The field momentum of a moving charge
Suppose an electron is moving at a uniform velocity through space, assuming for a moment that the velocity is low compared with the speed of light. Associated with this moving electron there is a momentum—even if the electron had no mass before it was charged—because of the momentum in the electromagnetic field. We can show that the field momentum is in the direction of the velocity $\FLPv$ of the charge and is, for small velocities, proportional to $v$. For a point $P$ at the distance $r$ from the center of the charge and at the angle $\theta$ with respect to the line of motion (see Fig. 28–1) the electric field is radial and, as we have seen, the magnetic field is $\FLPv\times\FLPE/c^2$. The momentum density, Eq. (27.21), is \begin{equation*} \FLPg=\epsO\FLPE\times\FLPB. \end{equation*}  It is directed obliquely toward the line of motion, as shown in the figure, and has the magnitude \begin{equation*} g=\frac{\epsO v}{c^2}\,E^2\sin\theta. \end{equation*}  The fields are symmetric about the line of motion, so when we integrate over space, the transverse components will sum to zero, giving a resultant momentum parallel to $\FLPv$. The component of $\FLPg$ in this direction is $g\sin\theta$, which we must integrate over all space. We take as our volume element a ring with its plane perpendicular to $\FLPv$, as shown in Fig. 28–2. Its volume is $2\pi r^2\sin\theta\,d\theta\,dr$. The total momentum is then \begin{equation*} \FLPp=\int\frac{\epsO\FLPv}{c^2}\,E^2\sin^2\theta\,2\pi r^2\sin\theta\, d\theta\,dr. \end{equation*}  Since $E$ is independent of $\theta$ (for $v\ll c$), we can immediately integrate over $\theta$; the integral is  \begin{equation*} \int\sin^3\theta\,d\theta=-\int(1-\cos^2\theta)\,d(\cos\theta)= -\cos\theta+\frac{\cos^3\theta}{3}. \end{equation*} 
 \begin{align*} \int\sin^3\theta\,d\theta&=-\int(1-\cos^2\theta)\,d(\cos\theta)\\ &=-\cos\theta+\frac{\cos^3\theta}{3}. \end{align*}  The limits of $\theta$ are $0$ and $\pi$, so the $\theta$-integral gives merely a factor of $4/3$, and \begin{equation*} \FLPp=\frac{8\pi}{3}\,\frac{\epsO\FLPv}{c^2} \int E^2r^2\,dr. \end{equation*}  The integral (for $v\ll c$) is the one we have just evaluated to find the energy; it is $q^2/16\pi^2\epsO^2a$, and \begin{equation} \FLPp=\frac{2}{3}\,\frac{q^2}{4\pi\epsO}\,\frac{\FLPv}{ac^2},\notag \end{equation}  or \begin{equation} \label{Eq:II:28:3} \FLPp=\frac{2}{3}\,\frac{e^2}{ac^2}\,\FLPv. \end{equation}  The momentum in the field—the electromagnetic momentum—is proportional to $\FLPv$. It is just what we should have for a particle with the mass equal to the coefficient of $\FLPv$. We can, therefore, call this coefficient the electromagnetic mass, $m_{\text{elec}}$, and write it as \begin{equation} \label{Eq:II:28:4} m_{\text{elec}}=\frac{2}{3}\,\frac{e^2}{ac^2}. \end{equation} 