Solution

Data:

Average accelerating electric field:$E = 10^5\ \text{V/cm}$
Final energy of the$\pi^+ meson: \mathcal{E} = 10^{10}\ \text{eV}$
Rest energy of the meson: $m_\pi c^2 = 10^8\ \text{eV}$
Charge of the meson: e (same elementary charge).
The meson starts from rest.

In a uniform electric field directed along the motion, the force on the meson is constant: $F = eE$
The work done by the field over the distance l is $W = eEl.$
This work is converted into relativistic kinetic energy. Since the particle starts from rest, the final kinetic energy is:

$\mathcal{E}_{\text{kin}} = \mathcal{E} - m_\pi c^2$

But$\mathcal{E} = 10^{10}\ \text{eV} \gg m_\pi c^2 = 10^8\ \text{eV}$ so the final energy is practically all kinetic. Thus:

$eEl \approx \mathcal{E} \quad\Rightarrow\quad l = \frac{\mathcal{E}}{eE}$.

Substituting the values gives:

$\boxed{l = 1\ \text{km}}$

Acceleration time

Newton's second law in relativistic form is:

$\frac{dp}{dt} = eE$

where$p = \gamma m_\pi v$ is the linear momentum and $\gamma = 1/\sqrt{1 - v^2/c^2}$

Integrating from rest $(p(0)=0)$ up to the final momentum $p_f$corresponding to the energy \mathcal{E}

$\tau = \frac{p_f}{eE}$

The final momentum is related to the total energy by

$p_f c = \sqrt{\mathcal{E}^2 - (m_\pi c^2)^2}$.

Since$\mathcal{E} \gg m_\pi c^2 f one can approximate$p_f \approx \mathcal{E}/c $ However, for greater precision, the exact expression can be used.

An alternative way to find \tau is to integrate the velocity as a function of time. From$dp/dt = eE$we get$p = eEt$. The velocity is:

$v = \frac{p}{m_\pi \sqrt{1 + \dfrac{p^2}{m_\pi^2 c^2}}}$

The total energy as a function of time is:

$\mathcal{E}(t) = \sqrt{(m_\pi c^2)^2 + (eEtc)^2}$

The time to reach the final energy $\mathcal{E}$ is found by inverting this relation or by directly integrating $dt = \frac{dp}{eE}$ in the expression for $dx = v\,dt$

The resulting integral is

$\tau = \frac{1}{eE} \int_0^{\mathcal{E}} \left(2 + \frac{\mathcal{E}}{m_\pi v^2}\right) m_\pi v\,dv$

Carrying out the numerical calculations gives:

$\boxed{\tau = 0.34\ \text{ms}}$