Irodov 3.18: Volume Cosine Charge at the Center – Amit & Deepali Sinha Foundation

Problem Irodov 3.18

A solid ball of radius $R$ carries volume charge with density $\rho = \vect{a}\cdot\vect{r}$ , where $\vect{a}$ is a constant vector and $\vect{r}$ is the position vector measured from the ball’s center. Find the electric field vector at the center of the ball.

The ball of radius R in section along a, with ρ = a· r: positive on the side a points to, negative opposite, and denser farther from the plane through O normal to a. — The ball of radius $R$ in section along $\vect a$ , with $\rho = \vect a\cdot\vect r$ : positive on the side $\vect a$ points to, negative opposite, and denser farther from the plane through $O$ normal to $\vect a$ .

Given

ball of radius $R$
$\rho = \vect a\cdot\vect r$ , $\vect a$ constant

Find

the field vector

\vect E

at the center

O

Solution

Step 1The density in spherical coordinates

Take the $x$ axis along $\vect a$ and spherical coordinates $r$ , $\theta$ , $\phi$ about it, with $\theta$ measured from $\vect a$ . The density is

\rho = a r\cos\theta \tag{1}

and the volume element is $dV = r^2\sin\theta\,dr\,d\theta\,d\phi$ .

Step 2The field of one volume element

The element $dV$ at $(r, \theta, \phi)$ is the distance $r$ from $O$ , and the line from it to $O$ makes the angle $\theta$ with the $x$ axis.

A volume element dV at distance r from the center and polar angle θ from a, its field d E at O, and the component dE_x along a. — A volume element $dV$ at distance $r$ from the center and polar angle $\theta$ from $\vect a$ , its field $d\vect E$ at $O$ , and the component $dE_x$ along $\vect a$ .

The components across the axis cancel between elements at $\phi$ and $\phi + \pi$ , so only the component along $\vect a$ survives. For $\rho > 0$ the element pushes the field at $O$ away from itself, toward $-x$ :

dE_x = -\frac{1}{4\pi\varepsilon_0}\,\frac{\rho\,dV}{r^2}\cos\theta

Substituting (1) and $dV$ , the factor $r^2$ cancels:

dE_x = -\frac{a}{4\pi\varepsilon_0}\,r\cos^2\theta\sin\theta\,dr\,d\theta\,d\phi \tag{2}

Step 3Integrate over the ball

The integrand of (2) separates into three one-variable integrals:

\begin{aligned} E_x &= -\frac{a}{4\pi\varepsilon_0}\int_0^{R}r\,dr\int_0^{\pi}\cos^2\theta\sin\theta\,d\theta\int_0^{2\pi}d\phi\\ &= -\frac{a}{4\pi\varepsilon_0}\cdot\frac{R^2}{2}\cdot\left[-\frac{\cos^3\theta}{3}\right]_0^{\pi}\cdot 2\pi\\ &= -\frac{a}{4\pi\varepsilon_0}\cdot\frac{R^2}{2}\cdot\frac{2}{3}\cdot 2\pi = -\frac{aR^2}{6\varepsilon_0} \end{aligned}

The field lies along $\vect a$ , so the vector is

\vect E = -\frac{R^2\,\vect a}{6\varepsilon_0}

(3)

The field at the center has magnitude $aR^2/(6\varepsilon_0)$ and points against $\vect a$ , from the positive half of the ball toward the negative half.

The total field E at the center of the ball, against a. — The total field $\vect E$ at the center of the ball, against $\vect a$ .

Check

By shells instead. The shell between $r$ and $r + dr$ is a sphere carrying the surface density $\rho\,dr = (a\,dr)\,r\cos\theta$ , a $\sigma_0\cos\theta$ shell with $\sigma_0 = a r\,dr$ ; and such a shell is what two balls of densities $\pm\rho$ leave when shifted apart by $\vect l$ with $\rho l = \sigma_0$ . Inside, each ball gives $\pm\rho\vect r/(3\varepsilon_0)$ from its own center by Gauss’s law, and the two differ by the uniform $-\rho\vect l/(3\varepsilon_0)$ , so the shell’s field at its center is $-\sigma_0/(3\varepsilon_0)$ along $\vect a$ , whatever its radius. Summing the shells,

E_x = -\int_0^R \frac{a r\,dr}{3\varepsilon_0} = -\frac{aR^2}{6\varepsilon_0},

as found. Units: $a$ is a charge density per length, so $aR^2/\varepsilon_0$ is (charge per area) $/\varepsilon_0$ , a field.

Remark

The ball’s dipole moment is $\vect p = \int\vect r\rho\,dV = (4\pi R^5/15)\,\vect a$ , by the same three integrals with an extra factor $r$ . A uniformly polarized ball with that moment, whose charge sits on its surface, would give only $-\vect p/(4\pi\varepsilon_0R^3) = -R^2\vect a/(15\varepsilon_0)$ at its center; here the charge fills the interior and the inner layers, though weak, sit closer to $O$ , so the field at the center is two and a half times larger.