How long will it take for an electron born without an initial velocity in an
+
cm(1cVm=300 CGS units of strength)
+
electric field with an intensity of E = 104V
+
to travel a distance of l = 1 m in this field?
+
### 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$as
In this expression, the distance l already appears as a function of time$\tau (through p_0)$ Now we isolate $\tau$ We move the term -1 to the other side and square:
+
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
How long will it take for an electron born without an initial velocity in an
cm(1cVm=300 CGS units of strength)
electric field with an intensity of E = 104V
to travel a distance of l = 1 m in this field?
### Solution
### 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:
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:
$\frac{dp}{dt} = eE \quad\Rightarrow\quad p = eEt$
$\frac{dp}{dt} = eE \quad\Rightarrow\quad p = eEt$
because the electron is born at rest $(p(0)=0)$.
because the electron is born at rest $(p(0)=0)$.
Now we need to relate momentum to velocity. In special relativity, this relation is
Now we need to relate momentum to velocity. In special relativity, this relation is
$p = \frac{mv}{\sqrt{1 - v^2/c^2}}\quad\Rightarrow\quad v = \frac{p/m}{\sqrt{1 + p^2/(m^2c^2)}}$
$p = \frac{mv}{\sqrt{1 - v^2/c^2}}\quad\Rightarrow\quad v = \frac{p/m}{\sqrt{1 + p^2/(m^2c^2)}}$
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
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$as
In this expression, the distance l already appears as a function of time$\tau (through p_0)$ Now we isolate $\tau$ We move the term -1 to the other side and square:
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
