Uniform field between parallel plates: $E = V/d$

Electric Fields & Potential — AQA A-Level Physics

E = \frac{V}{d}

$E$: electric field strengthThe force per unit positive chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). at a point in an electric field. Measured in N C⁻¹ or V m⁻¹. (V m^-1)
$V$: 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 plates (V)
$d$: separation between the plates (m)

This equation is only valid for a uniform field between parallel plates. Do not use it for a point chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C)..
Greater voltageThe energyThe capacity to do work. Measured in joules (J). transferred per unit charge between two points. Measured in volts (V). Informal term for potential difference. between the plates means a stronger field.
Greater plate separation means a weaker field.
The field is directed from the positive plate to the negative plate.
If one plate is earthed, its voltageThe energyThe capacity to do work. Measured in joules (J). transferred per unit charge between two points. Measured in volts (V). Informal term for potential difference. is 0 V.

Combining the two expressions for E

When a point charge moves between parallel plates, the two expressions for E can be equated:

\begin{aligned} E &= \frac{F}{Q} \\ &= \frac{V}{d} \end{aligned}

Rearranging: $F = QV/d. This gives$ the force on a charge Q between plates with p.d. V separated by distance d.
The work doneEnergy transferred when a force moves an object. In electrical circuits, W = QV (charge times potential difference). moving charge Q through p.d. V is: $W = QV$.

Worked Example

Two parallel plates separated by 3.5 cm have a potential difference of 7.9 kV. Calculate the electric force on a point charge of 2.6 x 10^-15 C placed between them.

Show Solution

List known values

$V = 7.9 \text{ kV} = 7900 \text{ V}$
$d = 3.5 \text{ cm} = 0.035 \text{ m}$
$Q = 2.6 \times 10^{-15} \text{ C}$

Use F = QV/d

$$F = \frac{QV}{d} = \frac{(2.6 \times 10^{-15}) \times 7900}{0.035}$$

Evaluate

$$F = 5.9 \times 10^{-10} \text{ N (2 s.f.)}$$

Answer

$F = 5.9 \times 10^{-10}$ N

Common Mistake MEDIUM

Students often: Using $E = V/d$ for the field around a point charge.
Instead: $E = V/d$ is only for uniform fields between parallel plates. For a point charge, use $E = Q / (4piepsilon0 r^2)$.