Work and energy on a charge in a field

Key ideas

Moving a charge through an electric field involves work — energy is transferred. If the field does the work, the charge speeds up, gaining kinetic energy.

Work done moving a charge

The work done (change in electric potential energy) moving a charge $q$ a distance $d$ through a uniform field $E$ :

$\Delta E_p = qEd$

$\Delta E_p$ — change in electric potential energy = work done (J)
$q$ — charge (C)
$E$ — field strength (N C⁻¹)
$d$ — distance moved along the field (m)

Because $E = V/d$ , this can also be written using the voltage moved through:

$\Delta E_p = qV$

Energy converted to motion

When a field does work on a charge that is free to move, that work becomes kinetic energy:

$E_k = \tfrac{1}{2}mv^2$

$m$ — mass of the charged particle (kg)
$v$ — final speed (m s⁻¹)

For a charge accelerated from rest, all the work done equals the kinetic energy gained:

$qEd = \tfrac{1}{2}mv^2$

:::tip "Distance" in $\Delta E_p = qEd$ must be the distance moved along the field direction. If a charge moves sideways (at right angles to the field), the field does no work on it and its potential energy does not change — only motion along the field lines counts. :::

Worked ExampleSpeed of a charge accelerated between plates

A charge of $2.0$ µC and mass $5.0 \times 10^{-8}$ kg starts from rest and is accelerated through a field of $4.0 \times 10^{4}$ N C⁻¹ across a plate gap of $3.0$ mm. Find its speed as it reaches the far plate.

Step 1 — Work done by the field

$\Delta E_p = qEd = (2.0 \times 10^{-6})(4.0 \times 10^{4})(3.0 \times 10^{-3}) = 2.4 \times 10^{-4} \text{ J}$

Step 2 — All of it becomes kinetic energy (started from rest)

$\tfrac{1}{2}mv^2 = 2.4 \times 10^{-4} \text{ J}$

Step 3 — Solve for $v$

$v^2 = \frac{2 \times 2.4 \times 10^{-4}}{5.0 \times 10^{-8}} = 9600 \;\Rightarrow\; v = \sqrt{9600} = 98 \text{ m s}^{-1}$

Practice question

A $3.0$ µC charge is moved $2.0$ mm along a field of $5.0 \times 10^{4}$ N C⁻¹. How much work is done?

Worked solution: $\Delta E_p = qEd = (3.0 \times 10^{-6})(5.0 \times 10^{4})(2.0 \times 10^{-3}) = 3.0 \times 10^{-4}$ J.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A charge of $5.0$ µC moves through a potential difference of $80$ V.

Calculate the work done on the charge.

Merit

A charge of $1.0$ µC and mass $2.0 \times 10^{-9}$ kg is accelerated from rest through a field of $3.0 \times 10^{4}$ N C⁻¹ over $5.0$ mm.

Calculate its final speed.

Excellence

Two charges of equal size but different mass are released from rest at the same plate and accelerated across an identical field. Explain fully which reaches the far plate faster, and whether they arrive with the same kinetic energy.

Work and energy on a charge in a field

Covers AS91173

Key ideas

Moving a charge through an electric field involves work — energy is transferred. If the field does the work, the charge speeds up, gaining kinetic energy.

Work done moving a charge

The work done (change in electric potential energy) moving a charge $q$ a distance $d$ through a uniform field $E$ :

$\Delta E_p = qEd$

$\Delta E_p$ — change in electric potential energy = work done (J)
$q$ — charge (C)
$E$ — field strength (N C⁻¹)
$d$ — distance moved along the field (m)

Because $E = V/d$ , this can also be written using the voltage moved through:

$\Delta E_p = qV$

Energy converted to motion

When a field does work on a charge that is free to move, that work becomes kinetic energy:

$E_k = \tfrac{1}{2}mv^2$

$m$ — mass of the charged particle (kg)
$v$ — final speed (m s⁻¹)

For a charge accelerated from rest, all the work done equals the kinetic energy gained:

$qEd = \tfrac{1}{2}mv^2$

Worked ExampleSpeed of a charge accelerated between plates

Step 1 — Work done by the field

$\Delta E_p = qEd = (2.0 \times 10^{-6})(4.0 \times 10^{4})(3.0 \times 10^{-3}) = 2.4 \times 10^{-4} \text{ J}$

Step 2 — All of it becomes kinetic energy (started from rest)

$\tfrac{1}{2}mv^2 = 2.4 \times 10^{-4} \text{ J}$

Step 3 — Solve for $v$

$v^2 = \frac{2 \times 2.4 \times 10^{-4}}{5.0 \times 10^{-8}} = 9600 \;\Rightarrow\; v = \sqrt{9600} = 98 \text{ m s}^{-1}$

Practice question

A $3.0$ µC charge is moved $2.0$ mm along a field of $5.0 \times 10^{4}$ N C⁻¹. How much work is done?

Worked solution: $\Delta E_p = qEd = (3.0 \times 10^{-6})(5.0 \times 10^{4})(2.0 \times 10^{-3}) = 3.0 \times 10^{-4}$ J.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A charge of $5.0$ µC moves through a potential difference of $80$ V.

Calculate the work done on the charge.

Merit

A charge of $1.0$ µC and mass $2.0 \times 10^{-9}$ kg is accelerated from rest through a field of $3.0 \times 10^{4}$ N C⁻¹ over $5.0$ mm.

Calculate its final speed.

Excellence

Work and energy on a charge in a field

Key ideas

Work done moving a charge

Energy converted to motion

Step 1 — Work done by the field

Step 2 — All of it becomes kinetic energy (started from rest)

Step 3 — Solve for vvv

Practice question

Test yourself

Practice by grade

Work and energy on a charge in a field

Key ideas

Work done moving a charge

Energy converted to motion

Step 1 — Work done by the field

Step 2 — All of it becomes kinetic energy (started from rest)

Step 3 — Solve for vvv

Practice question

Test yourself

Practice by grade

Step 3 — Solve for $v$

Step 3 — Solve for $v$