Work done on a charge equals qDeltaV

Electric Fields & Potential — AQA A-Level Physics

\Delta W = q \Delta V

$DeltaW$: work doneEnergy transferred when a force moves an object. In electrical circuits, W = QV (chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). times potential difference). (J)
$q$: chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). being moved (C)
$DeltaV$: potential differenceThe energyThe capacity to do work. Measured in joules (J). transferred per unit chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). between two points. Measured in volts (V). between the two points (V)

When a charge moves through an electric field, work is done on it.
The potential differenceThe energyThe capacity to do work. Measured in joules (J). transferred per unit charge between two points. Measured in volts (V). between two points: $DeltaV = V_{\text{final}} - V_{\text{initial}}$.
Work is done when a positive charge moves against the field lines (towards higher potential) or when a negative charge moves with the field lines.

Electric potential energy of two point charges

E_p = \frac{Q_1 Q_2}{4\pi\varepsilon_0 r}

$E_p$: electric potentialThe work doneEnergy transferred when a force moves an object. In electrical circuits, W = QV (charge times potential difference). per unit positive charge in bringing a small test charge from infinity to that point. energy (J)
$Q_1, Q_2$: magnitudes of the charges (C)
$r$: distance between centres of the charges (m)

The work done on a point charge equals the change in electric potentialThe work done per unit positive charge in bringing a small test charge from infinity to that point. energy.
When $V = 0 (at$ infinity), $E_{p} = 0$.

Worked Example

A point charge of +7.0 nC is located 150 mm from point S and 220 mm from point R. Calculate the work done when a +3.0 nC charge moves from R to S.

Show Solution

List known values

$r_S = 0.15$ m, $r_R = 0.22$ m
$Q = +7.0 \times 10^{-9}$ C (produces the field)
$q = +3.0 \times 10^{-9}$ C (being moved)

Calculate DeltaV between R and S

$$\Delta V = \frac{Q}{4\pi\varepsilon_0}\left(\frac{1}{r_S} - \frac{1}{r_R}\right) = \frac{7.0 \times 10^{-9}}{4\pi \times 8.85 \times 10^{-12}}\left(\frac{1}{0.15} - \frac{1}{0.22}\right) = 133.5 \text{ V}$$

Calculate work done

$$W = q \Delta V = (3.0 \times 10^{-9}) \times 133.5 = 4.0 \times 10^{-7} \text{ J}$$

Answer

$W = 4.0 \times 10^{-7}$ J

Common Mistake MEDIUM

Students often: Confusing Q (the charge producing the potential) with q (the charge being moved).
Instead: Q produces the field. q is the small charge being moved through that field. Use the correct one in each equation.