Motion of charged particles in an electric field

In a nutshell

When a charged particle enters an electric field, it experiences a force parallel to the direction of the electric field lines. If the electric field is uniform, then the charged particle will undergo uniform acceleration and move like a projectile.

Equations

Description	Equation
Electrostatic force	$F=EQ$
Newton's second law	$F=ma$

Variable definitions

Quantity Name	Symbol	Derived Unit	SI BASE Units
$force$	$F$	$N$	$kg\, m\,s^{-2}$
$electric \ field \ strength$	$E$	$N\,C^{-1}$	$kg\,m\,s^{-3}\,A^{-1}$
$charge$	$Q$	$C$	$A\,s$
$mass$	$m$	$kg$	$kg$
$acceleration$	$a$	$m\,s^{-2}$	$m\,s^{-2}$

Projectile motion

Charged particles in an electric field experience a force. When a charged particle moves though a uniform electric field, it will follow a trajectory similar to that of a mass in a gravitational field.

The charge travelling through the electric field will experience a force, which is dependent on its charge and the electric field strength.

The force is given as:

$F=EQ$

The force acts parallel to the direction of the electric field lines. A positive charge will move in the direction of the field lines and a negative charge will move in the opposite direction to the field lines.

Physics; Electric fields; KS5 Year 12; Motion of charged particles in an electric field

1	More massive, positively charged particle
2	Less massive, negatively charged particle

Horizontal motion

The component of velocity, which is perpendicular to the electric field, remains unchanged throughout electric field. This is because there are no forces acting perpendicular to the electric field. Therefore, there is no acceleration in the horizontal component of the particle.

Vertical motion

There will be a change of velocity in the vertical motion of the particle, as this is the direction of the force.

As the force is constant, the charged particle will accelerate at a constant rate in the direction of the force. The acceleration can be calculated using Newton's second law:

$F=ma$

This means that the equations of motion ( $suvat$ ) can be used as the acceleration is uniform.

Example

An electron is initially travelling horizontally at $5.4\times 10^5 \, m\,s^{-1}$ before passing between two charged parallel plates. The electric field strength between the plates is $0.40 \, N\, C^{-1}$ and the vertical distance between the path of the electron and the positive plate is $16\,mm$ . Calculate the time taken for the electron to reach the plate.

Physics; Electric fields; KS5 Year 12; Motion of charged particles in an electric field

Firstly, write down the known values:

$u_v=0 \,m\,s^{-1} \newline \\[0.1in]s_v=16\, mm=0.016\, m \newline \\[0.1in]E=0.40 \, N\,C^{-1} \newline \\[0.1in]m_e=9.11\times 10^{-31} \, kg \newline \\[0.1in]Q_e=1.6\times10^{-16} \, C$

Note: You only need the vertical components to work out the answer to this question. $u_v$ and $s_v$ are the vertical components for initial velocity and displacement respectively.

Next, write down the equations needed and rearrange if necessary:

$F=EQ \newline \\[0.1in]F=ma_v\rightarrow a_v=\frac Fm \newline \\[0.1in]s_v=u_vt+\frac 12 a_vt^2$

Find force:

$F=0.40 \times 1.6\times 10^{-19} \newline \\[0.1in]F=6.4\times 10^{-20} \, N$

Use the value of force to obtain a value for vertical acceleration:

$a_v=\frac {6.4\times10^{-20}}{9.11\times 10^{-31}} \newline \\[0.1in]a_v=7.02525...\times 10^{10} \, m\,s^{-2}$ 

Then use value for vertical acceleration to obtain a value for the time:

Note: As the initial vertical velocity is $0\, m\,s^{-1}$ , the first term ( $u_vt$ ) is $0$ .

$t^2=\frac {2s_v}{a} \rightarrow t=\sqrt {\frac {2s_v}{a}} \newline \\[0.1in]t=\sqrt {\frac {2 \times 0.016}{7.02525\times10^{10}}} \newline \\[0.1in]t=6.74907...\times 10^{-7} \, s \newline \\[0.1in]$

Make sure to round to the lowest number of significant figures of the values given by the question:

$time \ taken, t=6.7 \times 10^{-7} \, s$

The time taken for the electron to hit the positive plate is $\underline{6.7 \times 10^{-7} \, s}.$ 