TASC TCE 2019-style practice: A beam of electrons is accelerated through 2.4\ kV and allowed to pass into a magnetic field at right angles. Show that the speed of the electrons is about 3 × 10^7\ m s^-1.

The electric potential energy lost equals the kinetic energy gained, qV = 12 m v^2, so v = 2qVm. $v = 2(1.610^-19)(2400)9.1110^-31 = 8.43 × 10^14 = 2.90 × 10^7\ m s^-1.$ The electrons reach about 2.9 × 10^7\ m s^-1, confirming the "about 3 × 10^7" value. Markers want qV = 12 m v^2 and substitution of the electron mass and charge.

§-Syllabus dot point

TASPhysicsSyllabus dot point

How does an electric field change the energy and motion of a charge?

Analyse the work done on a charge and the motion of charged particles in uniform electric fields.

Work done moving a charge through a potential difference, the electronvolt, and the parabolic and accelerated motion of charged particles in uniform electric fields.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

This dot point connects fields to energy and motion: how much energy a charge gains crossing a potential difference, and how a charged particle moves once inside a field.

Work and potential difference

When a charge $q$ moves through a potential difference $V$ , the work done by the field is:

W = qV

If the charge starts at rest and is free to move, all this work becomes kinetic energy:

qV = \tfrac{1}{2}mv^2

This is how electron guns and particle accelerators give charges high speeds: a known voltage accelerates them to a calculable speed. Inside a uniform field the force is constant, so you can also find the work as $W = qEd$ over a distance $d$ along the field, and $V = Ed$ links the two views.

Electric potential $V$ at a point is the potential energy per unit charge placed there, measured in volts ( $\text{J C}^{-1}$ ). The potential difference between two points is what does work on a charge moving between them, which is why only differences in potential, not absolute values, appear in calculations. In a uniform field the potential drops steadily from the positive plate to the negative plate, and the field points from high potential to low potential.

Charges accelerated along the field

A charge released in a uniform field accelerates along the field lines (positive charges follow the field, negative charges go against it). The acceleration is constant:

a = \frac{F}{m} = \frac{qE}{m}

so the constant-acceleration equations of motion apply directly. This is the simplest case: straight-line acceleration, like free fall but driven by the electric force instead of gravity. Combined with $V = Ed$ , you can swap freely between an energy view ( $qV = \tfrac12 m v^2$ ) and a kinematic view ( $v^2 = u^2 + 2as$ ) of the same accelerated charge.

Charges fired across the field

If a charge enters a uniform field moving perpendicular to it, the motion splits into two independent parts, just like a projectile:

Along the original direction the velocity is constant (no force that way).
Across the field the charge accelerates uniformly under $F = qE$ .

The result is a parabolic path. This is exactly how the old cathode ray tube deflected electrons to paint a picture, and it is a favourite exam scenario because it combines fields with projectile analysis. The vertical deflection while the charge is between plates of length $\ell$ is $y = \tfrac12 \dfrac{qE}{m}\left(\dfrac{\ell}{u}\right)^2$ , found by using the time of flight $t = \dfrac{\ell}{u}$ from the constant horizontal motion.

Worked example: electron accelerated through a voltage

An electron starting from rest is accelerated through a potential difference of $500\ \text{V}$ . Find its final speed. Use $m_e = 9.1 \times 10^{-31}\ \text{kg}$ and $q = 1.6 \times 10^{-19}\ \text{C}$ .

Step 1: Apply conservation of energy to relate voltage to speed

When a charge moves through a potential difference, the electric field does work on it equal to $W = qV$ . Because the electron starts from rest and the only force acting on it is the electric force, all of this work converts into kinetic energy. Setting the work done equal to the kinetic energy gained gives the key equation:

qV = \tfrac{1}{2}mv^2

This energy method is far more efficient than calculating the force and acceleration separately, because it works regardless of how the field varies along the path.

Step 2: Rearrange for the speed and substitute

Solving for $v$ by dividing both sides by $\tfrac{1}{2}m$ and taking the square root, then substituting $q = 1.6\times10^{-19}\ \text{C}$ , $V = 500\ \text{V}$ , and $m = 9.1\times10^{-31}\ \text{kg}$ :

v = \sqrt{\frac{2qV}{m}} = \sqrt{\frac{2(1.6\times10^{-19})(500)}{9.1\times10^{-31}}} = \sqrt{1.76\times10^{14}} = 1.3\times10^{7}\ \text{m s}^{-1}

The factor of 2 in the numerator comes from rearranging $\tfrac{1}{2}mv^2$ , and it is a common source of error if omitted.

Final answer: The electron reaches a speed of $1.3 \times 10^7\ \text{m s}^{-1}$ .

In the exam, for a charge accelerated through a voltage use the energy method $qV = \tfrac12 mv^2$ . For a charge fired across a field, separate the motion into constant-velocity and constant-acceleration components and treat it as a projectile with acceleration $\dfrac{qE}{m}$ . Keep the geometry of the plates clearly in mind so you use the right length for the time of flight.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20244 marksIn a vacuum tube, an electron is accelerated between a cathode and anode with an accelerating voltage of

3000\ \text{V}

. Calculate the speed with which the electron reaches the anode (assume negligible initial speed). The electron then enters a uniform field midway between two plates

7.00\ \text{cm}

long and

5.00\ \text{cm}

apart at a potential difference of

2000\ \text{V}

. Calculate the electric field strength between the plates.

Show worked answer →

The work done by the accelerating voltage becomes kinetic energy, $qV = \tfrac12 m v^2$ :

(1.6\times10^{-19})(3000) = \tfrac12 (9.11\times10^{-31}) v^2.

v^2 = \frac{2(4.8\times10^{-16})}{9.11\times10^{-31}} = 1.054 \times 10^{15}, \quad v = 3.25 \times 10^7\ \text{m s}^{-1}.

The field between the parallel plates is uniform: $E = \dfrac{V}{d} = \dfrac{2000}{0.05} = 4.0 \times 10^4\ \text{V m}^{-1}$ (equivalently $\text{N C}^{-1}$ ).

So the electron reaches the anode at about $3.25 \times 10^7\ \text{m s}^{-1}$ and the deflecting field is $4.0 \times 10^4\ \text{V m}^{-1}$ . Markers want $qV = \tfrac12 m v^2$ for the speed and $E = \dfrac{V}{d}$ for the field.

TCE 20235 marksA charged ball of mass

3.00 \times 10^{-6}\ \text{kg}

is suspended by a cotton thread between two parallel plates

4.00\ \text{cm}

apart. When

500\ \text{V}

is placed across them, the thread hangs at

30^\circ

to the vertical. Calculate the electrostatic force on the ball, then the electric field strength between the plates and hence the charge on the ball.

Show worked answer →

The ball is in equilibrium under weight (down), tension (along the thread) and a horizontal electrostatic force.

Weight: $W = mg = (3.00\times10^{-6})(9.81) = 2.943 \times 10^{-5}\ \text{N}$ .

Resolving, the horizontal electrostatic force balances the horizontal component of tension while the vertical component balances weight, so $F = W\tan 30^\circ = (2.943\times10^{-5})(0.5774) = 1.70 \times 10^{-5}\ \text{N}$ .

Field strength: $E = \dfrac{V}{d} = \dfrac{500}{0.04} = 1.25 \times 10^4\ \text{V m}^{-1}$ .

Charge from $F = qE$ : $q = \dfrac{F}{E} = \dfrac{1.70\times10^{-5}}{1.25\times10^4} = 1.36 \times 10^{-9}\ \text{C}$ .

The ball carries about $1.4\ \text{nC}$ . Markers want $F = W\tan\theta$ , $E = \dfrac{V}{d}$ , then $q = \dfrac{F}{E}$ .

TCE 20192 marksA beam of electrons is accelerated through

2.4\ \text{kV}

and allowed to pass into a magnetic field at right angles. Show that the speed of the electrons is about

3 \times 10^7\ \text{m s}^{-1}

Show worked answer →

The electric potential energy lost equals the kinetic energy gained, $qV = \tfrac12 m v^2$ , so $v = \sqrt{\dfrac{2qV}{m}}$ .

v = \sqrt{\frac{2(1.6\times10^{-19})(2400)}{9.11\times10^{-31}}} = \sqrt{8.43 \times 10^{14}} = 2.90 \times 10^7\ \text{m s}^{-1}.

The electrons reach about $2.9 \times 10^7\ \text{m s}^{-1}$ , confirming the "about $3 \times 10^7$ " value. Markers want $qV = \tfrac12 m v^2$ and substitution of the electron mass and charge.