Solution

We start from Newton's second law in its relativistic form. Since the electric force is constant and equal to $eE$, the linear momentum p of the electron grows linearly with time:

$\frac{dp}{dt} = eE \quad\Rightarrow\quad p = eEt$

because the electron is born at rest $(p(0)=0)$.

Now we need to relate momentum to velocity. In special relativity, this relation is

$p = \frac{mv}{\sqrt{1 - v^2/c^2}} \quad\Rightarrow\quad v = \frac{p/m}{\sqrt{1 + p^2/(m^2c^2)}}$

The distance traveled x is obtained by integrating velocity over time. Using $p = eEt$ and the change of variable $dp = eE$,dt, we write the accumulated distance up to time$\tau$as

$l = \int_0^\tau v\,dt = \int_0^{p_0} \frac{p/m}{\sqrt{1 + p^2/(m^2c^2)}}\,\frac{dp}{eE}, \qquad p_0 = eE\tau$

This integral is standard. Making the substitution $u = 1 + p^2/(m^2c^2)$, it is solved directly and yields:

$l = \frac{mc^2}{eE}\left( \sqrt{1 + \frac{p_0^2}{m^2c^2}} - 1 \right)$

In this expression, the distance l already appears as a function of time$\tau$ (through $p_0)$
Now we isolate $\tau$ We move the term -1 to the other side and square

$\left(\frac{eEl}{mc^2} + 1\right)^2 = 1 + \frac{p_0^2}{m^2c^2}$

We subtract 1 and expand the square of the parenthesis

$\frac{p_0^2}{m^2c^2} = \left(\frac{eEl}{mc^2}\right)^2 + 2\frac{eEl}{mc^2}$

We substitute $p_0 = eE\tau$ and cancel the common factors

$\frac{e^2E^2\tau^2}{m^2c^2} = \frac{eEl}{mc^2}\left( \frac{eEl}{mc^2} + 2 \right)$

We multiply both sides by $m^2c^2/(e^2E^2)$ and simplify

$\tau^2 = \frac{ml}{eE}\left( 2 + \frac{eEl}{mc^2} \right)$

Finally, we take the positive square root, since time is positive, and obtain:

$\boxed{\tau = \sqrt{\frac{ml}{eE}\left(2 + \frac{eEl}{mc^2}\right)}}$

If we apply the numerical data whut the formula we have :

$\boxed{\tau \approx 4.7 \times 10^{-9}\ \text{s} \;(4.7\ \text{ns})}$

The electron travels 1 meter in approximately 4.7 nanoseconds, reaching speeds close to that of light due to the intense electric field.