A Charge Accelerated Through a Potential Difference

Push a charge through a potential difference and it gains qV of kinetic energy. That single line connects volts to joules to metres per second, and it is where the electronvolt comes from. This gives you the energy, the speed, the field in the gap and — importantly — the point at which the classical answer stops being true.

Last updated 11 Nov 2026

Energy, speed and the electronvolt bridge

qV = ½mv² · 1 eV per volt per e

Accelerating potential difference (V)The voltage across the gap. A cathode-ray tube ran at a few kilovolts, a dental X-ray tube at 60–70 kV, an electron microscope at 100–300 kV. The energy gained is qV whatever the shape of the field or the path taken, because the electric force is conservative.

Charge of the particle (multiples of e)1 for a proton or an electron, 2 for an alpha particle. The page takes this as positive and accelerates it from the positive plate to the negative one; a negative particle gains exactly the same energy going the other way, so the numbers below are unchanged.

Mass of the particle (atomic mass units, u)A proton is 1.007276 u, an electron 0.000548580 u, an alpha 4.001506 u. The atomic mass constant is 1.660 539 068 92(52) × 10⁻²⁷ kg (CODATA 2022). Try the electron at 2000 V: it comes out at nearly 9% of the speed of light, where the proton is at 0.2%.

Width of the accelerating gap (cm)The distance between the plates. It does not affect the energy gained or the final speed at all — only how strong the field is, how big the acceleration is and how long the crossing takes. Halve the gap and the field doubles, the acceleration doubles, and the particle arrives in half the time at the same speed.

The accelerating gap, drawn to the plate separation you enter. The particle is shown six times on its way across, at distances proportional to the square of the time, so the widening spacing is the acceleration itself. The green arrow at the exit is the speed it leaves with, on a logarithmic scale so that a slow proton and a fast electron both fit the frame. Widen the gap and the picture stretches, the field weakens and the exit arrow does not change by a pixel, because the gap width has no effect on the final speed.

618.99km/sExample

A proton (1.007276 u, charge e) accelerated through 2000 V across a 5.0 cm gap

Work done by the field becomes kinetic energy

qV = ½mv² · v = √(2qV/m) · 1 eV = 1.602 176 634 × 10⁻¹⁹ J (exact) · E = V/d · relativistic: K = (γ − 1)mc²

qV: the work the field does, and therefore the kinetic energy gained if the particle started from rest. It does not depend on the shape of the field or the route taken — only on the potential difference between the two ends, because the electric force is conservative
the electronvolt: the energy one elementary charge gains through one volt. Since 2019 the elementary charge is exactly 1.602 176 634 × 10⁻¹⁹ C by definition, so the electronvolt is exactly that many joules — an exact conversion with no uncertainty, which is unusual and worth knowing
v = √(2qV/m): the speed, and note the square root: to double the speed you need four times the voltage. A row above gives that voltage directly. The lighter the particle the faster it goes for the same volts, by 1/√m — an electron leaves a proton standing by a factor of 43
E = V/d: the field in a uniform gap. The gap width changes the field, the acceleration and the crossing time, and changes the final energy and speed not at all. Students often expect a narrower gap to give a faster particle; it gives a quicker one
when relativity bites: the true kinetic energy is (γ − 1)mc², and ½mv² is its first term. The classical speed runs about 0.375β² high, so it is out by 0.004% at 0.01c, 0.4% at 0.1c and a few per cent by 0.3c. The band under the answer and the percentage row above both track it

Worked example

A proton (1.007276 u, charge e) accelerated through 2000 V across a 5.0 cm gap
READ THE ENERGY STRAIGHT OFF THE VOLTAGE. A charge of 1 e through 2000 V gains 2000 eV = 2.000 keV. No arithmetic: that is what an electronvolt is for. In joules it is 2000 × 1.602 × 10⁻¹⁹ = 3.204 × 10⁻¹⁶ J.
GET THE MASS INTO KILOGRAMS. m = 1.007276 × 1.6605 × 10⁻²⁷ = 1.6726 × 10⁻²⁷ kg.
USE ½mv² = qV. v = √(2K/m) = √(2 × 3.204 × 10⁻¹⁶ / 1.6726 × 10⁻²⁷) = √(3.832 × 10¹¹) = 6.190 × 10⁵ m/s = 618.99 km/s.
CHECK WHETHER RELATIVITY MATTERS. v/c = 0.00206, so γ = 1.000002 and the classical answer is high by 0.00016%. Comfortably negligible, and the band under the result says so. Put the electron's mass in instead and the same 2000 V gives 26 524 km/s, which is 8.8% of c — still only 0.3% out, but heading that way.
NOW THE GAP, WHICH CHANGES NONE OF THE ABOVE. E = V/d = 2000/0.050 = 40 000 V/m = 40 kV/m, so a = qE/m = (1.602 × 10⁻¹⁹ × 40 000)/(1.6726 × 10⁻²⁷) = 3.83 × 10¹² m/s². The crossing takes t = v/a = 162 ns.
SANITY-CHECK THE GAP WITH SUVAT. If the acceleration really is 3.83 × 10¹² m/s² over 0.050 m, then v² = 2as = 2 × 3.83 × 10¹² × 0.050 = 3.83 × 10¹¹, giving v = 6.19 × 10⁵ m/s. The same answer by a completely different route, which is the useful check: energy methods and kinematics have to agree.

The same 2000 V applied to four different particles

The first three gain the same energy because they all carry one elementary charge, and differ in speed only through √m. The alpha gains twice the energy because it carries twice the charge — and ends up at almost exactly the deuteron’s speed, because it also has twice the mass, and the two factors of two cancel in √(2qV/m). The last column is the percentage by which ½mv² overstates the speed; it grows as roughly 0.375 × (v/c)².
Particle	Energy gained	Speed (classical)	v/c	Classical error
Electron	2.00 keV	26,524 km/s	0.0885	0.29%
Proton	2.00 keV	618.99 km/s	0.00206	0.00016%
Deuteron	2.00 keV	437.8 km/s	0.00146	0.00008%
Alpha particle (charge 2e)	4.00 keV	439.0 km/s	0.00146	0.00008%

One line of algebra that connects volts, joules and metres per second

qV = ½mv² is the whole page. The field does work qV on the charge, and if the charge started from rest that work is its kinetic energy. Nothing about the geometry matters: the gap can be wide or narrow, the field uniform or lumpy, the path straight or curved, and the energy gained is the same, because the electric force is conservative. Only the potential difference between the two ends appears.

That is where the electronvolt comes from, and it is a definition rather than a measurement. One elementary charge through one volt gains one electronvolt, and since the 2019 revision of the SI fixed e at exactly 1.602 176 634 × 10⁻¹⁹ C, one eV is exactly that many joules. No bracket, no uncertainty. It is also why accelerator energies are quoted in eV in the first place: the number is simply the machine’s voltage, read off a dial.

Watch the square root. v = √(2qV/m), so doubling the speed takes four times the voltage, and a particle a hundred times lighter goes ten times faster for the same volts. That is why an electron through 2000 V is already at nearly 9% of the speed of light while a proton through the same 2000 V is at 0.2%. If you expected the energies to be the same and the speeds to be the same, only the first half of that is right.

And here is where the classical answer stops being true. Kinetic energy is really (γ − 1)mc², of which ½mv² is the first term. The classical speed comes out high by about 0.375(v/c)² — four parts in a hundred thousand at 1% of c, four parts in a thousand at 10%, a few per cent by 30%. This page prints the exact relativistic speed beside the classical one and the percentage between them, and the band under the headline changes as you cross those thresholds. Push the voltage far enough and the classical formula will cheerfully return a speed greater than c, which is not a fault in the page but the clearest possible signal that the formula has run out of road. For what a magnetic field then does to the particle, see the cyclotron frequency page, and for the potential itself, electric potential and potential energy.

Frequently asked questions

Does the gap width change the final speed?

No, and this is the most useful surprise on the page. The gap changes the field (E = V/d), the acceleration and the time taken, but the work done is qV whatever the distance, so the exit speed is identical. Set the gap to 0.1 cm and then to 100 cm and watch the speed row sit still while the field row changes by a factor of a thousand.

Why do a proton and an electron gain the same energy but very different speeds?

Because the energy depends on the charge and the voltage, and both carry one elementary charge. The speed then depends on the mass through v = √(2qV/m), and the proton is 1836 times heavier, so it comes out √1836 = 43 times slower. Same energy, different speed.

When do I need to use the relativistic formula?

When the error matters more to you than the extra arithmetic. As a rule of thumb the classical speed is high by roughly 0.375(v/c)²: that is 0.004% at 1% of c, 0.4% at 10%, and about 3% at 30%. School questions almost never need the correction for protons and heavier particles; they often do for electrons above a few tens of kilovolts, which is why an electron microscope’s design cannot ignore it. The page gives you both figures and the percentage so you need not guess.

What if the particle is negative?

It gains the same energy, moving the other way. An electron accelerates from the negative plate towards the positive one and arrives with the same qV. This page takes the charge as a positive number for clarity; nothing in the physics changes except the direction, which the diagram draws left to right either way.

Why is the de Broglie wavelength row here?

Because it is the bridge to modern physics and it is calculated from exactly the momentum this page has just worked out: λ = h/p. It is the reason electron microscopes exist — 2000 V gives an electron a wavelength of about 27 pm, some twenty thousand times shorter than visible light, and resolution follows wavelength.

Related calculators

References

Fundamental Physical Constants — Frequently used constants (2022 CODATA wallet card), NIST Committee on Data for Science and Technology, physics.nist.gov/cuu/pdf/wallet_2022.pdf (read 11 November 2026). Every constant on this page comes from here, and the thing worth noticing is which numbers have a bracket after them. The elementary charge is printed 1.602 176 634 × 10⁻¹⁹ C and the speed of light 299 792 458 m s⁻¹, both with no bracket at all, because since 2019 they are exact by definition — the coulomb and the metre are now defined from them. But the vacuum electric permittivity is printed 8.854 187 8188(14) × 10⁻¹² F m⁻¹ and the vacuum magnetic permeability 1.256 637 061 27(20) × 10⁻⁶ N A⁻²: both now carry an uncertainty of about 1.6 parts in 10¹⁰, where before 2019 μ₀ was exactly 4π × 10⁻⁷ and ε₀ exactly 1/μ₀c². Almost every school book still prints the old exact forms. At three significant figures nothing changes; which constants are definitions changed completely.
The International System of Units (SI), 9th edition 2019, version V4.01 of June 2026, Bureau International des Poids et Mesures, bipm.org/documents/d/guest/si-brochure-9-en-pdf (read 11 November 2026). Cited for the units, from Table 4 of coherent derived units with special names: magnetic flux, weber (Wb) = kg m² s⁻² A⁻¹ = V s; magnetic flux density, tesla (T) = kg s⁻² A⁻¹ = Wb/m²; inductance, henry (H) = kg m² s⁻² A⁻² = Wb/A. Those three equalities are worth more than they look: a weber is a volt-second, so a flux change of one weber in one second is one volt — which is Faraday’s law written as a unit. And a henry is a weber per ampere, so an inductance is just how much flux linkage one ampere produces.
Annual Syllabus, Class XII, Physics, Code 42, Session 2025-26, Central Board of Secondary Education, edustud.nic.in/edu/SYLLABUS_2025_26/12/12_Physics_EM.pdf (read 11 November 2026). The board’s own document, cited and not endorsed by it. Unit I Chapter 1 asks for “Electric flux, statement of Gauss’s theorem and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell (field inside and outside)”. Unit III Chapter 4 asks for “force on a moving charge in uniform magnetic and electric fields”, “torque experienced by a current loop in uniform magnetic field” and “Current loop as a magnetic dipole and its magnetic dipole moment”. Unit IV Chapter 6 is one line and covers three of this batch’s records: “Electromagnetic induction; Faraday’s laws, induced EMF and current; Lenz’s Law, Self and mutual induction”. The Hall effect is not in this document.
University Physics Volume 2, chapters 11 and 14, OpenStax, Rice University (openstax.org/books/university-physics-volume-2, key-equations pages read 11 November 2026). An openly licensed university text, cited here because the coaching sites that dominate searches for this topic are not citable and mostly copy one another. Chapter 11 gives the radius of a charged particle’s path as r = mv/(qB), its period as T = 2πm/(qB), the Hall potential as V = IBl/(neA), the drift velocity in crossed fields as v_d = E/B, the magnetic dipole moment as μ = NIA and the torque as τ = μ × B. Chapter 14 gives L_solenoid = μ₀N²A/l, NΦ = LI, ε = −L dI/dt, M = N₂Φ₂₁/I₁ = N₁Φ₁₂/I₂ and U = ½LI². Neither key-equations page lists a cyclotron frequency or a field energy density, so those two are stated here from the definitions and not attributed to this source.