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Alexphysics edited Jun 23, 2026, 4:19 PM

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		+	### Statement
		+
		+	$14.4.14.$ [Insert the problem statement]
		+
		+	### 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:
		+
		+	$\boxed{\tau = 0.34\ \text{ms}}$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$14.4.14.$ [Insert the problem statement]

			### 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:

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

			#### Answer

			[Insert a concise answer or boxed result]