Charged Particles in Electric Fields

1. The Fundamental Equation of Motion

When a particle with charge $q$ and mass $m$ is in an electric field $\vec{E}$, it experiences an electric force given by:

According to Newton's second law, this net force will cause the particle to accelerate:

For a charged particle in an electric field (and ignoring gravity, which is typically negligible for atomic particles), the electric force is the net force. Combining these two fundamental laws gives us the acceleration of the particle:

This is the central equation for our entire lesson. It tells us that a charged particle in an electric field will accelerate. Crucially, if the electric field $\vec{E}$ is uniform (constant in magnitude and direction), then the acceleration $\vec{a}$ is also constant. This is a powerful result, as it means we can use the familiar constant-acceleration kinematic equations to describe the particle's motion.

The direction of the acceleration depends on the sign of the charge $q$:

• For a positive charge ($q>0$), $\vec{a}$ is in the same direction as $\vec{E}$.
• For a negative charge ($q<0$), $\vec{a}$ is in the opposite direction to $\vec{E}$.

To see how this is applied, let's start by working through a basic calculation.

2. Case 1: Motion Parallel to the Field

The simplest scenario is one-dimensional motion, which occurs when a particle's velocity is parallel to the uniform electric field. A common example is a particle released from rest between two charged plates.

Let's consider an electron (charge $-e$, mass $m_e$) released from rest at the negative plate in a uniform field $\vec{E}$ pointing to the right.

• The force $\vec{F}=(-e)\vec{E}$ is to the left, but that doesn't make sense. Let's make the field point from a positive plate on the left to a negative plate on the right, so $\vec{E}$ points right. If we release the electron from rest near the negative plate, it won't move. Let's adjust the problem.
• Corrected Scenario: Consider two plates creating a uniform field $\vec{E}$ pointing to the right. An electron is released from rest near the negative plate on the left. The electric force $\vec{F}=(-e)\vec{E}$ will be directed to the left, opposite the field. Oops, the electron is already at the negative plate.

Let's re-state this correctly, as it's a common setup. We have a uniform field $\vec{E}$ between two plates. An electron is released from rest at the surface of the negative plate. The field lines point from the positive plate to the negative plate. So, if the electron is at the negative plate, the field points towards it. The force on the electron ($\vec{F}=-e\vec{E}$) is opposite the field direction, meaning the force is directed away from the negative plate and towards the positive plate.

The electron will accelerate towards the positive plate with a constant acceleration of magnitude $a=eE/m_e$. We can then use kinematics to find its velocity and position at any time $t$, or its final velocity after traveling a distance $\Delta x$.

• Velocity at time $t$: $v_f=at=(\frac{eE}{m_e})t$
• Final velocity after distance $\Delta x$: $v_f^2=2a\Delta x=2(\frac{eE}{m_e})\Delta x$

The following video shows a detailed calculation for exactly this type of problem.

3. Case 2: Motion Perpendicular to the Field

Now for the more interesting case: what happens when a charged particle enters a uniform electric field with an initial velocity that is perpendicular to the field? This is the situation shown in the diagram of the electron's parabolic path above. This is a direct analog to projectile motion in a gravitational field.

Let's set up our coordinate system. A particle with charge $q$ and mass $m$ enters a region of uniform vertical field $\vec{E}=E_y\hat{j}$ with an initial horizontal velocity ${\vec{v}}_i=v_x\hat{i}$.

We analyze the motion by separating it into horizontal (x) and vertical (y) components:

• Horizontal Motion (x-direction):

  • There is no force in the x-direction ($F_x=0$).
  • Therefore, there is no acceleration in the x-direction ($a_x=0$).
  • The particle moves with a constant horizontal velocity, $v_x$.
  • The horizontal position is given by $x(t)=v_{x}t$.

• Vertical Motion (y-direction):

  • There is a constant vertical force $F_y=q{E}_y$.
  • This causes a constant vertical acceleration, $a_y=\frac{q{E}_y}{m}$.
  • The vertical velocity and position are given by the standard kinematic equations:
    • $v_y(t)=a_{y}t=(\frac{q{E}_y}{m})t$ (since initial vertical velocity is zero).
    • $y(t)=\frac{1}{2}{a}_y{t}^2=\frac{1}{2}(\frac{q{E}_y}{m})t^2$.

The particle's trajectory is the combination of these two motions. By solving the horizontal equation for time ($t=x/v_x$) and substituting it into the vertical position equation, we get the path:

This is the equation of a parabola ($y=k{x}^2$), confirming the shape of the trajectory.

This video provides an excellent, detailed walkthrough of a typical problem involving this type of motion.

4. A Comprehensive Example

Exam-style questions, like those for the ESSLCE, often combine these scenarios. A particle might be accelerated from rest (Case 1), then pass through a region with no field, and finally enter a second field for deflection (Case 2).

The video Charged Particles: Motion Parallel and Perpendicular to the Electric Field, which you watched a part of earlier, presents exactly this kind of multi-stage problem. Let's now look at the second stage.

Conclusion

Today we've seen how the simple application of Newton's second law to the electric force allows us to fully analyze the motion of charged particles in uniform electric fields. This analysis hinges on one key idea: a uniform field produces a constant acceleration, allowing us to use the standard equations of kinematics.

Key Takeaways:

• The acceleration of a charged particle in an electric field is given by $\vec{a}=q\vec{E}/m$.
• In a uniform field, this acceleration is constant.
• If a particle's initial velocity is parallel to the field, its motion is one-dimensional and can be solved with standard linear kinematics.
• If a particle's initial velocity is perpendicular to the field, its motion is two-dimensional and can be treated as projectile motion: constant velocity in the axis perpendicular to the field, and constant acceleration in the axis parallel to the field.
• The trajectory in the 2D case is a parabola.

In our next lesson, we will begin our study of Electric Potential and Energy. You will learn that analyzing these same problems from an energy perspective (using concepts like work and potential difference) can often be an even more powerful and efficient method. The final velocity you calculated using kinematics in today's problems can also be found by equating the work done by the field ($W=qEd$) to the change in kinetic energy ($\Delta K$). This powerful connection is what we will explore next.