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Electric Field Between Parallel Plates

Calculates the (approximately uniform) electric field strength between two parallel conducting plates from the voltage across them and their separation.

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Electric Field CalculatorParallel Plates

E = V / d
E = electric field (V/m)  ·  V = potential difference (V)  ·  d = plate separation (m)
⟹ SolveE, V, d
V/m
V
m
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Electric Field
E: V: d:
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Field Magnitude
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E = V / d  ·  Uniform field between infinite parallel plates. Units: V/m, V, m.

Interpretation

Electric field between parallel plates: E = V/d, where V is potential difference, d is plate separation. It is uniform in ideal plates. Example: 100 V across 0.01 m → E = 10,000 V/m.

E = V / d
Electric Field Between Parallel Plates

Variables

SymbolQuantityUnit
EElectric field strength between the platesV/m
VVoltage (potential difference) across the platesV
dSeparation distance between the platesm

What it means

For two parallel conducting plates with a potential difference V and separation d, the electric field between them is uniform (neglecting edge effects) and given by E = V/d. This is derived from the relation E = –dV/dx for a uniform field. This configuration is used in capacitors, where the capacitance C = ε₀A/d. The uniform field makes it ideal for accelerating charged particles (as in cathode ray tubes) and for studying electron motion. In engineering, parallel plates are used in sensors, actuators, and as a model for more complex geometries. Understanding this field is essential for capacitor design and for electrostatic applications.

Worked example

Electric Field Between Parallel Plates – Two Examples

Real‑World
Scenario: A 100 V battery is connected across plates 0.01 m apart. Find the electric field.
ParameterValue
V100 V
d0.01 m
1E = V/d = 100 / 0.01 = 10,000 V/m
Result 10,000 V/m ✓ Uniform field
Scenario: A 12 V battery across plates 0.002 m apart. Find E.
ParameterValue
V12 V
d0.002 m
1E = 12 / 0.002 = 6,000 V/m
Result 6,000 V/m ✓ Lower
Key insight: E = V/d – the electric field is uniform between parallel plates.

Common mistakes

  • Parallel plates: Assumes uniform field between infinite plates – edge effects are ignored.
  • Voltage V: Potential difference between the plates.
  • Distance d: Separation between plates – in metres.
  • Units: V in volts, d in m → E in V/m.
  • Direction: Field points from higher to lower potential (from positive to negative plate).

Applications

The electric field between parallel plates, E = V/d, is uniform and depends only on the voltage and separation distance. This simple formula is used extensively in the design of capacitors, particle deflectors, and electrostatic actuators. Engineers use it to calculate the force on charged particles, to design ion thrusters, and to model field‑effect transistors. In instrumentation, it is applied in electrometer design and in calibration of electric field sensors. The uniform field also serves as a model for studying charged particle motion in a controlled environment. By using this relation, professionals can design devices that rely on precise electric fields for operation.

  • Design of parallel‑plate capacitors and supercapacitors
  • Electrostatic deflection in cathode ray tubes and mass spectrometers
  • Ion thruster and plasma propulsion systems
  • Electric field sensors and calibration standards
  • Educational demonstrations of uniform fields

Frequently Asked Questions

Q01What is the electric field between two parallel plates?
A01

For two large, flat, parallel plates with a potential difference V and separation d, the electric field between them is uniform and given by E = V / d. The field is perpendicular to the plates and points from the positive to the negative plate.

Q02What are the assumptions for the uniform field formula?
A02

  • The plates are large enough that fringing effects at the edges are negligible.
  • The plates are perfectly conducting.
  • The field is uniform in the central region.

Q03What is the common mistake when applying this formula?
A03

Using it near the plate edges where the field is not uniform (fringing). Also, forgetting that V is the potential difference, not the absolute potential.

Q04How do you calculate the force on a charge between the plates?
A04

The force is F = q·E = q·V/d. This is used in deflection tubes and electrostatic actuators.

Q05What is the capacitance of a parallel‑plate capacitor?
A05

Capacitance is C = ε₀·A/d, where A is the plate area and ε₀ is the permittivity of free space. If a dielectric is present, multiply by the relative permittivity ε_r.

Q06How does the field change if a dielectric is inserted?
A06

The field between the plates decreases for the same voltage, because the dielectric reduces the effective electric field. The formula E = V/d still holds (with V constant), but the charge distribution changes.

Q07What is the energy stored in a parallel‑plate capacitor?
A07

The energy is U = ½CV² = ½ε₀E²·Ad (energy density = ½ε₀E²). This is used in energy storage applications.

Q08What are some real‑world applications?
A08

  • Capacitive touch screens.
  • Electrostatic speakers.
  • Accelerometers.
  • High‑voltage transmission lines (field near conductors).