Topics

Work and electric potential energy

Covers AS91173

Work done on a charge

Moving a charge through an electric field means a force acts over a distance, so work is done and energy is transferred.

$\Delta E_p = qEd$

$\Delta E_p$ — change in electric potential energy, or the work done (J)
$q$ — size of the charge (C)
$E$ — electric field strength (N C⁻¹)
$d$ — distance moved parallel to the field (m)
Note this is just $W = Fd$ with $F$ replaced by $qE$ .
Because $E = V/d$ , an equivalent form is:

$\Delta E_p = qV$

where $V$ is the potential difference between the start and end points of the movement.
This second form is often quicker when a charge moves all the way from one plate to the other.

The distance must be along the field

$d$ is the distance moved parallel to the field lines — for parallel plates, the distance moved toward or away from a plate.
Moving a charge sideways, at right angles to the field, does no work at all and changes its potential energy by zero.
- The force is perpendicular to the motion, exactly as with a centripetal force in mechanics.
If a charge moves diagonally, use only the component of the displacement along the field.

Gaining and losing potential energy

The direction of the movement decides the sign:

A charge moving in the direction of the force on it (a positive charge moving toward the negative plate):
- the field does work on the charge,
- it loses electric potential energy,
- it gains kinetic energy and speeds up.
A charge moved against the force on it:
- work must be done by an external agent,
- it gains electric potential energy,
- it is like lifting a mass against gravity.
The parallel with gravity is exact and worth using in explanation answers:

Gravity	Electric field
$F = mg$	$F = qE$
$E_p = mgh$	$\Delta E_p = qEd$
Mass falls, losing $E_p$ , gaining $E_k$	Charge accelerates along the force, losing $E_p$ , gaining $E_k$

Voltage as energy per coulomb

$V = \frac{\Delta E}{q}$

$V$ — potential difference (V)
$\Delta E$ — energy transferred (J)
$q$ — charge moved (C)
This says that one volt is one joule per coulomb.
It is the same relationship as $\Delta E_p = qV$ , rearranged, and it is the bridge between the static-electricity and circuits halves of this standard.

Worked ExampleWork done moving a charge

Two parallel plates are $6.0$ mm apart with $300$ V across them. A charge of $2.0$ µC is moved from the positive plate to the negative plate. Find (a) the field strength, (b) the work done on the charge, and (c) check the answer using $\Delta E_p = qV$ .

Step 1 — Field strength

$E = \frac{V}{d} = \frac{300}{6.0 \times 10^{-3}} = 5.0 \times 10^{4} \text{ N C}^{-1}$

Step 2 — Work done

The charge moves the full gap, so $d = 6.0 \times 10^{-3}$ m:

$\Delta E_p = qEd = (2.0 \times 10^{-6})(5.0 \times 10^{4})(6.0 \times 10^{-3})$ $\Delta E_p = 6.0 \times 10^{-4} \text{ J}$

Step 3 — Check with the voltage form

$\Delta E_p = qV = (2.0 \times 10^{-6})(300) = 6.0 \times 10^{-4} \text{ J} \;\checkmark$

$\Delta E_p = 6.0 \times 10^{-4} \text{ J}$

Worked ExampleA partial and a sideways movement

In a uniform field of $4.0 \times 10^{4}$ N C⁻¹ between horizontal plates, a $5.0$ nC charge is moved $2.0$ mm directly toward the lower plate, and then $8.0$ mm horizontally. Find the total work done on the charge.

Step 1 — The movement along the field

$d = 2.0 \times 10^{-3} \text{ m}, \qquad q = 5.0 \times 10^{-9} \text{ C}$ $\Delta E_p = qEd = (5.0 \times 10^{-9})(4.0 \times 10^{4})(2.0 \times 10^{-3})$ $\Delta E_p = 4.0 \times 10^{-7} \text{ J}$

Step 2 — The horizontal movement

The field between parallel plates is vertical, so a horizontal movement is perpendicular to the field. The force is at right angles to the motion, so:

$W = 0 \text{ J}$

Step 3 — Total

$\Delta E_p = 4.0 \times 10^{-7} + 0 = 4.0 \times 10^{-7} \text{ J}$