An electron entering an extended stationary and uniform electric field at a ve-
+
locity βc penetrates into this field to a depth l. The electron velocity is directed
+
along the field. To what depth will the electrons penetrate, if, on the contrary,
+
an electric field hits the stationary electrons with the same speed? Solve this
+
problem in two ways, using: a) the effect of relativistic distance reduction, b)
+
the relationship between the work of A and the change in the particle mass
+
∆m : A = c2∆m.
+
### Solution
a) Length contraction method
In the first case (moving electron, field at rest), the depth l is the distance the electron travels inside the field until it stops. In the reference frame S' that moves with the electron before it enters the field, the electron is at rest and it is the field that moves toward it. In S', the field has a proper length l' (the extension of the region where the field is nonzero). The length measured in the laboratory S is contracted:
In the second case (electron at rest, moving field), the situation in the laboratory is the reverse. The field moves with speed \beta c and its length in S is$ l = l'/\gamma = (\gamma l)/\gamma = l$. The electron remains inside the field while it completely passes over it. The distance traveled by the electron in the laboratory during that interval is precisely the length of the field in S, that is, l. Thus,
$\boxed{l_1 = l}$
b) Work–energy method
First case: the electron enters with speed$\beta c $and slows down to rest under the electric force eE. The initial kinetic energy is$ (\gamma-1)m_e c^2$. The work done by the field is eE l. Equating,
$eE l = (\gamma-1)m_e c^2 \quad\Longrightarrow\quad\boxed{l = \frac{(\gamma-1)m_e c^2}{eE}}$
Second case: the electron, initially at rest, is accelerated by the moving field. The field takes a time$ T = l/(\beta c) $to pass over the electron (since its length in S is l) During that time, the electron is subjected to a constant force eE. The momentum acquired is
$p = eE\,T = \frac{eE l}{\beta c}$
The final kinetic energy K and the momentum p satisfy the relativistic relation
$(K + m_e c^2)^2 = (pc)^2 + (m_e c^2)^2$
Substituting p and using the expression for l from the first case$ (eE l = (\gamma-1)m_e c^2)$, one obtains that the final kinetic energy is exactly $(\gamma-1)m_e c^2$. Therefore, the electron reaches the same speed $\beta c $with which the field was moving.
@@ -34,13 +42,15 @@Solution
The distance l_1 traveled by the electron during time T can be calculated by integrating the relativistic velocity $v(t) = (p/m_e)/\sqrt{1+(p/m_e c)^2} with p = eE t$. The result is
Substituting$ T = l/(\beta c) $and "eE l = (\gamma-1)m_e c^2$ one arrives at
+
Substituting$ T = l/(\beta c) $and $eE l = (\gamma-1)m_e c^2$ one arrives at
$\boxed{l_1 = \frac{(\gamma-1)m_e c^2}{eE} = l}$
#### Answer
−
[Insert a concise answer or boxed result]
+
$\boxed{l_1 = l}$
+
+
$\boxed{l_1 = \frac{(\gamma-1)m_e c^2}{eE} = l}$
@@ -1,7 +1,15 @@
### Statement
### Statement
$14.4.11.$[Insert the problem statement]
$14.4.11.$
An electron entering an extended stationary and uniform electric field at a ve-
locity βc penetrates into this field to a depth l. The electron velocity is directed
along the field. To what depth will the electrons penetrate, if, on the contrary,
an electric field hits the stationary electrons with the same speed? Solve this
problem in two ways, using: a) the effect of relativistic distance reduction, b)
the relationship between the work of A and the change in the particle mass
∆m : A = c2∆m.
### Solution
### Solution
a) Length contraction method
a) Length contraction method
In the first case (moving electron, field at rest), the depth l is the distance the electron travels inside the field until it stops. In the reference frame S' that moves with the electron before it enters the field, the electron is at rest and it is the field that moves toward it. In S', the field has a proper length l' (the extension of the region where the field is nonzero). The length measured in the laboratory S is contracted:
In the first case (moving electron, field at rest), the depth l is the distance the electron travels inside the field until it stops. In the reference frame S' that moves with the electron before it enters the field, the electron is at rest and it is the field that moves toward it. In S', the field has a proper length l' (the extension of the region where the field is nonzero). The length measured in the laboratory S is contracted:
In the second case (electron at rest, moving field), the situation in the laboratory is the reverse. The field moves with speed \beta c and its length in S is$ l = l'/\gamma = (\gamma l)/\gamma = l$. The electron remains inside the field while it completely passes over it. The distance traveled by the electron in the laboratory during that interval is precisely the length of the field in S, that is, l. Thus,
In the second case (electron at rest, moving field), the situation in the laboratory is the reverse. The field moves with speed \beta c and its length in S is$ l = l'/\gamma = (\gamma l)/\gamma = l$. The electron remains inside the field while it completely passes over it. The distance traveled by the electron in the laboratory during that interval is precisely the length of the field in S, that is, l. Thus,
$\boxed{l_1 = l}$
$\boxed{l_1 = l}$
b) Work–energy method
b) Work–energy method
First case: the electron enters with speed$\beta c $and slows down to rest under the electric force eE. The initial kinetic energy is$ (\gamma-1)m_e c^2$. The work done by the field is eE l. Equating,
First case: the electron enters with speed$\beta c $and slows down to rest under the electric force eE. The initial kinetic energy is$ (\gamma-1)m_e c^2$. The work done by the field is eE l. Equating,
$eE l = (\gamma-1)m_e c^2 \quad\Longrightarrow\quad\boxed{l = \frac{(\gamma-1)m_e c^2}{eE}}$
$eE l = (\gamma-1)m_e c^2 \quad\Longrightarrow\quad\boxed{l = \frac{(\gamma-1)m_e c^2}{eE}}$
Second case: the electron, initially at rest, is accelerated by the moving field. The field takes a time$ T = l/(\beta c) $to pass over the electron (since its length in S is l) During that time, the electron is subjected to a constant force eE. The momentum acquired is
Second case: the electron, initially at rest, is accelerated by the moving field. The field takes a time$ T = l/(\beta c) $to pass over the electron (since its length in S is l) During that time, the electron is subjected to a constant force eE. The momentum acquired is
$p = eE\,T = \frac{eE l}{\beta c}$
$p = eE\,T = \frac{eE l}{\beta c}$
The final kinetic energy K and the momentum p satisfy the relativistic relation
The final kinetic energy K and the momentum p satisfy the relativistic relation
$(K + m_e c^2)^2 = (pc)^2 + (m_e c^2)^2$
$(K + m_e c^2)^2 = (pc)^2 + (m_e c^2)^2$
Substituting p and using the expression for l from the first case$ (eE l = (\gamma-1)m_e c^2)$, one obtains that the final kinetic energy is exactly $(\gamma-1)m_e c^2$. Therefore, the electron reaches the same speed $\beta c $with which the field was moving.
Substituting p and using the expression for l from the first case$ (eE l = (\gamma-1)m_e c^2)$, one obtains that the final kinetic energy is exactly $(\gamma-1)m_e c^2$. Therefore, the electron reaches the same speed $\beta c $with which the field was moving.
@@ -34,13 +42,15 @@Solution
The distance l_1 traveled by the electron during time T can be calculated by integrating the relativistic velocity $v(t) = (p/m_e)/\sqrt{1+(p/m_e c)^2} with p = eE t$. The result is
The distance l_1 traveled by the electron during time T can be calculated by integrating the relativistic velocity $v(t) = (p/m_e)/\sqrt{1+(p/m_e c)^2} with p = eE t$. The result is
