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Accelerating charged particles

Covers AS91173

The energy route

The most reliable way to find the speed of a charged particle accelerated through a field is to follow the energy:

Electric potential energy lost = kinetic energy gained:

$qV = \tfrac{1}{2}mv^2 \qquad \text{or} \qquad qEd = \tfrac{1}{2}mv^2$

$q$ — charge (C), $V$ — potential difference moved through (V)
$E$ — field strength (N C⁻¹), $d$ — distance along the field (m)
$m$ — mass of the particle (kg), $v$ — final speed (m s⁻¹)
This assumes the particle starts from rest. If it does not, use the change:

$qV = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2$

Rearranged for the speed from rest:

$v = \sqrt{\frac{2qV}{m}}$

The force route

The alternative is to work through the force and the equations of motion:

Find the field: $E = V/d$ .
Find the force: $F = qE$ .
Find the acceleration: $a = F/m$ .
Use the equations of motion, e.g. $v_f^2 = v_i^2 + 2ad$ .
Both routes give the same answer. Use the energy route when the question asks only for a speed — it is shorter and avoids two chances of error.
Use the force route when the question asks for the acceleration, the time taken, or the path of the particle.

Particle masses and charges

These are on the resource sheet, but the pattern is worth knowing:

Particle	Charge	Mass
Electron	$-1.6 \times 10^{-19}$ C	$9.11 \times 10^{-31}$ kg
Proton	$+1.6 \times 10^{-19}$ C	$1.67 \times 10^{-27}$ kg

The charges are equal in size and opposite in sign.
The proton is about 1800 times more massive than the electron.
Accelerated through the same potential difference, both gain the same energy — but the electron reaches a speed about $\sqrt{1800} \approx 43$ times greater, because $v \propto 1/\sqrt{m}$ .

Charged particles fired across the field

A particle fired into the gap perpendicular to the field behaves exactly like a projectile:
- along the original direction of motion: no force, so constant velocity,
- along the field: constant force, so constant acceleration,
- the two are independent and share only the time.
The path is therefore a parabola, and it is solved with the same two-column method as projectile motion in mechanics — with $a = qE/m$ in place of $g$ .
This is how the deflection plates in an oscilloscope or an inkjet printer steer a beam.

Worked ExampleAccelerating an electron

An electron is accelerated from rest through a potential difference of $2000$ V. Find its final speed. (Electron: $q = 1.6 \times 10^{-19}$ C, $m = 9.11 \times 10^{-31}$ kg.)

Step 1 — Energy gained

$\Delta E_p = qV = (1.6 \times 10^{-19})(2000) = 3.2 \times 10^{-16} \text{ J}$

Step 2 — All of it becomes kinetic energy

Starting from rest:

$\tfrac{1}{2}mv^2 = 3.2 \times 10^{-16}$

Step 3 — Solve for the speed

$v^2 = \frac{2(3.2 \times 10^{-16})}{9.11 \times 10^{-31}} = \frac{6.4 \times 10^{-16}}{9.11 \times 10^{-31}} = 7.02 \times 10^{14}$ $v = \sqrt{7.02 \times 10^{14}} = 2.65 \times 10^{7} \approx 2.6 \times 10^{7} \text{ m s}^{-1}$

$v = 2.6 \times 10^{7} \text{ m s}^{-1}$

Worked ExampleComparing an electron and a proton

An electron and a proton are each accelerated from rest through the same potential difference of $500$ V. Compare the energy each gains and the speed each reaches.

Step 1 — Energy gained

Both carry the same size of charge, so both gain the same energy:

$\Delta E_p = qV = (1.6 \times 10^{-19})(500) = 8.0 \times 10^{-17} \text{ J each}$

Step 2 — Speed of the electron

$v = \sqrt{\frac{2\Delta E_p}{m}} = \sqrt{\frac{2(8.0 \times 10^{-17})}{9.11 \times 10^{-31}}} = \sqrt{1.756 \times 10^{14}} = 1.33 \times 10^{7} \text{ m s}^{-1}$

Step 3 — Speed of the proton

$v = \sqrt{\frac{2(8.0 \times 10^{-17})}{1.67 \times 10^{-27}}} = \sqrt{9.58 \times 10^{10}} = 3.10 \times 10^{5} \text{ m s}^{-1}$

Step 4 — Compare

$\frac{1.33 \times 10^{7}}{3.10 \times 10^{5}} = 43$

$\text{Same energy } (8.0 \times 10^{-17} \text{ J}); \text{ the electron is } 43 \times \text{ faster}$

Worked ExampleA charge fired across the field

An electron travelling horizontally at $4.0 \times 10^{6}$ m s⁻¹ enters midway between two horizontal plates $40$ mm long, which produce a uniform field of $3.0 \times 10^{3}$ N C⁻¹ pointing downward. Find the vertical deflection as it leaves the plates.

Step 1 — Set out the two directions, as for a projectile

Horizontal: no force, so constant velocity $v_x = 4.0 \times 10^{6}$ m s⁻¹. Vertical: constant force $F = qE$ , so constant acceleration.

Step 2 — Time between the plates, from the horizontal motion

$t = \frac{d_x}{v_x} = \frac{40 \times 10^{-3}}{4.0 \times 10^{6}} = 1.0 \times 10^{-8} \text{ s}$

Step 3 — Vertical acceleration

$F = qE = (1.6 \times 10^{-19})(3.0 \times 10^{3}) = 4.8 \times 10^{-16} \text{ N}$ $a = \frac{F}{m} = \frac{4.8 \times 10^{-16}}{9.11 \times 10^{-31}} = 5.27 \times 10^{14} \text{ m s}^{-2}$

Step 4 — Vertical deflection, starting with zero vertical velocity

$d_y = v_{iy}t + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(5.27 \times 10^{14})(1.0 \times 10^{-8})^2$ $d_y = 2.6 \times 10^{-2} \text{ m} = 26 \text{ mm}$

$d_y = 26 \text{ mm (upward, since the electron is negative and the field points down)}$

Accelerating charged particles

Covers AS91173

The energy route

The most reliable way to find the speed of a charged particle accelerated through a field is to follow the energy:

Electric potential energy lost = kinetic energy gained:

$qV = \tfrac{1}{2}mv^2 \qquad \text{or} \qquad qEd = \tfrac{1}{2}mv^2$

$q$ — charge (C), $V$ — potential difference moved through (V)
$E$ — field strength (N C⁻¹), $d$ — distance along the field (m)
$m$ — mass of the particle (kg), $v$ — final speed (m s⁻¹)
This assumes the particle starts from rest. If it does not, use the change:

$qV = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2$

Rearranged for the speed from rest:

$v = \sqrt{\frac{2qV}{m}}$

The force route

The alternative is to work through the force and the equations of motion:

Find the field: $E = V/d$ .
Find the force: $F = qE$ .
Find the acceleration: $a = F/m$ .
Use the equations of motion, e.g. $v_f^2 = v_i^2 + 2ad$ .
Both routes give the same answer. Use the energy route when the question asks only for a speed — it is shorter and avoids two chances of error.
Use the force route when the question asks for the acceleration, the time taken, or the path of the particle.

Particle masses and charges

These are on the resource sheet, but the pattern is worth knowing:

Particle	Charge	Mass
Electron	$-1.6 \times 10^{-19}$ C	$9.11 \times 10^{-31}$ kg
Proton	$+1.6 \times 10^{-19}$ C	$1.67 \times 10^{-27}$ kg

The charges are equal in size and opposite in sign.
The proton is about 1800 times more massive than the electron.
Accelerated through the same potential difference, both gain the same energy — but the electron reaches a speed about $\sqrt{1800} \approx 43$ times greater, because $v \propto 1/\sqrt{m}$ .

Charged particles fired across the field

A particle fired into the gap perpendicular to the field behaves exactly like a projectile:
- along the original direction of motion: no force, so constant velocity,
- along the field: constant force, so constant acceleration,
- the two are independent and share only the time.
The path is therefore a parabola, and it is solved with the same two-column method as projectile motion in mechanics — with $a = qE/m$ in place of $g$ .
This is how the deflection plates in an oscilloscope or an inkjet printer steer a beam.

Worked ExampleAccelerating an electron

An electron is accelerated from rest through a potential difference of $2000$ V. Find its final speed. (Electron: $q = 1.6 \times 10^{-19}$ C, $m = 9.11 \times 10^{-31}$ kg.)

Step 1 — Energy gained

$\Delta E_p = qV = (1.6 \times 10^{-19})(2000) = 3.2 \times 10^{-16} \text{ J}$

Step 2 — All of it becomes kinetic energy

Starting from rest:

$\tfrac{1}{2}mv^2 = 3.2 \times 10^{-16}$

Step 3 — Solve for the speed

$v = 2.6 \times 10^{7} \text{ m s}^{-1}$

Worked ExampleComparing an electron and a proton

An electron and a proton are each accelerated from rest through the same potential difference of $500$ V. Compare the energy each gains and the speed each reaches.

Step 1 — Energy gained

Both carry the same size of charge, so both gain the same energy:

$\Delta E_p = qV = (1.6 \times 10^{-19})(500) = 8.0 \times 10^{-17} \text{ J each}$

Step 2 — Speed of the electron

$v = \sqrt{\frac{2\Delta E_p}{m}} = \sqrt{\frac{2(8.0 \times 10^{-17})}{9.11 \times 10^{-31}}} = \sqrt{1.756 \times 10^{14}} = 1.33 \times 10^{7} \text{ m s}^{-1}$

Step 3 — Speed of the proton

$v = \sqrt{\frac{2(8.0 \times 10^{-17})}{1.67 \times 10^{-27}}} = \sqrt{9.58 \times 10^{10}} = 3.10 \times 10^{5} \text{ m s}^{-1}$

Step 4 — Compare

$\frac{1.33 \times 10^{7}}{3.10 \times 10^{5}} = 43$

$\text{Same energy } (8.0 \times 10^{-17} \text{ J}); \text{ the electron is } 43 \times \text{ faster}$

Worked ExampleA charge fired across the field

Step 1 — Set out the two directions, as for a projectile

Horizontal: no force, so constant velocity $v_x = 4.0 \times 10^{6}$ m s⁻¹. Vertical: constant force $F = qE$ , so constant acceleration.

Step 2 — Time between the plates, from the horizontal motion

$t = \frac{d_x}{v_x} = \frac{40 \times 10^{-3}}{4.0 \times 10^{6}} = 1.0 \times 10^{-8} \text{ s}$

Step 3 — Vertical acceleration

$F = qE = (1.6 \times 10^{-19})(3.0 \times 10^{3}) = 4.8 \times 10^{-16} \text{ N}$ $a = \frac{F}{m} = \frac{4.8 \times 10^{-16}}{9.11 \times 10^{-31}} = 5.27 \times 10^{14} \text{ m s}^{-2}$

Step 4 — Vertical deflection, starting with zero vertical velocity

$d_y = v_{iy}t + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(5.27 \times 10^{14})(1.0 \times 10^{-8})^2$ $d_y = 2.6 \times 10^{-2} \text{ m} = 26 \text{ mm}$

$d_y = 26 \text{ mm (upward, since the electron is negative and the field points down)}$