoulombs per cubic metre. A dashed circle of radius r is drawn inside it.

What is the total charge of the sphere?
What is the electric field at $r = 5.0\ \text{cm}$ ?
The field inside grows from zero at the centre, then falls. At what distance from the centre is it greatest?
What is that greatest field?
A small charge is now fixed at the centre of the sphere, so that the field at $r = 5.0\ \text{cm}$ becomes zero. What is that charge, with its sign?
With that charge in place at the centre, what is the field just outside the surface of the sphere?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Use Gauss's law with a sphere of radius $r$ . The only work is the charge inside it, which needs an integral because the density changes with $r$ .

Step 1. Charge inside $r$ .

\displaystyle q_{\text{in}}(r) = \int_0^r \rho_0\left(1 - \frac{s}{R}\right)4\pi s^2\,ds = 4\pi\rho_0\left(\frac{r^3}{3} - \frac{r^4}{4R}\right)

So inside the sphere $E(r) = \dfrac{\rho_0}{\varepsilon_0}\left(\dfrac{r}{3} - \dfrac{r^2}{4R}\right)$ .

Step 2. Total charge (a). At $r = R$ : $Q = \dfrac{\pi\rho_0R^3}{3} = \dfrac{\pi \times 9.0 \times 10^{-6}\ \text{C/m}^3 \times (0.10\ \text{m})^3}{3} = 9.42\ \text{nC}$ .

Step 3. Field at $R/2$ (b). $E = \dfrac{\rho_0R}{\varepsilon_0}\left(\dfrac16 - \dfrac1{16}\right) = \dfrac{5}{48}\cdot\dfrac{\rho_0R}{\varepsilon_0} = 1.06 \times 10^{4}\ \text{N/C}$ .

Step 4. The greatest field (c, d). $\dfrac{dE}{dr} = \dfrac{\rho_0}{\varepsilon_0}\left(\dfrac13 - \dfrac{r}{2R}\right) = 0$ at $r = \dfrac{2R}{3} = 6.67\ \text{cm}$ . There

\displaystyle E_{\max} = \frac{\rho_0}{\varepsilon_0}\left(\frac{2R}{9} - \frac{R}{9}\right) = \frac{\rho_0R}{9\varepsilon_0} = 1.13 \times 10^{4}\ \text{N/C}

It is larger than the field at the surface, $\dfrac{\rho_0R}{12\varepsilon_0} = 8.47 \times 10^{3}\ \text{N/C}$ .

Step 5. The charge at the centre (e). The field at $R/2$ is zero when the total charge inside that radius is zero. The sphere's own charge there is $4\pi\rho_0R^3\left(\dfrac1{24} - \dfrac1{64}\right) = 2.95\ \text{nC}$ , so the added charge is $-2.95\ \text{nC}$ .

Step 6. Just outside (f). All the charge now acts as if at the centre: $E = \dfrac{(9.42 - 2.95) \times 10^{-9}\ \text{C}}{4\pi\varepsilon_0(0.10\ \text{m})^2} = 5.83 \times 10^{3}\ \text{N/C}$ .

Answer: (a) $9.42\ \text{nC}$ . (b) $1.06 \times 10^{4}\ \text{N/C}$ . (c) $6.67\ \text{cm}$ . (d) $1.13 \times 10^{4}\ \text{N/C}$ . (e) $-2.95\ \text{nC}$ . (f) $5.83 \times 10^{3}\ \text{N/C}$ .

Watch out. For a uniform sphere the field grows all the way to the surface. Here the outer layers hold little charge, so the field peaks inside. Using the uniform rule $E = \dfrac{Q}{4\pi\varepsilon_0R^3}\,r$ at $R/2$ gives $4.24 \times 10^{3}\ \text{N/C}$ , which is wrong.

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