9478 / 2027
Electric Fields overview

Lesson 4 of 7

Uniform fields between plates

  • I can calculate a uniform parallel-plate field from potential difference and perpendicular separation.Syllabus

    Syllabus 9478, 14(g). Calculate a uniform parallel-plate field from potential difference and perpendicular separation.

  • I can calculate the force on a charge in a uniform electric field.Syllabus

    Syllabus 9478, 14(h). Calculate the force on a charge in a uniform electric field.

Between ideal broad parallel plates, potential changes uniformly across the perpendicular gap. That gap determines the field, while the charge sign determines the force direction.

In the ideal interior region, neglecting edge effects, field magnitude is:

E = |ΔV|/d

d is the perpendicular separation between plates. The field points from higher to lower potential. Plate length along a particle's path is a different distance and does not replace d.

Worked plate field

Upper plate +120 V, lower plate 0 V

Two horizontal plates are separated by 0.0300 m. Their interior field has magnitude:

E = 120/0.0300 = 4000 V/m

The direction is downward, from the upper plate to the lower. The chosen 0 V label is a reference; it does not by itself mean that plate is physically connected to Earth.

The field follows the perpendicular potential gradient

Use the ideal interior of broad parallel plates and neglect fringing there. The upper plate is at +120 V and the lower one at 0 V; that zero is a chosen reference, not an earth connection.

Field downward; electron force upward

A 120-volt difference across 0.0300 metres gives a downward fieldTwo horizontal plates have upper potential positive 120 volts and lower potential zero volts. A vertical dimension marks their perpendicular 0.0300-metre separation. Three teal field arrows in the interior point downward from higher to lower potential. A positive test charge has a separate purple force arrow downward. An electron has a purple force arrow upward, opposite the field. Both force recipients are identified by their charge symbols. These are direction comparisons, not equal-force magnitudes for arbitrary charges or a drawing of forces caused by their velocities. The plate length is schematic in this panel; no edge-field shape or earth wire is shown.+120 V0 V (reference)E+qF-eFPerpendicular gap: 0.0300 m

E = 120/0.0300 = 4000 V/m downward. Purple arrows show the force on the named charges: +q downward and the electron upward. Their lengths are schematic; their direction is determined by qE, even when a charge is instantaneously stationary.

Upward position gives increasing potential

The uniform plate field is the negative slope of the straight potential-position graphThe horizontal graph coordinate is upward position y in metres, with zero at the middle of the plate gap. It runs from minus 0.015 to plus 0.015 metres. The vertical coordinate is potential in volts. The exact straight line passes through minus 0.015 metres and zero volts, zero metres and sixty volts, and plus 0.015 metres and 120 volts. The model is V equals sixty plus four thousand y. Its positive gradient is four thousand volts per metre, so the upward-positive electric-field component is negative four thousand volts per metre. The plotted position coordinate runs rightward on the graph even though positive physical y is upward in the plate view.-0.0150+0.0150306090120V / VUpward position y / m

V(y) = 60 + 4000y, with y in metres. Therefore Ey = -dV/dy = -4000 V/m. The negative component means downward, not a negative field magnitude.

The field is perpendicular to the plates. The potential-position graph uses upward y from the midpoint, so its positive gradient gives a negative, downward field component. Force arrows identify their charged recipients.

With y positive upward from the midpoint, the lower plate is at y = -0.0150 m and the upper at +0.0150 m. The midpoint potential is 60.0 V:

V(y) = 60.0 V + (4000 V/m)y
Ey = -dV/dy = -4000 V/m

For an electron, q = -1.60 × 10-19 C:

Fy = qEy
= (-1.60 × 10-19)(-4000)
= +6.40 × 10-16 N

Its force is upward. A positive charge would be forced downward. These statements remain true even if the particle is instantaneously stationary.

Try it step by step

Work it out in steps: Field strength E = V / d in V/m, then Acceleration of an electron a = eE / m, in units of 1015 m/s2.

  1. Problem 1. Two parallel plates 5.0 mm apart have a p.d. of 120 V between them. Take e = 1.60 x 10-19 C and the electron mass as 9.11 x 10-31 kg. Find the field strength and the acceleration of an electron between the plates.
  2. Problem 2. Two parallel plates 20 mm apart have a p.d. of 500 V between them. Use the same e and electron mass. Find the field strength and the acceleration of an electron.
  3. Problem 3. Two parallel plates 3.0 mm apart have a p.d. of 90 V between them. Use the same e and electron mass. Find the field strength and the acceleration of an electron.
Show the answer

Problem 1.

Field strength E = V / d in V/m: E = 120 / 0.0050 = 24000 V/m

Acceleration of an electron a = eE / m, in units of 1015 m/s2: a = 1.60 x 10-19 x 24000 / (9.11 x 10-31) = 4.22 x 1015 m/s2

Problem 2.

Field strength E = V / d in V/m: E = 500 / 0.020 = 25000 V/m

Acceleration of an electron a = eE / m, in units of 1015 m/s2: a = 1.60 x 10-19 x 25000 / (9.11 x 10-31) = 4.39 x 1015 m/s2

Problem 3.

Field strength E = V / d in V/m: E = 90 / 0.0030 = 30000 V/m

Acceleration of an electron a = eE / m, in units of 1015 m/s2: a = 1.60 x 10-19 x 30000 / (9.11 x 10-31) = 5.27 x 1015 m/s2

Watch out for this

A student says: "Moving two charged plates further apart at a fixed p.d. makes the field between them stronger." What is wrong with this?

Show the answer

For parallel plates E = V/d. At a fixed p.d., a larger separation d gives a weaker field: doubling d halves E.

Check your understanding

Horizontal plates are 0.0300 m apart and 0.0600 m long. The upper plate is at +120 V and the lower at 0 V. What are the interior field and the force direction on an electron?

  1. Field 4000 V/m downward; electron force upward.
  2. Field 2000 V/m downward; electron force downward.
  3. Field 4000 V/m upward; electron force downward.
Show the answer

Field 4000 V/m downward; electron force upward.

The field magnitude is 120/0.0300 = 4000 V/m and points from higher to lower potential. An electron is forced in the opposite direction.

The force is set by q and E, not by the velocity direction. If this is the only significant force on a particle of constant mass, it produces constant signed acceleration. Motion parallel to E and entry perpendicular to E therefore require different component descriptions.