## Electron flying into an electric field

### Task number: 2140

An electron flies with a certain velocity in an area where two electric fields are acting simultanenously: a horizontal field *E*_{h} = 400 Vm^{-1} and a vertical field *E*_{v} = 300 Vm^{-1}. The electron flies into the combined field along with its field lines and its velocity increases twice on a distance of 2.7 mm. Asses the final velocity of the electron.

#### Two possible solutions

The task can be approached in two ways. We show both of them.

**A**. Studying the*movement*of the electron**B**. Studying the laws of*energy conservation*#### Hint for both solutions

The electric field is assigned as two components. What is the total electric field?

#### Option A – Hint

What type of motion does the electron perform after flying into the homogeneous electric field along with the field lines?

#### Option A – Solution: Electron motion

The homogeneous electric field exerts a constant force on the electron in the direction of the field lines. As the electron flied along the field lines, the force changes only the magnitude of its velocity, not its direction. The electron thus moves along a straight line with a uniform acceleration and a non-zero initial velocity. It follows that we can limit the calculation to one dimension. The following relationships apply to this type of motion

\[v=v_0 + at,\] \[s=v_0 t + \frac{1}{2}at^{2}.\]Let us now calculate the final velocity

\[2v_0 = v_0 + at\qquad \Rightarrow \qquad v_0 = at.\]*v*_{f}of the electron. We start by calculating*v*_{0}. Because the speed increases twice on the given distance, we can write*v*_{f}= 2*v*_{0}We now put this expression into the formula for the distance travelled:

\[s=v_0t+\frac{1}{2}at^{2}= at^{2}+\frac{1}{2}at^{2}=\frac{3}{2}at^{2},\] \[t=\sqrt{\frac{2s}{3a}}.\]We derived the equation for the time

\[v_0 = a \sqrt{\frac{2s}{3a}}=\sqrt{\frac{2sa}{3}}.\]*t*, that it takes the electron to double its velocity. To determine the initial speed*v*_{0}, we use the relationship*v*_{0}=*at*and we substitute for*t*We get acceleration

\[m_\mathrm e a = E e\qquad \Rightarrow \qquad a=\frac{E e}{m_\mathrm e},\]*a*from the 2^{nd}Newton’s law. For this we need the electrical force*F*_{e}exerted by the field on the electron, which is given by \(F_e=Eq\). This relationship is one-dimensional, and*q*is the charge of the electron. We then get the equation for the accelerationwhere

*m*_{ e }is the electron mass and*e*is the charge of the electron, which is equal to the elementary charge. We find these constants in the tables.We substitute the acceleration into the formula for initial velocity

\[v_0 = \sqrt{\frac{2sa}{3}}=\sqrt{\frac{2sE e}{3m_\mathrm e}}.\]We now determine the magnitude of the total electric field

\[E = \sqrt{E^{2}_\mathrm {v}+E^{2}_\mathrm {h}}.\]*E*. We use the Pythagoras theorem because the horizontal*E*_{h}and the vertical*E*_{v}components of the field are perpendicular to each otherAccording to the assignment, the final speed is twice the initial speed

\[v_\mathrm f = 2v_0=2\sqrt{\frac{2sE e}{3m_\mathrm e}}=2\sqrt{\frac{2se\sqrt{E_\mathrm {v}^2+E_\mathrm {h}^2}}{3m_\mathrm e}}.\]#### Option B – Hint

What can we say about the kinetic energy at the beginning and at the end of the motion? Was there any work done during the motion?

#### Option B – Solution

For the kinetic energy of the electron, we have the equation

\[E_\mathrm {k1}+W_\mathrm e=E_\mathrm {k2},\]

\[\frac{1}{2}m_\mathrm e{v_0}^2+ F_\mathrm es=\frac{1}{2}m_\mathrm e {v_\mathrm k}^2,\]*E*_{ k1 }is the kinetic energy of the electron when it entered the field,*W*_{e}is the work done by the electric force*F*_{e}on the distance*s*, and*E*_{k2}is the kinetic energy of the electron at the end of the movement. The electric force acts in the direction of motion, and the work that it does thus can be calculated as \(W_\mathrm e=F_\mathrm es\). We express the energy and workwhere

*m*_{e}is the electron mass and*e*is the elementary charge - we find these last two constants in the tables. We calculate the electric force*F*_{e}as \(F_\mathrm e=Ee\).Now, we just put in the assigned fact that the final velocity of the electron

\[\frac{1}{2}m_\mathrm e{v_0}^2+Ees=\frac{1}{2}m_\mathrm e {(2v_0)}^2\] \[E e s = \frac{3}{2} m_\mathrm e {v_0}^2.\]*v*_{f}is double its initial velocity*v*_{0}(\(v_\mathrm f=2\,v_0\)) . We substitute and rearrange the formulaWe express

\[v_0 = \sqrt{\frac{2E es}{3m_\mathrm e}}.\]*v*_{0}from this equation:Last, we only need to assess the magnitude of the total electric field

\[E = \sqrt{E^{2}_\mathrm {v}+E^{2}_\mathrm {h}}.\]*E*and then we calculate the final veloctity of the electron*v*_{f}. The total field that is composed of two perpendicular components*E*_{h}and*E*_{v}is given by the Pythagorean theorem:The final velocity

\[v_\mathrm f = 2v_0=2\sqrt{\frac{2E es}{3m_\mathrm e}}=2\sqrt{\frac{2se\sqrt{E_\mathrm v^2+E_\mathrm h^2}}{3m_e}}.\]*v*_{f}is then#### Notation and numerical calculation

*E*_{h}= 400 Vm^{-1}Horizontal electric field *E*_{v}= 300 Vm^{-1}Vertical electric field *s*= 2.7 mm = 2.7·10^{-3}mDisplacement of the electron *v*_{k}= ? (ms^{-1})Final velocity of the electron *From the tables:**e*= 1.6·10^{-19}CElementary charge *m*_{e}= 9.1·10^{-31}kgElectron mass

\[v_\mathrm k=\sqrt{\frac{8se\sqrt{E_\mathrm v^2+E_\mathrm h^2}}{3m_\mathrm e}}= \sqrt{\frac{8{\cdot} 2.7{\cdot} 10^{-3}\cdot 1.6{\cdot} 10^{-19}\sqrt{300^2+400^2}}{3{\cdot} 9.1{\cdot} 10^{-31}}}\,\mathrm{ms^{-1}}\,\dot{=}\,8{\cdot} 10^{5}\,\mathrm{ms^{-1}}\,\dot{=}\,800\,\mathrm {km \cdot s^{-1}}\]#### Answer

The final velocity of the electron is 800 km·s

^{-1}.#### What if the electron did not fly into the field along the field lines?

The electron in this task moves in a homogeneous field. The gravitational field of the Earth is also a homogeneous field (under displacements smaller than several kilometers). We can thus use the analogy of a body (a ball) that was thrown down towards the Earth. If the electron flied into the assigned electric field at an angle, it would undergo a similar motion as a ball that was thrown at the same angle towards the Earth.

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