Field Generated by Macroscopic Charge Density

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Theorem

Let $B$ be a body of matter.

Let $P$ be a point inside $B$ whose position vector is $\mathbf r$.


The electric field at $P$ generated by the macroscopic charge density within $B$ is given by:

$\ds \map {\mathbf E} {\mathbf r} = \dfrac 1 {4 \pi \varepsilon_0} \int_{\text {all space} } \dfrac {\paren {\mathbf r - \mathbf r'} \map \rho {\mathbf r'} } {\size {\mathbf r - \mathbf r'}^3} \rd V'$

where:

$\d V'$ is an infinitesimal volume element
$\mathbf r'$ is the position vector of $\d V'$
$\map \rho {\mathbf r'}$ is the macroscopic charge density of the macroscopic electric field at $\mathbf r'$
$\varepsilon_0$ denotes the vacuum permittivity.


Proof

From Electric Field Strength from Assemblage of Point Charges, the electric field strength caused by an assemblage of point charges $q_1, q_2, \ldots, q_n$ is given by:

$\ds \map {\mathbf E} {\mathbf r} = \dfrac 1 {4 \pi \epsilon_0} \sum_i \dfrac {\paren {\mathbf r - \mathbf r_i} q_i} {\size {\mathbf r - \mathbf r_i}^3}$

where $\mathbf r_1, \mathbf r_2, \ldots, \mathbf r_n$ are the position vectors of $q_1, q_2, \ldots, q_n$ respectively.

We apply the same principle to the macroscopic charge density and convert the summation into a definite integral, as follows:


Consider a volume element $\d V'$ which is smaller than the scale used for a macroscopic electric field, but still large enough to contain many atoms.

Charge-from-volume-element.png

The electric field strength caused by $\d V'$ is:

$\map {\mathbf E_{\text {atomic} } } {\mathbf r'} = \dfrac 1 {4 \pi \epsilon_0} \dfrac {\paren {\mathbf r - \mathbf r'} \rd q} {\size {\mathbf r - \mathbf r'}^3}$

where $\d q$ is the electric charge on $\d V'$.

By definition of macroscopic charge density:

$\map \rho {\mathbf r'} = \dfrac {\d q} {\d V'}$

and so:

$\map {\mathbf E} {\mathbf r'} = \dfrac 1 {4 \pi \epsilon_0} \dfrac {\paren {\mathbf r - \mathbf r'} \map \rho {\mathbf r'} } {\size {\mathbf r - \mathbf r'}^3} \rd V'$


Integrating over all space:

$\ds \map {\mathbf E} {\mathbf r} = \dfrac 1 {4 \pi \varepsilon_0} \int_{\text {all space} } \dfrac {\paren {\mathbf r - \mathbf r'} \map \rho {\mathbf r'} } {\size {\mathbf r - \mathbf r'}^3} \rd V'$

Hence the result.

$\blacksquare$


Sources