Physics · Electrodynamics

Electric field strength: force, direction, parallel plates

Electric field is a vector specifying electric force per signed test charge. For a uniform plate field, its magnitude also equals voltage magnitude divided by perpendicular separation.

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Formula

Calculate

E E_{hom} = \frac{F _{el}}{q} (q \neq = 0) = \frac{∣ U ∣}{d}

E in V/m = N/C · F in N · q in C · U in V · d in m

Variables & units

Variables & units
Symbol	Meaning	Unit
$E$	Electric field vector; its magnitude in the plate relation	V/m = N/C
$F$	Electric force vector on the probe	N
$q$	Signed nonzero test charge	C
$U$	Potential difference between plates; magnitude in the plate relation	V
$d$	Perpendicular plate separation	m

LaTeX

\begin{aligned}\mathbf E&=\frac{\mathbf F_{\mathrm{el}}}{q}\quad(q\ne0)\\E_{\mathrm{hom}}&=\frac{|U|}{d}\end{aligned}

QUANTA STUDYquanta-study.de

Positive left and negative right plates, field arrows pointing right and electron force left. Potential drops by 360 V toward the right. — In the idealized interior plate region, 360 V across 6 mm gives a field of 60 kV/m to the right. Force on an electron acts leftward. Plate geometry is schematic.

Electric field strength calculator

Find the field strength from force and test charge, or in a parallel-plate capacitor from voltage and plate separation, away from the edges.

Calculation model

Find

Voltage U

Voltage between the two plates as a magnitude, e.g. 200 V from a power supply.

Plate separation d

Perpendicular distance between the plates, e.g. 2 cm.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for E, substitute, evaluate

1. Rearrange for E
$E=\frac U d$
2. Substitute values (in SI units)
- d = 2 cm = 0.02 m
$\frac{200}{0.02}=10000$
3. Result
E = 10 kV/m

Result: Field strength E

E = 10 kV/m

With U = 200 V and d = 2 cm, E = 10 kV/m.

Quick tips

Between parallel plates E = U/d: twice the gap at the same voltage halves the field strength. 200 V across 2 cm gives 10,000 V/m.
A negative test charge reverses the force, not the field: in E = F/q both F and q change sign together. N/C and V/m are the same unit.

Assumptions and limits

Large parallel plates away from edges; uniform electrostatic field. Positive normal separation and voltage/field magnitudes, with no direction information. This is not a breakdown model.

Derivation & background

Michael Faraday shaped the development of electromagnetic field theory. The test-charge model describes the source field independently of the chosen probe. Vector relation $\mathbf F_{\mathrm{el}}=q\mathbf E$ accounts for charge sign. For uniform parallel plates away from edges, magnitude obeys $E=\frac{|U|}{d}$ . A single point source in vacuum has magnitude $E=\frac{|Q|}{4\pi\varepsilon_0r^2}$ , radially outward for positive sources and inward for negative ones. $\frac{V}{m}$ and $\frac{N}{C}$ are equivalent units.

Exam blueprint

Validity range

The force definition uses electric force only and negligible probe backreaction. The magnitude plate relation assumes a uniform electrostatic field, perpendicular separation and neglected fringe fields.

Derivation steps

Electric force defines the field vector; the potential integral explains the additional uniform plate relation.

1Choose a small test charge with known sign and negligible influence on the sources.
2Separate electric force from magnetic, mechanical or gravitational contributions.
3The definition is $\mathbf E=\frac{\mathbf F_{\mathrm{el}}}{q}$ ; negative charge reverses force direction.
4In a conservative electrostatic field, potential difference is the negative field integral along the path.
5Between large parallel plates, field magnitude is approximately constant away from the edges.
6For the perpendicular plate path, $|U|=Ed$ , hence $E=\frac{|U|}{d}$ .
7Check equivalent units $\frac{V}{m}$ and $\frac{N}{C}$ ; magnitude alone contains no direction.

Rearrangements

Force on a charge

\mathbf F_{\mathrm{el}}=q\mathbf E

q carries charge sign. A positive probe has force along field, a negative probe opposite; balance other forces separately.

Voltage

|\Delta V|=Ed

Uniform electrostatic field along the normal plate path. Signed potential change along field is negative.

Plate separation

d=\frac{|\Delta V|}{E}

For $E>0$ in a uniform normal plate field. Breakdown also depends on gas state and electrode geometry; a rule of thumb is not a safe operating limit.

Task variant

Large parallel plates have perpendicular separation 5 mm and potential difference magnitude 100 V. Away from edges, uniform field points along +x. What electric force acts on a positive $q=2 \mu{}C$ probe?

Approach: $E=\frac{|\Delta V|}{d}$ , then $F_x=qE_x$ .
With $d=0.005 m$ , $E=\frac{100\,\mathrm V}{0.005\,\mathrm m}=20000\,\mathrm{V\,m^{-1}}$ and $F_x=(2\cdot10^{-6}\,\mathrm C)(20000\,\mathrm{N\,C^{-1}})=+0.040\,\mathrm N$ .

Check: $\frac{F_x}{q}=20000\,\mathrm{N\,C^{-1}}$ .

A model calculation explicitly assumes Emax=3 kV/mm as a rough guide for a uniform 1 cm air gap. What voltage follows mathematically? Why is this not a universal actual breakdown voltage?

Approach: $|\Delta V|=E_{\max}d$ .
Substitution gives $(3\cdot10^6\,\mathrm{V\,m^{-1}})(0.010\,\mathrm m)=30000\,\mathrm V=30\,\mathrm{kV}$ .
This is only the specified uniform rule-of-thumb model’s result.
Gas pressure, temperature, humidity, gap width and electrode tips affect actual breakdown; voltage alone does not determine spark length.

Common mistakes

Using $E=\frac Ud$ in the field of a point charge.

For a single vacuum point source, $E=\frac{k_e|Q|}{r^2}$ is magnitude; direction follows charge sign. Voltage divided by a finite path length gives a path average there.

Not converting millimetres to metres.

Insert d in metres, otherwise E is off by a factor of 1000.

Getting the force direction wrong for negative charges.

Negative charges experience the force opposite to the field direction.

Exam context

Classics: electron in a parallel-plate capacitor (acceleration, deflection in an oscilloscope), Millikan experiment and combination with the energy relation $W=qU$ .

Formula cluster

Electric fields

The field concept links the Coulomb force and the capacitor.

F = ke·|q₁q₂|/r²

Coulomb's Law

C = Q/U

Capacitance of a Capacitor

Worked example

Large parallel plates: $|U|=200\,\mathrm V$ , $d=4\,\mathrm{mm}=0.004\,\mathrm m$ . Thus $E=\frac{200\,\mathrm V}{0.004\,\mathrm m}=50000\,\mathrm{V/m}$ . An electron experiences $|F_{\mathrm{el}}|=eE\approx8.0\times10^{-15}\,\mathrm N$ , opposite to the field.

Applications

Parallel-plate capacitors and the Millikan experiment, electron deflection, photocopiers and laser printers. Air breakdown additionally requires gas and geometric data; approximately $\frac{3 kV}{mm}$ is only a conditional guide value.

How do you calculate with Electric Field Strength?

Here is how to work through a typical Electric Field Strength (E⃗ = F⃗el/q; Ehom = |U|/d) task step by step:

1
Task
Large parallel plates have perpendicular separation 5 mm and potential difference magnitude 100 V. Away from edges, uniform field points along +x. What electric force acts on a positive $q=2 \mu{}C$ probe?
Solution path
1. Approach: $E=\frac{|\Delta V|}{d}$ , then $F_x=qE_x$ .
2. With $d=0.005 m$ , $E=\frac{100\,\mathrm V}{0.005\,\mathrm m}=20000\,\mathrm{V\,m^{-1}}$ and $F_x=(2\cdot10^{-6}\,\mathrm C)(20000\,\mathrm{N\,C^{-1}})=+0.040\,\mathrm N$ .
Check: $\frac{F_x}{q}=20000\,\mathrm{N\,C^{-1}}$ .
2
Task
A model calculation explicitly assumes Emax=3 kV/mm as a rough guide for a uniform 1 cm air gap. What voltage follows mathematically? Why is this not a universal actual breakdown voltage?
Solution path
1. Approach: $|\Delta V|=E_{\max}d$ .
2. Substitution gives $(3\cdot10^6\,\mathrm{V\,m^{-1}})(0.010\,\mathrm m)=30000\,\mathrm V=30\,\mathrm{kV}$ .
3. This is only the specified uniform rule-of-thumb model’s result.
4. Gas pressure, temperature, humidity, gap width and electrode tips affect actual breakdown; voltage alone does not determine spark length.

Understand and apply the equation

An electric field describes the electric force a charge would experience at a location. The source sets the field; the test charge’s sign sets its force direction. Voltage gives a local field magnitude only under additional assumptions, such as between large parallel plates away from their edges.

A probe reveals the field; it does not define the source

Imagine a small positive charge between charged plates. Its electric force indicates the field direction at that point. With unchanged sources, a negative probe experiences the opposite force. Dividing force by signed charge gives the same field vector in both cases. We can therefore describe the source arrangement first, then apply it to different particles.[1][2]

Prior knowledge: a vector has magnitude and direction. Dividing by a negative number reverses direction.
Use the electric force component only; gravity or magnetic force does not enter the electric-field definition.
Potential is energy per charge. Its change along a path depends on the field along that path.

[1][2]

How do you calculate electric field from force?

\begin{aligned}\mathbf F_{\mathrm{el}}&=q\mathbf E\\\mathbf E&=\frac{\mathbf F_{\mathrm{el}}}{q}\quad(q\ne0)\end{aligned}

Divide vectors by signed charge. A negative test particle experiences force opposite to the field.[1]

The test charge must not significantly alter the source configuration. Its backreaction is neglected in the ideal test-charge limit. Field direction follows force on a positive test charge. With force magnitudes alone, $E=\frac{|F_{\mathrm{el}}|}{|q|}$ ; direction cannot be reconstructed from those magnitudes. Total measured force may also include gravity, friction or magnetic effects that must not be divided by q and called electric field.[1]

Source fields superpose as vectors

\mathbf E(\mathbf r)=\frac1{4\pi\varepsilon_0}\sum_i Q_i\frac{\mathbf r-\mathbf r_i}{|\mathbf r-\mathbf r_i|^3}

For fixed point sources in vacuum, add their individual Coulomb fields, not their magnitudes.[1]

Each displacement vector points from its source to the observation point. Positive sources contribute along that vector and negative sources opposite to it. Superposition is a physical statement about the linear field model; $E=\frac{F}{q}$ alone does not prove it. The field vector can be evaluated away from ideal point sources. At a source itself the point expression is singular and cannot model internal charge structure. The individual source contributions follow from Coulomb’s law.[1]

We use CODATA 2022 $\varepsilon_0\approx8.8541878188\cdot10^{-12}\,\mathrm{F/m}$ . Unlike exactly defined elementary charge e, $\varepsilon_0$ has measurement uncertainty in the present SI. It is small relative to example precision; extra digits serve checks rather than claims of equally precise measurements.[6][7]

Michael Faraday’s historical field concept is part of the origin of this description. The field represents source configuration through a local property that can subsequently be used for different small test charges. The mathematical vector form and distinct source and probe roles retain direction and signs explicitly.[1][9]

When does the plate formula apply to voltage and separation?

Voltage compares two locations; field strength describes one location. A small potential drop over a small distance gives the local field component in the limit. For constant field you may instead use the entire path. When field varies along it, voltage divided by path length gives a mean tangential component rather than an arbitrary local value. Electric potential energy connects this potential difference to the signed test charge.[2][3]

\begin{aligned}\Delta V&=V_B-V_A=-\int_A^B\mathbf E\cdot d\boldsymbol\ell\\\mathbf E&=-\nabla V\end{aligned}

In a conservative electrostatic field, potential difference is a path integral with a minus sign.[3][2]

Between large parallel plates, the field far from edges is approximately uniform and normal to the plates. Across normal separation $d>0$ , $|\Delta V|=Ed$ and $E=\frac{|\Delta V|}{d}$ . Field points from higher to lower potential. With a nonuniform field, $\frac{|\Delta V|}{d}$ gives only the magnitude of the averaged field projection along the straight path, not automatically the local magnitude everywhere. An oblique path requires vector projection.[3][4]

A local field is not a path average

QUANTA STUDYquanta-study.de

For the third example: read local field from the sloping curve. The dashed line denotes the mean over the entire plotted path. Areas under both curves over that path are equal.[2][3]

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Units and valid rearrangements

\begin{aligned}1\,\frac{\mathrm V}{\mathrm m}&=1\,\frac{\mathrm J}{\mathrm C\,\mathrm m}=1\,\frac{\mathrm N}{\mathrm C}\\|F_{\mathrm{el}}|&=|q|E\end{aligned}

Voltage and force definitions lead to the same field-strength unit.

Convert millimetres to metres before the plate calculation. For $E>0$ , $|q|=\frac{|F_{\mathrm{el}}|}{E}$ determines charge magnitude but not sign. A known force component and matching field component give $q=\frac{F_{\mathrm{el}, x}}{E_x}$ when $E_x\ne0$ . Both components must use the same coordinates. Rearranging $d=\frac{|\Delta V|}{E}$ assumes the uniform normal plate model, not merely compatible units.

Uniform field and oblique source vector

Electron between model plates

Given: Large parallel plates have $|\Delta V|=360\,\mathrm V$ and $d=6.00\,\mathrm{mm}$ ; edge regions are excluded. Field points along +x. An electron test charge $q=-e$ with $e=1.602176634\cdot10^{-19}\,\mathrm C$ is in the field.
Find: Find field and electric force including direction.

Approach: In a uniform normal plate field, potential difference gives field magnitude; signed charge then determines the force component.
$\begin{aligned}E&=\frac{|\Delta V|}{d}\\F_x&=qE_x\end{aligned}$
Convert separation to metres and calculate field magnitude.
$E=\frac{360\,\mathrm V}{0.00600\,\mathrm m}=6.00\times10^4\,\mathrm{V/m}$
Use the negative charge sign.
$F_{\mathrm{el},x}=-eE=-9.613059804\times10^{-15}\,\mathrm N$

Result: $E=60.0\,\mathrm{kV\,m^{-1}}$ along +x. Electric force has magnitude 9.613×10⁻¹⁵ N and points along −x, toward the higher-potential plate.

Check: $\frac{F_x}{q}$ returns +60000 N/C. Potential decreases by 360 V over 6 mm along +x: $-E_xd=-360\,\mathrm V$ . A positive particle would experience the opposite force direction.

Field at a point outside a fixed source

Given: A fixed point source $Q=+1.50\,\mathrm{nC}$ is at the origin. $P=(0.0300\,\mathrm m,0.0400\,\mathrm m)$ . Vacuum model; test charge $q=-2.00\,\mathrm{pC}$ with negligible backreaction.
Find: Find field components and force on the probe.

Approach: The fixed point source produces a radial Coulomb field. Actual position components and the charge sign determine field and test force.
$\begin{aligned}\mathbf E&=\frac{Q}{4\pi\varepsilon_0}\frac{\mathbf r}{r^3}\\\mathbf F&=q\mathbf E\end{aligned}$
Construct distance and direction from geometry.
$\begin{aligned}r&=\sqrt{(0.0300\,\mathrm m)^2+(0.0400\,\mathrm m)^2}\\&=0.0500\,\mathrm m\\\hat{\mathbf r}&=(0.600,0.800)\end{aligned}$
Evaluate Coulomb field magnitude and components.
$\begin{aligned}E&\approx5392.53107\,\mathrm{N/C}\\E_x&\approx3235.51864\,\mathrm{N/C}\\E_y&\approx4314.02486\,\mathrm{N/C}\end{aligned}$
Multiply both components by the same signed charge.
$\mathbf F_{\mathrm{el}}\approx(-6.47103729,-8.62804971)\times10^{-9}\,\mathrm N$

Result: Field points obliquely away from the origin. The negative probe experiences force toward the origin; force magnitude is approximately 1.0785062×10⁻⁸ N.

Check: $\sqrt{E_x^2+E_y^2}$ equals the separately computed field magnitude. Likewise $|q|E=1.0785062\cdot10^{-8}\,\mathrm N$ gives force-vector magnitude. Normalized components 0.6 and 0.8 satisfy $0.6^2+0.8^2=1$ .

Distinguish mean and local field

Given: Along x, an electrostatic potential is $V(x)=V_0-ax^2$ with $a=200\,\mathrm{V\,m^{-2}}$ . Compare x=0 with x=3.00 cm.
Find: Find potential difference, mean field component and the field component at the right endpoint.

Approach: the negative derivative gives local field. Negative potential difference divided by displacement gives its mean.
$\begin{aligned}E_x(x)&=-\frac{\mathrm dV}{\mathrm dx}=2ax\\\Delta V&=-aL^2\\\overline E_x&=-\frac{\Delta V}{L}=aL\end{aligned}$
Substitute $L=0.0300\,\mathrm m$ .
$\begin{aligned}\Delta V&=-(200)(0.0300)^2\,\mathrm V=-0.180\,\mathrm V\\\overline E_x&=(200)(0.0300)\,\mathrm{V\,m^{-1}}=6.00\,\mathrm{V\,m^{-1}}\\E_x(L)&=2(200)(0.0300)\,\mathrm{V\,m^{-1}}=12.0\,\mathrm{V\,m^{-1}}\end{aligned}$

Result: The field at the right endpoint is twice the mean field. Potential decreases by 0.180 V.

Check: The area under the linearly rising field is a triangle: $\frac12(0.0300\,\mathrm m)(12.0\,\mathrm{V\,m^{-1}})=0.180\,\mathrm V$ . Including the potential integral’s minus sign recovers −0.180 V.

Explore further: Motion from electric force

An electron may already move transversely to the field. Its electric acceleration still points opposite E. Without other forces and at nonrelativistic speed, $\mathbf a=\frac{q\mathbf E}{m}$ : a constant field component produces constant acceleration in that component. Initial transverse motion persists in the ideal plate model while the field component curves the path. Initial position, velocity, field region and elapsed time are all needed to determine the actual trajectory. For this component, the equations of motion with constant acceleration apply.[1][2]

At 200 V across a normal plate separation of 4 mm, $E=\frac{200\,\mathrm V}{0.004\,\mathrm m}=50000\,\mathrm{V/m}$ . Electron-force magnitude is about $8\cdot10^{-15}\,\mathrm N$ , directed opposite to the field. Electron mass and elementary charge are different: charge links field to electric force, while mass links total force to acceleration.[6][7]

Five different field questions

Rearrange · Foundation

A positive probe $q=+3.00\,\mathrm{nC}$ experiences only electric force $F_x$ =−12.0 μN. Find $E_x$ .

Show hint

Use signed components and charge.

Learning goal: Infer signed field component from electric probe force.

Show solution and calculation

Model and approach: Only electric force counts; the probe is positive.
$E_x=\frac{F_{\mathrm{el},x}}q$ $E_x=\frac{-12.0\cdot10^{-6}\,\mathrm N}{3.00\cdot10^{-9}\,\mathrm C}=-4000\,\mathrm{N/C}$ .
Field points along −x; a positive charge follows its force direction.

Check: $qE_x=-12.0\,\mathrm{\mu N}$ .

E_x=-4.00\times10^3\,\mathrm{N/C}

[1]

Rearrange · Foundation

A uniform normal plate field must have magnitude 80.0 kV/m with $d=2.50\,\mathrm{mm}$ . What voltage magnitude is needed?

Show hint

Use $|\Delta V|=Ed$ , not $\frac Ed$ .

Learning goal: Infer plate voltage from uniform field.

Show solution and calculation

Model and approach: Uniform normal plate field; separation must be in metres.
$|\Delta V|=Ed$ $|\Delta V|=80000\,\mathrm{V/m}\cdot0.00250\,\mathrm m=200\,\mathrm V$ .
Which plate is at higher potential sets field direction.

Check: $\frac{200\,\mathrm V}{0.00250\,\mathrm m}=80000\,\mathrm{V\,m^{-1}}$ .

|\Delta V|=200\,\mathrm V

[2]

Transfer · Application

In an electrostatic field, $V(x)=V_0-ax^2$ with $a=200\,\mathrm{V/m^2}$ . Find $E_x$ at $x=0.0300\,\mathrm m$ and explain why dividing voltage by separation is not a general local-field rule here.

Show hint

Differentiate potential and keep the minus sign.

Learning goal: Derive local field from a nonlinear potential function.

Show solution and calculation

Model and approach: Local field is the negative potential derivative.
$E_x=-\frac{\mathrm dV}{\mathrm dx}=2ax$ $E_x=-\frac{\mathrm dV}{\mathrm dx}=2ax=12.0\,\mathrm{V/m}$ .
Field depends on position.
Potential difference across a finite path determines its averaged field projection, not an identical value everywhere.

Check: $E_x(0)=0$ .

E_x=2ax=12.0\,\mathrm{V/m}

[3]

Interpret · Application

A local electrostatic potential map is $V(x,y)=V_0+(3\,\mathrm{V/m})x+(4\,\mathrm{V/m})y$ . Find field components, magnitude and force direction on a negative probe. Are equipotentials parallel to the field?

Show hint

Use $\vec E=-\nabla V$ ; negative charge reverses force direction.

Learning goal: Read field direction from a two-dimensional potential map.

Show solution and calculation

Approach: field components are (−3,−4) V/m and magnitude is 5 V/m.
A negative probe experiences force toward (+x,+y), opposite the field.
Lines $3x+4y=\mathrm{constant}$ are perpendicular to E.

Check: along an equipotential, $\mathbf E\cdot\mathrm d\boldsymbol\ell=0$ and potential does not change. The constant V₀ affects neither field nor force.

\begin{aligned}\mathbf E&=-\nabla V=(-3,-4)\,\mathrm{V/m}\\|\mathbf E|&=\sqrt{3^2+4^2}\,\mathrm{V/m}=5\,\mathrm{V/m}\end{aligned}

[3][1]

Error analysis · Application

A +2 μC probe has weight 4 mN downward and measured net force 2 mN upward. Only gravity and electric force act. A solution infers $E=1000\,\mathrm{N\,C^{-1}}$ upward from net force. Correct it.

Show hint

$\vec F_{\mathrm{el}}=q\vec E$ refers to electric force, not automatically net force.

Learning goal: Separate measured net force before inferring an electric field.

Show solution and calculation

Approach: taking upward as positive, $F_{\mathrm{el}}-4\,\mathrm{mN}=2\,\mathrm{mN}$ gives $F_{\mathrm{el}}=6\,\mathrm{mN}$ , hence $E=3000\,\mathrm{N\,C^{-1}}$ upward.

Check: $qE-4\,\mathrm{mN}$ reproduces the measured 2 mN. The incorrect field gives electric force 2 mN and downward net force. Measured net force can be divided directly by q only when other forces are absent.

\begin{aligned}F_{\mathrm{el}}&=F_{\mathrm{net}}-F_g=2\,\mathrm{mN}-(-4\,\mathrm{mN})=6\,\mathrm{mN}\\E&=\frac{6\cdot10^{-3}\,\mathrm N}{2\cdot10^{-6}\,\mathrm C}=3000\,\mathrm{N\,C^{-1}}\end{aligned}

[1]

Materials and time variation require additional assumptions

Electric field as a force quantity can also be defined beyond electrostatic problems. The time-independent potential description here nevertheless assumes a conservative electrostatic field. With magnetic induction, a scalar potential difference is not generally a path-independent electric voltage integral. Material polarization changes the source description; a number $\varepsilon_r$ cannot be inserted into every vacuum point-source expression without appropriate geometry and material modelling.[5][3][8]

Large fields can make a gas or dielectric conductive. One nominal air-field threshold is not a universal breakdown rule: gas conditions, electrode shape, separation and local field enhancement matter. Ideal plates do not model spark channels or resulting charge redistribution. These limits concern the model rather than the identity $1\,\mathrm{V/m}=1\,\mathrm{N/C}$ .[5]

What to take away

The definition is a vector ratio using signed test charge. A negative probe reverses force, not the source field.
Voltage is a potential difference. Voltage divided by separation directly gives magnitude only for a uniform normal plate field.
For a varying electrostatic field, use the negative potential derivative or field integral.

[1][2][3]

Sources for this explanation

Quanta · Updated October 2, 2026

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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October 2, 2026Refined question headings and added scientifically relevant paragraph links; retained existing derivations, examples, diagrams, exercises and sources.
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Quanta. Electric Field Strength. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/elektrisches-feld

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Frequently asked questions about Electric Field Strength

How do you calculate the electric field strength in a parallel-plate capacitor?+

Between large parallel plates, the magnitude of the potential difference divided by perpendicular separation gives the approximately uniform field magnitude. At 200 V and 4 mm this is $\frac{50000 V}{m}$ . Convert separation to metres first and check that it is small compared with both lateral plate dimensions. The field is nonuniform near edges, where simple division does not give an accurate local value. Field direction points from higher to lower potential. A negative charge experiences electric force opposite to it. Breakdown and permissible operating voltage require additional information about the actual material and configuration.

What is the difference between

E =\frac{ F}{q}

and

E =\frac{ U}{d}

The force definition determines the electric field vector at a location: divide electric force by test charge, including its sign. The probe should scarcely alter the source configuration. Dividing voltage by distance instead follows from the potential integral only for a uniform field and a suitable path along its direction. Magnitude calculations use positive force or voltage magnitudes respectively. A point-charge field varies spatially; a voltage over an extended interval initially determines a path average there. Magnetic or mechanical force contributions must also be excluded from the electric force used in the definition.

Are

\frac{V}{m}

and

\frac{N}{C}

really the same unit?+

Yes. A volt is a joule per coulomb, and a joule is a newton metre. Consequently volts per metre and newtons per coulomb are the same unit. The forms emphasise different measurement routes: voltage and separation, or force and test charge. In $\frac{a 50000 V}{m}$ field, a positive 2 nC charge experiences 100 µN of electric force along the field. An equally large negative charge experiences the same force magnitude in the opposite direction. Unit equivalence does not justify applying the plate formula to every field; geometry, field variation and signs still require checking.

How does an electron move in a uniform electric field?+

An electron is negatively charged, so its electric force points opposite to the field. In a nonrelativistic model with electric force alone, acceleration is constant when the field is uniform and steady. An initially stationary electron accelerates towards higher potential. If it initially moves along the field, it first slows and can turn around. A transverse initial velocity remains as a transverse component without additional forces. At $\frac{60000 V}{m}$ , the force magnitude is approximately $9.61\times{}10^{-15} N$ . Collisions, magnetic forces, boundary surfaces or relativistic speeds require an appropriately extended account of its motion.

What happens when the field strength in air becomes too large?+

A strong field can accelerate existing free charge carriers and cause further ionisation. A previously largely insulating gas then becomes conducting, and a discharge can develop. The commonly quoted approximate air value of $\frac{3 MV}{m}$ applies only under particular conditions and is not a universal switching threshold. Pressure, temperature, humidity, gap width and electrode geometry affect breakdown. Near sharp tips the local field can greatly exceed voltage divided by average separation. A capacitor model therefore provides an ideal field estimate, while assessing breakdown requires additional material and geometric information about the actual configuration.

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Electric field strength: force, direction, parallel plates

Electric field is a vector specifying electric force per signed test charge. For a uniform plate field, its magnitude also equals voltage magnitude divided by perpendicular separation.

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Symbol

Meaning

Unit

E

Electric field vector; its magnitude in the plate relation

V/m = N/C

F

Electric force vector on the probe

q

Signed nonzero test charge

U

Potential difference between plates; magnitude in the plate relation

d

Perpendicular plate separation

Derivation & background

Exam blueprint

Validity range

Derivation steps

Electric force defines the field vector; the potential integral explains the additional uniform plate relation.

1Choose a small test charge with known sign and negligible influence on the sources.
2Separate electric force from magnetic, mechanical or gravitational contributions.
3The definition is $\mathbf E=\frac{\mathbf F_{\mathrm{el}}}{q}$ ; negative charge reverses force direction.
4In a conservative electrostatic field, potential difference is the negative field integral along the path.
5Between large parallel plates, field magnitude is approximately constant away from the edges.
6For the perpendicular plate path, $|U|=Ed$ , hence $E=\frac{|U|}{d}$ .
7Check equivalent units $\frac{V}{m}$ and $\frac{N}{C}$ ; magnitude alone contains no direction.

How do you calculate with Electric Field Strength?

Here is how to work through a typical Electric Field Strength (E⃗ = F⃗el/q; Ehom = |U|/d) task step by step:

Task

Solution path

Approach: $E=\frac{|\Delta V|}{d}$ , then $F_x=qE_x$ .
With $d=0.005 m$ , $E=\frac{100\,\mathrm V}{0.005\,\mathrm m}=20000\,\mathrm{V\,m^{-1}}$ and $F_x=(2\cdot10^{-6}\,\mathrm C)(20000\,\mathrm{N\,C^{-1}})=+0.040\,\mathrm N$ .

Check: $\frac{F_x}{q}=20000\,\mathrm{N\,C^{-1}}$ .

Task

A model calculation explicitly assumes Emax=3 kV/mm as a rough guide for a uniform 1 cm air gap. What voltage follows mathematically? Why is this not a universal actual breakdown voltage?

Solution path

Approach: $|\Delta V|=E_{\max}d$ .
Substitution gives $(3\cdot10^6\,\mathrm{V\,m^{-1}})(0.010\,\mathrm m)=30000\,\mathrm V=30\,\mathrm{kV}$ .
This is only the specified uniform rule-of-thumb model’s result.
Gas pressure, temperature, humidity, gap width and electrode tips affect actual breakdown; voltage alone does not determine spark length.

A probe reveals the field; it does not define the source

Prior knowledge: a vector has magnitude and direction. Dividing by a negative number reverses direction.
Use the electric force component only; gravity or magnetic force does not enter the electric-field definition.
Potential is energy per charge. Its change along a path depends on the field along that path.

[1][2]

How do you calculate electric field from force?

\begin{aligned}\mathbf F_{\mathrm{el}}&=q\mathbf E\\\mathbf E&=\frac{\mathbf F_{\mathrm{el}}}{q}\quad(q\ne0)\end{aligned}

Divide vectors by signed charge. A negative test particle experiences force opposite to the field.[1]

Source fields superpose as vectors

\mathbf E(\mathbf r)=\frac1{4\pi\varepsilon_0}\sum_i Q_i\frac{\mathbf r-\mathbf r_i}{|\mathbf r-\mathbf r_i|^3}

For fixed point sources in vacuum, add their individual Coulomb fields, not their magnitudes.[1]

When does the plate formula apply to voltage and separation?

\begin{aligned}\Delta V&=V_B-V_A=-\int_A^B\mathbf E\cdot d\boldsymbol\ell\\\mathbf E&=-\nabla V\end{aligned}

In a conservative electrostatic field, potential difference is a path integral with a minus sign.[3][2]

A local field is not a path average

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For the third example: read local field from the sloping curve. The dashed line denotes the mean over the entire plotted path. Areas under both curves over that path are equal.[2][3]

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Units and valid rearrangements

\begin{aligned}1\,\frac{\mathrm V}{\mathrm m}&=1\,\frac{\mathrm J}{\mathrm C\,\mathrm m}=1\,\frac{\mathrm N}{\mathrm C}\\|F_{\mathrm{el}}|&=|q|E\end{aligned}

Voltage and force definitions lead to the same field-strength unit.

Uniform field and oblique source vector

Electron between model plates

Given: Large parallel plates have $|\Delta V|=360\,\mathrm V$ and $d=6.00\,\mathrm{mm}$ ; edge regions are excluded. Field points along +x. An electron test charge $q=-e$ with $e=1.602176634\cdot10^{-19}\,\mathrm C$ is in the field.
Find: Find field and electric force including direction.

Approach: In a uniform normal plate field, potential difference gives field magnitude; signed charge then determines the force component.
$\begin{aligned}E&=\frac{|\Delta V|}{d}\\F_x&=qE_x\end{aligned}$
Convert separation to metres and calculate field magnitude.
$E=\frac{360\,\mathrm V}{0.00600\,\mathrm m}=6.00\times10^4\,\mathrm{V/m}$
Use the negative charge sign.
$F_{\mathrm{el},x}=-eE=-9.613059804\times10^{-15}\,\mathrm N$

Result: $E=60.0\,\mathrm{kV\,m^{-1}}$ along +x. Electric force has magnitude 9.613×10⁻¹⁵ N and points along −x, toward the higher-potential plate.

Check: $\frac{F_x}{q}$ returns +60000 N/C. Potential decreases by 360 V over 6 mm along +x: $-E_xd=-360\,\mathrm V$ . A positive particle would experience the opposite force direction.

Field at a point outside a fixed source

Given: A fixed point source $Q=+1.50\,\mathrm{nC}$ is at the origin. $P=(0.0300\,\mathrm m,0.0400\,\mathrm m)$ . Vacuum model; test charge $q=-2.00\,\mathrm{pC}$ with negligible backreaction.
Find: Find field components and force on the probe.

Approach: The fixed point source produces a radial Coulomb field. Actual position components and the charge sign determine field and test force.
$\begin{aligned}\mathbf E&=\frac{Q}{4\pi\varepsilon_0}\frac{\mathbf r}{r^3}\\\mathbf F&=q\mathbf E\end{aligned}$
Construct distance and direction from geometry.
$\begin{aligned}r&=\sqrt{(0.0300\,\mathrm m)^2+(0.0400\,\mathrm m)^2}\\&=0.0500\,\mathrm m\\\hat{\mathbf r}&=(0.600,0.800)\end{aligned}$
Evaluate Coulomb field magnitude and components.
$\begin{aligned}E&\approx5392.53107\,\mathrm{N/C}\\E_x&\approx3235.51864\,\mathrm{N/C}\\E_y&\approx4314.02486\,\mathrm{N/C}\end{aligned}$
Multiply both components by the same signed charge.
$\mathbf F_{\mathrm{el}}\approx(-6.47103729,-8.62804971)\times10^{-9}\,\mathrm N$

Result: Field points obliquely away from the origin. The negative probe experiences force toward the origin; force magnitude is approximately 1.0785062×10⁻⁸ N.

Distinguish mean and local field

Given: Along x, an electrostatic potential is $V(x)=V_0-ax^2$ with $a=200\,\mathrm{V\,m^{-2}}$ . Compare x=0 with x=3.00 cm.
Find: Find potential difference, mean field component and the field component at the right endpoint.

Approach: the negative derivative gives local field. Negative potential difference divided by displacement gives its mean.
$\begin{aligned}E_x(x)&=-\frac{\mathrm dV}{\mathrm dx}=2ax\\\Delta V&=-aL^2\\\overline E_x&=-\frac{\Delta V}{L}=aL\end{aligned}$
Substitute $L=0.0300\,\mathrm m$ .
$\begin{aligned}\Delta V&=-(200)(0.0300)^2\,\mathrm V=-0.180\,\mathrm V\\\overline E_x&=(200)(0.0300)\,\mathrm{V\,m^{-1}}=6.00\,\mathrm{V\,m^{-1}}\\E_x(L)&=2(200)(0.0300)\,\mathrm{V\,m^{-1}}=12.0\,\mathrm{V\,m^{-1}}\end{aligned}$

Result: The field at the right endpoint is twice the mean field. Potential decreases by 0.180 V.

Explore further: Motion from electric force

Five different field questions

Rearrange · Foundation

A positive probe $q=+3.00\,\mathrm{nC}$ experiences only electric force $F_x$ =−12.0 μN. Find $E_x$ .

Show hint

Use signed components and charge.

Learning goal: Infer signed field component from electric probe force.

Show solution and calculation

Model and approach: Only electric force counts; the probe is positive.
$E_x=\frac{F_{\mathrm{el},x}}q$ $E_x=\frac{-12.0\cdot10^{-6}\,\mathrm N}{3.00\cdot10^{-9}\,\mathrm C}=-4000\,\mathrm{N/C}$ .
Field points along −x; a positive charge follows its force direction.

Check: $qE_x=-12.0\,\mathrm{\mu N}$ .

E_x=-4.00\times10^3\,\mathrm{N/C}

[1]

Rearrange · Foundation

A uniform normal plate field must have magnitude 80.0 kV/m with $d=2.50\,\mathrm{mm}$ . What voltage magnitude is needed?

Show hint

Use $|\Delta V|=Ed$ , not $\frac Ed$ .

Learning goal: Infer plate voltage from uniform field.

Show solution and calculation

Model and approach: Uniform normal plate field; separation must be in metres.
$|\Delta V|=Ed$ $|\Delta V|=80000\,\mathrm{V/m}\cdot0.00250\,\mathrm m=200\,\mathrm V$ .
Which plate is at higher potential sets field direction.

Check: $\frac{200\,\mathrm V}{0.00250\,\mathrm m}=80000\,\mathrm{V\,m^{-1}}$ .

|\Delta V|=200\,\mathrm V

[2]

Transfer · Application

In an electrostatic field, $V(x)=V_0-ax^2$ with $a=200\,\mathrm{V/m^2}$ . Find $E_x$ at $x=0.0300\,\mathrm m$ and explain why dividing voltage by separation is not a general local-field rule here.

Show hint

Differentiate potential and keep the minus sign.

Learning goal: Derive local field from a nonlinear potential function.

Show solution and calculation

Model and approach: Local field is the negative potential derivative.
$E_x=-\frac{\mathrm dV}{\mathrm dx}=2ax$ $E_x=-\frac{\mathrm dV}{\mathrm dx}=2ax=12.0\,\mathrm{V/m}$ .
Field depends on position.
Potential difference across a finite path determines its averaged field projection, not an identical value everywhere.

Check: $E_x(0)=0$ .

E_x=2ax=12.0\,\mathrm{V/m}

[3]

Interpret · Application

A local electrostatic potential map is $V(x,y)=V_0+(3\,\mathrm{V/m})x+(4\,\mathrm{V/m})y$ . Find field components, magnitude and force direction on a negative probe. Are equipotentials parallel to the field?

Show hint

Use $\vec E=-\nabla V$ ; negative charge reverses force direction.

Learning goal: Read field direction from a two-dimensional potential map.

Show solution and calculation

Approach: field components are (−3,−4) V/m and magnitude is 5 V/m.
A negative probe experiences force toward (+x,+y), opposite the field.
Lines $3x+4y=\mathrm{constant}$ are perpendicular to E.

Check: along an equipotential, $\mathbf E\cdot\mathrm d\boldsymbol\ell=0$ and potential does not change. The constant V₀ affects neither field nor force.

\begin{aligned}\mathbf E&=-\nabla V=(-3,-4)\,\mathrm{V/m}\\|\mathbf E|&=\sqrt{3^2+4^2}\,\mathrm{V/m}=5\,\mathrm{V/m}\end{aligned}

[3][1]

Error analysis · Application

A +2 μC probe has weight 4 mN downward and measured net force 2 mN upward. Only gravity and electric force act. A solution infers $E=1000\,\mathrm{N\,C^{-1}}$ upward from net force. Correct it.

Show hint

$\vec F_{\mathrm{el}}=q\vec E$ refers to electric force, not automatically net force.

Learning goal: Separate measured net force before inferring an electric field.

Show solution and calculation

Approach: taking upward as positive, $F_{\mathrm{el}}-4\,\mathrm{mN}=2\,\mathrm{mN}$ gives $F_{\mathrm{el}}=6\,\mathrm{mN}$ , hence $E=3000\,\mathrm{N\,C^{-1}}$ upward.

\begin{aligned}F_{\mathrm{el}}&=F_{\mathrm{net}}-F_g=2\,\mathrm{mN}-(-4\,\mathrm{mN})=6\,\mathrm{mN}\\E&=\frac{6\cdot10^{-3}\,\mathrm N}{2\cdot10^{-6}\,\mathrm C}=3000\,\mathrm{N\,C^{-1}}\end{aligned}

[1]

Materials and time variation require additional assumptions

What to take away

The definition is a vector ratio using signed test charge. A negative probe reverses force, not the source field.
Voltage is a potential difference. Voltage divided by separation directly gives magnitude only for a uniform normal plate field.
For a varying electrostatic field, use the negative potential derivative or field integral.

[1][2][3]

Sources for this explanation

Quanta · Updated October 2, 2026

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October 2, 2026Refined question headings and added scientifically relevant paragraph links; retained existing derivations, examples, diagrams, exercises and sources.
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Frequently asked questions about Electric Field Strength

How do you calculate the electric field strength in a parallel-plate capacitor?+

What is the difference between

E =\frac{ F}{q}

and

E =\frac{ U}{d}

Are

\frac{V}{m}

and

\frac{N}{C}

really the same unit?+

How does an electron move in a uniform electric field?+

What happens when the field strength in air becomes too large?+

Retain Electric Field Strength for exams

Create a curated FSRS exam set for E⃗ = F⃗el/q; Ehom = |U|/d: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Electric field strength: force, direction, parallel plates

Variables & units

Electric field strength calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electric fields

Worked example

Applications

How do you calculate with Electric Field Strength?

A probe reveals the field; it does not define the source

How do you calculate electric field from force?

Source fields superpose as vectors

When does the plate formula apply to voltage and separation?

A local field is not a path average

Units and valid rearrangements

Uniform field and oblique source vector

Electron between model plates

Field at a point outside a fixed source

Distinguish mean and local field

Explore further: Motion from electric force

Five different field questions

A positive probe q=+3.00 nCq=+3.00\,\mathrm{nC}q=+3.00nC experiences only electric force FxF_xFx​=−12.0 μN. Find ExE_xEx​.

A uniform normal plate field must have magnitude 80.0 kV/m with d=2.50 mmd=2.50\,\mathrm{mm}d=2.50mm. What voltage magnitude is needed?

In an electrostatic field, V(x)=V0−ax2V(x)=V_0-ax^2V(x)=V0​−ax2 with a=200 V/m2a=200\,\mathrm{V/m^2}a=200V/m2. Find ExE_xEx​ at x=0.0300 mx=0.0300\,\mathrm mx=0.0300m and explain why dividing voltage by separation is not a general local-field rule here.

A local electrostatic potential map is V(x,y)=V0+(3 V/m)x+(4 V/m)yV(x,y)=V_0+(3\,\mathrm{V/m})x+(4\,\mathrm{V/m})yV(x,y)=V0​+(3V/m)x+(4V/m)y. Find field components, magnitude and force direction on a negative probe. Are equipotentials parallel to the field?

A +2 μC probe has weight 4 mN downward and measured net force 2 mN upward. Only gravity and electric force act. A solution infers E=1000 N C−1E=1000\,\mathrm{N\,C^{-1}}E=1000NC−1 upward from net force. Correct it.

Materials and time variation require additional assumptions

What to take away

Sources for this explanation

Frequently asked questions about Electric Field Strength

Subject references and further reading

Related formulas

More formulas: Electricity and magnetism

Retain Electric Field Strength for exams

Electric field strength: force, direction, parallel plates

Variables & units

Electric field strength calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electric fields

Worked example

Applications

How do you calculate with Electric Field Strength?

A probe reveals the field; it does not define the source

How do you calculate electric field from force?

Source fields superpose as vectors

When does the plate formula apply to voltage and separation?

A local field is not a path average

Units and valid rearrangements

Uniform field and oblique source vector

Electron between model plates

Field at a point outside a fixed source

Distinguish mean and local field

Explore further: Motion from electric force

Five different field questions

A positive probe q=+3.00 nCq=+3.00\,\mathrm{nC}q=+3.00nC experiences only electric force FxF_xFx​=−12.0 μN. Find ExE_xEx​.

A uniform normal plate field must have magnitude 80.0 kV/m with d=2.50 mmd=2.50\,\mathrm{mm}d=2.50mm. What voltage magnitude is needed?

In an electrostatic field, V(x)=V0−ax2V(x)=V_0-ax^2V(x)=V0​−ax2 with a=200 V/m2a=200\,\mathrm{V/m^2}a=200V/m2. Find ExE_xEx​ at x=0.0300 mx=0.0300\,\mathrm mx=0.0300m and explain why dividing voltage by separation is not a general local-field rule here.

A local electrostatic potential map is V(x,y)=V0+(3 V/m)x+(4 V/m)yV(x,y)=V_0+(3\,\mathrm{V/m})x+(4\,\mathrm{V/m})yV(x,y)=V0​+(3V/m)x+(4V/m)y. Find field components, magnitude and force direction on a negative probe. Are equipotentials parallel to the field?

A +2 μC probe has weight 4 mN downward and measured net force 2 mN upward. Only gravity and electric force act. A solution infers E=1000 N C−1E=1000\,\mathrm{N\,C^{-1}}E=1000NC−1 upward from net force. Correct it.

Materials and time variation require additional assumptions

What to take away

Sources for this explanation

Frequently asked questions about Electric Field Strength

Subject references and further reading

Related formulas

A positive probe $q=+3.00\,\mathrm{nC}$ experiences only electric force $F_x$ =−12.0 μN. Find $E_x$ .

A uniform normal plate field must have magnitude 80.0 kV/m with $d=2.50\,\mathrm{mm}$ . What voltage magnitude is needed?

In an electrostatic field, $V(x)=V_0-ax^2$ with $a=200\,\mathrm{V/m^2}$ . Find $E_x$ at $x=0.0300\,\mathrm m$ and explain why dividing voltage by separation is not a general local-field rule here.

A local electrostatic potential map is $V(x,y)=V_0+(3\,\mathrm{V/m})x+(4\,\mathrm{V/m})y$ . Find field components, magnitude and force direction on a negative probe. Are equipotentials parallel to the field?

A +2 μC probe has weight 4 mN downward and measured net force 2 mN upward. Only gravity and electric force act. A solution infers $E=1000\,\mathrm{N\,C^{-1}}$ upward from net force. Correct it.

A positive probe $q=+3.00\,\mathrm{nC}$ experiences only electric force $F_x$ =−12.0 μN. Find $E_x$ .

A uniform normal plate field must have magnitude 80.0 kV/m with $d=2.50\,\mathrm{mm}$ . What voltage magnitude is needed?

In an electrostatic field, $V(x)=V_0-ax^2$ with $a=200\,\mathrm{V/m^2}$ . Find $E_x$ at $x=0.0300\,\mathrm m$ and explain why dividing voltage by separation is not a general local-field rule here.

A local electrostatic potential map is $V(x,y)=V_0+(3\,\mathrm{V/m})x+(4\,\mathrm{V/m})y$ . Find field components, magnitude and force direction on a negative probe. Are equipotentials parallel to the field?

A +2 μC probe has weight 4 mN downward and measured net force 2 mN upward. Only gravity and electric force act. A solution infers $E=1000\,\mathrm{N\,C^{-1}}$ upward from net force. Correct it.