2.8 Find the electric field a distance \(z\) from the centre of a solid spherical charge of radius \(R\) and a uniform charge density of \( \mu \). Also, express the answers in terms of the total charge.
Let the charge be centered at the origin and its boundary defined by
$$x^2 + y^2 + z^2 = R^2$$
Let \( \vec{r} \) be the position vector of an arbitrary differential volume of charge.
$$\vec{r} = x \, \hat{i} + y \, \hat{j} + \sqrt{ \rho^2 – x^2 – y^2 } \, \hat{j}$$
Let \( \vec{H} \) be the position of the point at which we wish to measure the electric field.
$$\vec{H} = z \, \hat{k}$$
$$\vec{R} = \vec{H} – \vec{r} = -x \, \hat{i} – y \, \hat{j} + \left( z – \sqrt{ \rho^2 – x^2 – y^2 } \, \right)\hat{j}$$
$$\left| \left| \vec{R} \right| \right|^2 = \rho^2 + z^2 – 2 z \sqrt{ \rho^2 – x^2 – y^2 }$$
$$d\vec{E} = \frac{ 1 }{4 \pi \epsilon_0} \frac{ dq }{ \left( \rho^2 + z^2 – 2 z \sqrt{ \rho^2 – x^2 – y^2 } \right)^{3/2} } \left( – x \, \hat{i} – y \, \hat{j} + ( z – \sqrt{ \rho^2 – x^2 – y^2 }) \, \hat{k} \right)$$
Converting to spherical coordinates
$$d\vec{E} = \frac{ 1 }{4 \pi \epsilon_0} \frac{ dq}{ \left( \rho^2 + z^2 – 2 z \rho \cos \phi \right)^{3/2} } \left( – \rho \sin \phi \cos \theta \, \hat{i} – \rho \sin \phi \cos \theta \, \hat{j} + ( z – \rho \cos \phi ) \, \hat{k} \right)$$
$$d\vec{E} = \frac{ 1 }{4 \pi \epsilon_0} \frac{ \mu \rho^2 \sin \phi \, d\theta \, d\phi \, d\rho}{ \left( \rho^2 + z^2 – 2 z \rho \cos \phi \right)^{3/2} } \left( – \rho \sin \phi \cos \theta \, \hat{i} – \rho \sin \phi \cos \theta \, \hat{j} + ( z – \rho \cos \phi ) \, \hat{k} \right)$$
$$\vec{E} = \frac{ \mu }{4 \pi \epsilon_0} \int_0^ R \int_0^{\pi} \int_0^{2 \pi} \frac{ \rho^2 \sin \phi \, d\theta \, d\phi \, d\rho}{ \left( \rho^2 + z^2 – 2 z \rho \cos \phi \right)^{3/2} } \left( – \rho \sin \phi \cos \theta \, \hat{i} – \rho \sin \phi \cos \theta \, \hat{j} + ( z – \rho \cos \phi ) \, \hat{k} \right)$$
$$\vec{E} = \frac{ 2 \pi \mu }{4 \pi \epsilon_0} \int_0^ R \left( \int_0^{\pi} \frac{ ( z – \rho \cos \phi ) \sin \phi \, d\phi}{ \left( \rho^2 + z^2 – 2 z \rho \cos \phi \right)^{3/2} } \right) \rho^2 \, d\rho\, \hat{k}$$
From spherical shell solution obtained with Mathematica
$$\vec{E} = \frac{ 2 \pi \mu }{4 \pi \epsilon_0} \int_0^ R \left( \frac{ z – \rho }{z^2 | \, z – \rho \, |} + \frac{ z + \rho }{ z^2 | \, z + \rho \, |} \right) \rho^2 \, d\rho \, \hat{k}$$
\( z > R \)
$$\vec{E} = \frac{ 2 \pi \mu }{4 z^2 \pi \epsilon_0} \int_0^R 2 \rho^2 \, d\rho \, \hat{k} = \frac{ 2 \pi \mu }{4 z^2 \pi \epsilon_0} \frac{2 R^3}{3} \, \hat{k} = \frac{\mu R^3 }{3 z^2 \epsilon_0} \, \hat{k} = \frac{ Q_{total} }{4 z^2 \pi \epsilon_0} \, \hat{k}$$
\( z < R \) $$\vec{E} = \frac{ 2 \pi \mu }{4 z^2 \pi \epsilon_0} \int_0^z 2 \rho^2 \, d\rho \, \hat{k} = \frac{ 2 \pi \mu }{4 z^2 \pi \epsilon_0} \frac{2 z^3}{3} \, \hat{k} = \frac{\mu z }{3 \epsilon_0} \, \hat{k} = \frac{ z Q_{total} }{4 \pi \epsilon_0 R^3} \, \hat{k}$$
Due to the symmetry of the problem it is much more efficient to use Gauss’ law (question 2.12).
Gauss’ law in integral form is
$$\oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$$
Use spherical coordinates and let \( \rho \) be the radial distance from the origin.
for \( \rho < R \)
$$Q_{enc} = \mu \left(\frac{4}{3} \pi \rho^3\right)$$
$$\left| \left| \vec{E} \right| \right| (4 \pi \rho^2) = \mu \left( \frac{4}{3} \pi \rho^3 \right)$$
$$\left| \left| \vec{E} \right| \right| = \frac{\mu \left( \frac{4}{3} \pi \rho^3 \right)}{4 \pi \epsilon_0 \rho^2} = \frac{\mu \rho}{ 3 \epsilon_0 }$$
$$\vec{E} = \frac{\mu \rho}{ 3 \epsilon_0 } \hat{r}$$
In terms of total charge
$$ Q_{enc} = \frac{\rho^3}{R^3} Q_{total}$$
$$\left| \left| \vec{E} \right| \right| = \frac{\rho^3}{R^3} Q_{total} \frac{1}{4 \pi \epsilon_0 \rho^2} = \frac{ \rho Q_{total} }{4 \pi \epsilon_0 R^3}$$
For \(\rho > R\)
$$Q_{enc} = \mu \left(\frac{4}{3} \pi R^3 \right)$$
$$\left| \left| \vec{E} \right| \right| = \frac{\mu ( \frac{4}{3} \pi R^3)}{4 \pi \epsilon_0 \rho^2} = \frac{\mu R^3}{ 3 \epsilon_0 \rho^2 }$$
In order to compute the answers in terms of the total charge notice
$$Q_{enc} = \alpha Q_{total} = \alpha \mu \frac{4 \pi R^3}{3}$$
For \( \rho < R \)
$$Q_{enc} = \mu \frac{4 \pi \rho^3}{3}$$
Equating the two
$$\mu \frac{4 \pi \rho^3}{3} = \alpha \mu \frac{4 \pi R^3}{3}$$
$$\alpha = \frac{\rho^3}{R^3}$$