= Solution
Assume $qE>0$ and put
$$
\alpha=\frac{qE}{mc}.
$$
The nonzero four-velocity equations are
$$
\dot u^0=\alpha u^x,\qquad \dot u^x=\alpha u^0,
$$
with $u^0(0)=c$ and $u^x(0)=0$. Hence
$$
u^0=c\cosh(\alpha\tau),\qquad
u^x=c\sinh(\alpha\tau).
$$
Integrating from the origin gives the uniformly accelerated trajectory
$$
\boxed{
ct=\frac c\alpha\sinh(\alpha\tau),\qquad
x=\frac c\alpha\{\cosh(\alpha\tau)-1\},\qquad y=z=0.
}
$$
The light ray has $x_L=-h+ct$. At interception, $x_L=x$, so
$$
h=ct-x=\frac c\alpha\left(1-e^{-\alpha\tau}\right).
$$
A finite solution exists exactly when
$$
h<h_c=\frac c\alpha=\frac{mc^2}{qE}.
$$
Then
$$
\tau_c=-\frac1\alpha\log\left(1-\frac{\alpha h}{c}\right),
$$
and, writing $H=\alpha h/c$,
$$
t_c=\frac1\alpha\sinh(\alpha\tau_c)
=\frac{H(2-H)}{2\alpha(1-H)}.
$$
At $h=h_c$ interception is approached only as $t\to\infty$, and for $h>h_c$ it never occurs. The limiting backward light ray is the <Rindler horizon> of the accelerated particle.
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