Physics · Electrostatics

Electric potential-energy change from voltage difference

The change in electric potential energy of an unchanged test charge equals its signed charge multiplied by the difference between destination and starting potential. Electric field work has the opposite sign.

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Formula

Calculate

Δ U = q Δ V

ΔU in J; q in C; ΔV in V=J/C.

Variables & units

Variables & units
Symbol	Meaning	Unit
$ΔU$	Change in electric potential energy	J
$q$	Moved charge, including its sign	C
$ΔV$	Potential difference V_final−V_initial	V

LaTeX

\Delta U=q\,\Delta V

QUANTA STUDYquanta-study.de

Two lines through the origin: positive charge rises with potential difference, negative charge falls. Energy change is normalized to charge magnitude times 200 V. — For equally large positive and negative test charges, ΔU = qΔV. Normalized energy changes have opposite slopes; the graph is not a time trajectory.

Electric potential energy calculator

Find the energy change of a charge between two potentials, with its sign, in an unchanged electrostatic field.

Calculation model

Find

Signed charge q

Charge with its sign, e.g. −1 e for an electron or +2 µC for a positive test charge.

Potential difference destination minus start ΔV

Potential at the end minus potential at the start, e.g. +100 V when the charge moves to higher potential.

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

Working · rearrange for ΔU, substitute, evaluate

1. Rearrange for ΔU
$\Delta U=q\Delta V$
2. Substitute values (in SI units)
- q = -1 e = -1.60218 · 10⁻¹⁹ C
$(-1.60218\times10^{-19})\cdot 100=-1.60218\times10^{-17}$
3. Result
ΔU = -100 eV

Result: Energy change ΔU

ΔU = -100 eV

With q = -1 e and ΔV = 100 V, ΔU = -100 eV.

Quick tips

Work in elementary charges and volts and the energy comes out directly in eV: q = −1 e and ΔV = +100 V give ΔU = −100 eV.
Sign check: positive charges lose potential energy moving to lower potential, negative charges moving to higher potential. The field’s work is always −ΔU.

Assumptions and limits

Fixed test charge in an unchanged conservative electrostatic field, unchanged sources and negligible back-reaction. ΔV=Vdestination−Vstart using the same reference; ΔU is a change, not absolute energy without a specified zero or the full energy required to charge a capacitor.

Derivation & background

Electric potential is potential energy per unit charge, so $\Delta V=\frac{\Delta U}{q}$ and therefore $\Delta U=q\Delta V$ . The sign of q matters: a negative charge changes energy in the opposite direction to a positive charge across the same potential difference. Choosing the zero of potential changes U but not ΔU.

Exam blueprint

Validity range

Unchanged test charge in a conservative electrostatic source field, with negligible backreaction. $\Delta V=V_B-V_A$ is destination minus start. Electric work alone gives $\Delta K=-\Delta U$ . Quasistatic lossless transport with unchanged kinetic energy instead gives $W_{\mathrm{ext}}=\Delta U$ .

Derivation steps

Electric work and the conservative potential integral give the correctly signed energy change.

1Specify start A, destination B and an unchanged signed test charge q.
2Electric force is $\mathbf F_{\mathrm{el}}=q\mathbf E$ .
3Electric work is the line integral of force along the actual path.
4In a conservative field, destination potential minus start potential equals the negative electric field integral.
5Therefore $W_{\mathrm{el}}=-q\Delta V$ .
6Potential-energy change is negative field work: $\Delta U=q\Delta V$ .
7A common shift of potential reference cancels; check charge sign and unit C·V=J.

Rearrangements

Potential difference from energy change

\Delta V=\frac{\Delta U}{q}

For unchanged nonzero charge; retain its sign.

Task variant

A charge of $-3\,\mathrm{\mu C}$ crosses $\Delta V=4\,\mathrm V$ . Find its energy change.

$\Delta U=q\Delta V=-12\,\mathrm{\mu J}$ ; potential energy decreases.

Common mistakes

Equating potential V with potential energy U or forgetting the arbitrary reference zero.

$U=qV$ depends on q; the measurable, reference-independent change is $\Delta U=q\Delta V$ .

Exam context

Charged particles between plates, capacitor discharge and energy conversion in fields.

Formula cluster

Electric potential and energy

Voltage expresses energy change per charge.

E⃗ = F⃗el/q; Ehom = |U|/d

Electric Field Strength

F = ke·|q₁q₂|/r²

Coulomb's Law

E = ½CU²

Energy Stored in a Capacitor

Ekin = hf − WA

Photoelectric Effect (Einstein)

Worked example

For $q=+2.0\,\mathrm{\mu C}$ and $\Delta V=+5.0\,\mathrm V$ , $\Delta U=+10\,\mathrm{\mu J}$ . An electron over the same potential increase has $\Delta U=-5.0\,\mathrm{eV}$ , with field work $+5.0\,\mathrm{eV}$ . Unit check: C·V=J.

Applications

Accelerating charged particles, capacitors, electrical work and energy balances in electric fields.

How do you calculate with Change in electric potential energy?

Here is how to work through a typical Change in electric potential energy (ΔU = qΔV) task step by step:

1
Task
A charge of $-3\,\mathrm{\mu C}$ crosses $\Delta V=4\,\mathrm V$ . Find its energy change.
Solution path
$\Delta U=q\Delta V=-12\,\mathrm{\mu J}$ ; potential energy decreases.

Understand and apply the equation

Voltage is not a particle’s energy but a potential difference per charge. Only signed charge determines whether transport increases or decreases electric potential energy. Relation $\Delta U=q\Delta V$ connects this potential description to electric work and possible change in kinetic energy.

The same potential difference affects charges differently

Positive and negative probes at the same two locations experience the same potential difference but opposite changes in potential energy. Voltage describes the source field; charge connects that field to a particular probe’s energy. Begin with destination minus start and the charge sign.[1]

Distinguish potential-energy change, field work and kinetic-energy change. Field work is the negative potential-energy change. It becomes entirely kinetic energy only when no other force does work. In slow guided transport, an external mechanism supplies or receives the balancing contribution.[1]

How do you calculate signed change in electric potential energy?

\Delta U=U_B-U_A=q(V_B-V_A)=q\Delta V

V is potential and U here potential energy. In other articles U may denote voltage; do not confuse these quantities.[1]

A positive potential change increases a positive charge’s potential energy but decreases a negative charge’s energy. Field belongs to the fixed source configuration and both endpoints must be specified. The same expression cannot simply be assigned to an uncharged body with an unspecified changing charge; charge must remain constant during the transport.[1]

Potential follows from electric work

\begin{aligned}W_{\mathrm{el}}&=\int_A^Bq\mathbf E\cdot d\boldsymbol\ell=-\Delta U\\\Delta V&=-\int_A^B\mathbf E\cdot d\boldsymbol\ell\end{aligned}

For fixed charge and a conservative electrostatic field, the same path relates energy and potential.[1]

Multiply the definition of potential difference by q. This gives $-q\int_A^B\mathbf E\cdot\mathrm d\boldsymbol\ell=\Delta U$ and hence $\Delta U=q\Delta V$ . A quasistatic external transport with negligible losses has $W_{\mathrm{ext}}=\Delta U$ because kinetic energy is unchanged. Electric field work has the opposite sign. These are two views of the same energy balance, not an additional source of energy. Electric field strength and the selected displacement projection give the same work balance as the potential difference.[1]

\Delta K=W_{\mathrm{el}}=-q\Delta V

This additional equation applies when only the electric field force does work and no other energy change is omitted.[1]

A potential reference shift does not change energy differences

Replace V everywhere by $V+V_0$ . Then $qV$ shifts by constant $qV_0$ while $q(V_B-V_A)$ stays unchanged. Potential-energy differences therefore do not depend on a common reference zero. For multiple sources, first sum scalar potentials at each observation point and then take the difference. Potential can be nonzero at a point where vector field cancels by symmetry.[1][2]

In the test-charge model sources are fixed and backreaction neglected. Building an entire charge configuration requires accounting for work during incremental transport. Thus a linear capacitor stores total energy $\frac12QU$ rather than $QU$ because voltage grows while charging. The test charge considered here instead crosses an already established potential difference.[3]

How do you convert between joules and electronvolts?

\begin{aligned}1\,\mathrm{C\,V}&=1\,\mathrm J\\1\,\mathrm{eV}&=1.602176634\times10^{-19}\,\mathrm J\end{aligned}

One electronvolt is the energy unit e times 1 V, not a charge or predetermined sign of energy change.[4]

\begin{aligned}\Delta V&=\frac{\Delta U}{q}\quad(q\ne0)\\q&=\frac{\Delta U}{\Delta V}\quad(\Delta V\ne0)\end{aligned}

Retain signs; zero potential difference and zero energy change do not determine q.

For charge $q=ze$ , energy change in eV is numerically z times potential change in volts, including sign. A probe with $q=+2.0\,\mathrm{\mu C}$ across $\Delta V=+5.0\,\mathrm V$ gives $\Delta U=+10\,\mathrm{\mu J}$ ; an electron in the same potential rise instead has −5.0 eV potential-energy change. Any claimed energy gain or loss must therefore specify whether it concerns potential or kinetic energy.[4][1]

Negative probe and accelerated model ion

A negative test charge rises in potential

Given: A fixed test charge $q=-4.50\,\mathrm{pC}$ moves from $V_A=-60.0\,\mathrm V$ to $V_B=+120\,\mathrm V$ . Sources unchanged, conservative field; equal endpoint kinetic energies for the quasistatic additional balance.
Find: Find $\Delta U$ and electric versus quasistatic external work.

Approach: Destination minus starting potential is multiplied by unchanged signed charge. Electric work and quasistatic external work have opposite signs.
$\begin{aligned}\Delta V&=V_B-V_A\\\Delta U&=q\Delta V\\W_{\mathrm{el}}&=-\Delta U\end{aligned}$
Subtract initial potential from final potential.
$\Delta V=120\,\mathrm V-(-60\,\mathrm V)=180\,\mathrm V$
Multiply by negative charge.
$\Delta U=(-4.50\times10^{-12}\,\mathrm C)(180\,\mathrm V)=-8.10\times10^{-10}\,\mathrm J$
Distinguish work conventions for equal endpoint kinetic energies.
$\begin{aligned}W_{\mathrm{el}}&=+810\,\mathrm{pJ}\\W_{\mathrm{ext}}&=-810\,\mathrm{pJ}\end{aligned}$

Result: Potential energy decreases by 810 pJ despite increasing potential. Field does positive work; the quasistatic external mechanism receives the same energy.

Check: A common +1000 V offset changes endpoints to 940 V and 1120 V but retains $\Delta V=180\,\mathrm V$ . The quotient $\frac{\Delta U}{q}$ also returns 180 V. Signs do not depend on an arbitrary potential reference.

A doubly positive model ion starts at rest

Given: $q=+2e$ and specified model mass $m=6.64\times10^{-27}\,\mathrm{kg}$ . The ion starts at rest at 450 V and reaches 330 V; only electric work, no collisions, no relativistic correction needed.
Find: Find energy changes and final speed.

Approach: Starting at rest with electric work alone converts potential-energy loss entirely to kinetic energy, giving nonrelativistic speed.
$\begin{aligned}K_f&=-q\Delta V\\v&=\sqrt{\frac{2K_f}{m}}\end{aligned}$
Potential change is negative.
$\begin{aligned}\Delta V&=330\,\mathrm V-450\,\mathrm V=-120\,\mathrm V\\\Delta U&=-240\,\mathrm{eV}\\&=-3.8452239216\times10^{-17}\,\mathrm J\end{aligned}$
Field work increases kinetic energy by the same magnitude.
$\begin{aligned}K_f&=240\,\mathrm{eV}\\v&=\sqrt{\frac{2K_f}{m}}\approx1.07620\times10^5\,\mathrm{m/s}\end{aligned}$

Result: $\Delta U=-240\,\mathrm{eV}$ , $\Delta K=+240\,\mathrm{eV}$ and $v\approx107620\,\mathrm{m/s}$ . The 120 V drop is not itself the 240 eV energy; double charge provides factor 2.

Check: Using unrounded v, $\frac12mv^2$ returns $3.8452239216\cdot10^{-17}\,\mathrm J$ . $\frac vc\approx3.59\cdot10^{-4}$ is small, consistent with the nonrelativistic model. Initial and final $K+U$ sums agree.

A charge cannot always reach a potential endpoint

A positive body moving against field force toward higher potential loses kinetic energy. Without an extra drive it can reach only a potential rise $\Delta V\le\frac{K_0}{q}$ . Negative charge reverses the relevant potential trend. Formula $\Delta U=q\Delta V$ calculates energy change for specified endpoints; it does not guarantee that initial motion can actually reach them. Kinetic energy must therefore remain nonnegative along the trajectory actually reached.[1]

For a uniform electrostatic field, $\Delta V=-\mathbf E\cdot\Delta\mathbf r$ links work to projected displacement. Transverse motion along an equipotential changes no potential energy despite possibly nonzero field. At high particle energies the electric work balance remains, but $K=\frac12mv^2$ must be replaced by an appropriate relativistic energy expression.[1][2]

Example: a proposed endpoint cannot be reached

A positive model particle slows down

Given: A test charge $q=+e$ starts at $x=0$ with $K_0=180\,\mathrm{eV}$ moving toward $+x$ . For $x\ge0$ , $V(x)-V(0)=(600\,\mathrm{V/m})x$ , so the field points toward $-x$ . Only electric work; straight trajectory and nonrelativistic motion.
Find: Check a target at 0.400 m and find the turning point.

Approach: Kinetic energy must remain nonnegative along the accessible trajectory. It vanishes at the turning point.
$\begin{aligned}K(x)&=K_0-q\,[V(x)-V(0)]\\x_{\mathrm{turn}}&=\frac{K_0}{q\,(600\,\mathrm{V/m})}\end{aligned}$
Substitution: For one positive elementary charge, potential rise in volts has the same numerical energy increase in electronvolts.
$\begin{aligned}\Delta V(0.400\,\mathrm m)&=240\,\mathrm V\\K_{\mathrm{formal}}(0.400\,\mathrm m)&=(180-240)\,\mathrm{eV}=-60\,\mathrm{eV}\\x_{\mathrm{turn}}&=0.300\,\mathrm m\end{aligned}$

Result: The target is inaccessible under these conditions. The particle turns at $0.300\,\mathrm m$ ; formal negative kinetic energy is not a physical state.

Check: Field force opposes initial motion. Its work to the turning point is $-180\,\mathrm{eV}$ , exactly the negative initial kinetic energy. Additional external driving would change the assumed balance.

Energy partition up to the turning point

QUANTA STUDYquanta-study.de

The plotted domain ends at the accessible turning point.

K+\Delta U=180\,\mathrm{eV}

stays constant.

\Delta U

is the increase relative to the start, not an absolute potential-energy reference.[1]

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Five sign and energy checks

Calculate · Foundation

A charge +3.00 μC crosses $\Delta V=-20.0\,\mathrm V$ . Find $\Delta U$ and field work.

Show hint

Field work is negative potential-energy change.

Learning goal: Calculate signed energy change and field work.

Show solution and calculation

Approach: $\Delta U=q\Delta V,\quad W_{\mathrm{el}}=-\Delta U$ .
$\Delta U=-60.0\,\mathrm{\mu J}$ and $W_{\mathrm{el}}=+60.0\,\mathrm{\mu J}$ .
With only electric work, K increases by 60.0 µJ.

\Delta U=-60.0\,\mathrm{\mu J}

Rearrange · Application

A fixed charge $q=-2.00\,\mathrm{nC}$ has $\Delta U=+6.00\,\mathrm{\mu J}$ . Find $\Delta V$ .

Show hint

Divide by negative charge.

Learning goal: Infer potential difference from negative charge and energy change.

Show solution and calculation

Approach: $\Delta V=\frac{\Delta U}{q}$ .
$\Delta V=\frac{6.00\cdot10^{-6}\,\mathrm J}{-2.00\cdot10^{-9}\,\mathrm C}=-3000\,\mathrm V$ .
A potential drop increases negative-charge potential energy.

\Delta V=-3.00\times10^3\,\mathrm V

Transfer · Application

An electron initially has 100 eV kinetic energy and must cross $\Delta V=-75.0\,\mathrm V$ without other work. Is the endpoint energetically reachable, and what K remains?

Show hint

$q=-e$ gives positive $\Delta U$ for a potential drop.

Learning goal: Check final electron energy and energetic accessibility.

Show solution and calculation

Approach: $K_f=K_i-q\Delta V$ .
$\Delta U=+75.0\,\mathrm{eV}$ , so $K_f=(100-75)\,\mathrm{eV}=25.0\,\mathrm{eV}$ .
Positive remaining energy does not rule out the endpoint based on its energy.
It alone does not establish accessibility: a larger potential-energy barrier en route could turn the probe back earlier, and trajectory geometry must also allow passage.
On a monotonic decelerating path, the potential drop at the turning point would have magnitude 100 V.

K_f=25.0\,\mathrm{eV}

Interpret · Application

Two small probes +2 μC and −2 μC are each moved quasistatically from $V_A=10\,\mathrm V$ to $V_B=40\,\mathrm V$ in the same unchanged electrostatic field. No dissipative forces occur. Compare ΔU, field work and external work.

Show hint

Both have the same ΔV but opposite charge signs.

Learning goal: Compare signed field work and energy changes for opposite charges.

Show solution and calculation

Approach: $\Delta U=q(V_B-V_A)$ .
$\Delta U=q\Delta V$ gives +60 μJ for the positive probe and −60 μJ for the negative one.
Field work is $-\Delta U$ ; quasistatic external work equals $+\Delta U$ .

Check: external and electric work cancel, consistent with unchanged kinetic energy. Higher potential therefore does not mean higher potential energy for every charge. Small probes leave the sources unchanged.

\begin{aligned}\Delta U_+&=(2\,\mathrm{\mu C})(30\,\mathrm V)=60\,\mathrm{\mu J}\\\Delta U_-&=-60\,\mathrm{\mu J}\\W_{\mathrm{el}}&=-\Delta U,\qquad W_{\mathrm{ext}}=\Delta U\end{aligned}

[1]

Error analysis · Foundation

For $q=+3\,\mathrm{\mu C}$ , $V_A=5\,\mathrm V$ and $V_B=-15\,\mathrm V$ . A solution adds 100 V to both and claims field work has changed. Check using both calculations.

Show hint

The same offset cancels in $V_B-V_A$ .

Learning goal: Separate a constant potential offset from potential difference.

Show solution and calculation

Approach: $(V_B+V_0)-(V_A+V_0)=V_B-V_A$ . ΔV stays −20 V before and after the shift, leaving $\Delta U=-60\,\mathrm{\mu J}$ and field work +60 μJ.
Individual $qV_A$ and $qV_B$ values each increase by 300 μJ, but their difference stays fixed.

Check: a spatially constant offset also leaves the gradient and electric field unchanged. A chosen zero cannot change force or measured work difference.

\begin{aligned}\Delta V_{\mathrm{shift}}&=(V_B+100\,\mathrm V)-(V_A+100\,\mathrm V)=\Delta V\\W_{\mathrm{el}}&=-q\Delta V=60\,\mathrm{\mu J}\end{aligned}

[1][2]

When a scalar potential difference is insufficient

Path independence assumes a conservative electrostatic field and unchanged charge. With time-dependent magnetic induction, electric work is not generally determined solely by a scalar potential difference between endpoints. Moving sources, radiation, contact reactions or additional forces require fuller energy and field accounting. The test-charge relation does not calculate total self-energy of the source body or universal energy of arbitrary charging processes.[1][3][6]

In brief

Form $\Delta V=V_B-V_A$ , multiply by signed charge and identify the energy type. $\Delta U=q\Delta V$ , field work $-\Delta U$ and, where applicable, $\Delta K=-\Delta U$ describe one balance. Then check whether the assumed motion can actually reach the endpoint.[1]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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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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Change history

October 2, 2026Highlighted signed potential-energy change and energy units in question headings; added paragraph connections to field strength and kinetic reachability checks.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Clarified introduction, model connections and solution steps; added an original third example, quantitative diagram and concise summary.

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Quanta. Change in electric potential energy. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/elektrische-potentialenergie-spannung

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Frequently asked questions about Change in electric potential energy

What does the change in electric potential energy describe physically?+

The relation describes the change in potential energy of a fixed test charge moving between two locations in a conservative electric field. Multiply the destination potential minus the starting potential by the charge, including its sign. Positive charges gain potential energy when potential increases; negative charges lose it. Electric field work has the opposite sign. For −4.5 pC and a potential increase of 180 V, the energy change is −810 pJ. This is a difference for the probe in a prescribed source field, not total source self-energy or the general charging energy of a capacitor.

How can I check a calculation involving the change in electric potential energy?+

First record charge, starting potential and destination potential with their signs. Calculate destination minus start, then multiply. In the first example, −60 V and +120 V give an increase of 180 V; combined with −4.5 pC, this yields −810 pJ. Independently check that electric field work is +810 pJ. Slow motion with unchanged kinetic energy therefore requires −810 pJ of external work. Adding the same constant to both potentials must leave the result unchanged. Using magnitudes alone would remove this sign check, so given potentials and charge sign belong explicitly in the setup.

What assumptions and limits apply to the change in electric potential energy?+

The simple endpoint relation assumes unchanged charge and a conservative electrostatic field. The source field is prescribed, with a small probe producing negligible backreaction. With time-dependent magnetic induction, a scalar potential difference does not generally describe electric work completely. Moving sources, additional forces or radiation may require extra accounting terms. Only when electric force alone does work is the kinetic-energy change the negative of the potential-energy change. During guided slow motion, an external force instead supplies the corresponding work contribution. These different conditions of motion must be established before interpreting the resulting energy difference or using it to calculate speed.

How do I check units and scale for the change in electric potential energy?+

Coulomb multiplied by volt gives joule. Convert microcoulombs and picocoulombs carefully before substitution; a missing factor of a million can drastically change small-charge energies. Electronvolts are useful for individual particles: one positive elementary charge over a one-volt potential increase gains one electronvolt of potential energy. An electron over that same increase instead loses one electronvolt. For model charge +2e and a 120 V potential drop, the change is −240 eV. Check both sign and magnitude. A negative result here is an allowed energy change, rather than an incorrectly negative stored energy for a passive capacitor.

When should I combine the change in electric potential energy with another relation?+

To calculate speed, combine potential-energy change with the work–energy theorem and the appropriate kinetic-energy expression. With electric work alone, initial kinetic energy plus field work gives final kinetic energy. An initially stationary model charge +2e therefore gains 240 eV of kinetic energy across a 120 V potential drop. Given its mass and nonrelativistic motion, obtain speed from the square-root relation next. An electron moving towards lower potential can instead reach a turning point. An energy condition alone does not prove every spatial trajectory; field direction, initial velocity and possible additional forces determine the actual path.

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Subject references and further reading

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[1]OpenStax, University Physics Vol. 2, §7.2: Electric Potential and Potential Difference

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\oint_{\partial V} \mathbf{E}\cdot d\mathbf{A} = \frac{Q_{\mathrm{innen}}}{\varepsilon_0}

Gauss's Law for Electrostatics

U = R \cdot I

Ohm's Law

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Electric potential-energy change from voltage difference

BasicExam-relevantUpper secondaryUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

ΔU

Change in electric potential energy

q

Moved charge, including its sign

ΔV

Potential difference V_final−V_initial

Derivation & background

Exam blueprint

Validity range

Derivation steps

Electric work and the conservative potential integral give the correctly signed energy change.

1Specify start A, destination B and an unchanged signed test charge q.
2Electric force is $\mathbf F_{\mathrm{el}}=q\mathbf E$ .
3Electric work is the line integral of force along the actual path.
4In a conservative field, destination potential minus start potential equals the negative electric field integral.
5Therefore $W_{\mathrm{el}}=-q\Delta V$ .
6Potential-energy change is negative field work: $\Delta U=q\Delta V$ .
7A common shift of potential reference cancels; check charge sign and unit C·V=J.

How do you calculate with Change in electric potential energy?

Here is how to work through a typical Change in electric potential energy (ΔU = qΔV) task step by step:

Task

A charge of $-3\,\mathrm{\mu C}$ crosses $\Delta V=4\,\mathrm V$ . Find its energy change.

Solution path

$\Delta U=q\Delta V=-12\,\mathrm{\mu J}$ ; potential energy decreases.

Understand and apply the equation

Voltage is not a particle’s energy but a potential difference per charge. Only signed charge determines whether transport increases or decreases electric potential energy. Relation

\Delta U=q\Delta V

connects this potential description to electric work and possible change in kinetic energy.

The same potential difference affects charges differently

How do you calculate signed change in electric potential energy?

\Delta U=U_B-U_A=q(V_B-V_A)=q\Delta V

V is potential and U here potential energy. In other articles U may denote voltage; do not confuse these quantities.[1]

Potential follows from electric work

\begin{aligned}W_{\mathrm{el}}&=\int_A^Bq\mathbf E\cdot d\boldsymbol\ell=-\Delta U\\\Delta V&=-\int_A^B\mathbf E\cdot d\boldsymbol\ell\end{aligned}

For fixed charge and a conservative electrostatic field, the same path relates energy and potential.[1]

\Delta K=W_{\mathrm{el}}=-q\Delta V

This additional equation applies when only the electric field force does work and no other energy change is omitted.[1]

A potential reference shift does not change energy differences

How do you convert between joules and electronvolts?

\begin{aligned}1\,\mathrm{C\,V}&=1\,\mathrm J\\1\,\mathrm{eV}&=1.602176634\times10^{-19}\,\mathrm J\end{aligned}

One electronvolt is the energy unit e times 1 V, not a charge or predetermined sign of energy change.[4]

\begin{aligned}\Delta V&=\frac{\Delta U}{q}\quad(q\ne0)\\q&=\frac{\Delta U}{\Delta V}\quad(\Delta V\ne0)\end{aligned}

Retain signs; zero potential difference and zero energy change do not determine q.

Negative probe and accelerated model ion

A negative test charge rises in potential

Given: A fixed test charge $q=-4.50\,\mathrm{pC}$ moves from $V_A=-60.0\,\mathrm V$ to $V_B=+120\,\mathrm V$ . Sources unchanged, conservative field; equal endpoint kinetic energies for the quasistatic additional balance.
Find: Find $\Delta U$ and electric versus quasistatic external work.

Approach: Destination minus starting potential is multiplied by unchanged signed charge. Electric work and quasistatic external work have opposite signs.
$\begin{aligned}\Delta V&=V_B-V_A\\\Delta U&=q\Delta V\\W_{\mathrm{el}}&=-\Delta U\end{aligned}$
Subtract initial potential from final potential.
$\Delta V=120\,\mathrm V-(-60\,\mathrm V)=180\,\mathrm V$
Multiply by negative charge.
$\Delta U=(-4.50\times10^{-12}\,\mathrm C)(180\,\mathrm V)=-8.10\times10^{-10}\,\mathrm J$
Distinguish work conventions for equal endpoint kinetic energies.
$\begin{aligned}W_{\mathrm{el}}&=+810\,\mathrm{pJ}\\W_{\mathrm{ext}}&=-810\,\mathrm{pJ}\end{aligned}$

Result: Potential energy decreases by 810 pJ despite increasing potential. Field does positive work; the quasistatic external mechanism receives the same energy.

A doubly positive model ion starts at rest

Given: $q=+2e$ and specified model mass $m=6.64\times10^{-27}\,\mathrm{kg}$ . The ion starts at rest at 450 V and reaches 330 V; only electric work, no collisions, no relativistic correction needed.
Find: Find energy changes and final speed.

Approach: Starting at rest with electric work alone converts potential-energy loss entirely to kinetic energy, giving nonrelativistic speed.
$\begin{aligned}K_f&=-q\Delta V\\v&=\sqrt{\frac{2K_f}{m}}\end{aligned}$
Potential change is negative.
$\begin{aligned}\Delta V&=330\,\mathrm V-450\,\mathrm V=-120\,\mathrm V\\\Delta U&=-240\,\mathrm{eV}\\&=-3.8452239216\times10^{-17}\,\mathrm J\end{aligned}$
Field work increases kinetic energy by the same magnitude.
$\begin{aligned}K_f&=240\,\mathrm{eV}\\v&=\sqrt{\frac{2K_f}{m}}\approx1.07620\times10^5\,\mathrm{m/s}\end{aligned}$

Result: $\Delta U=-240\,\mathrm{eV}$ , $\Delta K=+240\,\mathrm{eV}$ and $v\approx107620\,\mathrm{m/s}$ . The 120 V drop is not itself the 240 eV energy; double charge provides factor 2.

A charge cannot always reach a potential endpoint

Example: a proposed endpoint cannot be reached

A positive model particle slows down

Given: A test charge $q=+e$ starts at $x=0$ with $K_0=180\,\mathrm{eV}$ moving toward $+x$ . For $x\ge0$ , $V(x)-V(0)=(600\,\mathrm{V/m})x$ , so the field points toward $-x$ . Only electric work; straight trajectory and nonrelativistic motion.
Find: Check a target at 0.400 m and find the turning point.

Approach: Kinetic energy must remain nonnegative along the accessible trajectory. It vanishes at the turning point.
$\begin{aligned}K(x)&=K_0-q\,[V(x)-V(0)]\\x_{\mathrm{turn}}&=\frac{K_0}{q\,(600\,\mathrm{V/m})}\end{aligned}$
Substitution: For one positive elementary charge, potential rise in volts has the same numerical energy increase in electronvolts.
$\begin{aligned}\Delta V(0.400\,\mathrm m)&=240\,\mathrm V\\K_{\mathrm{formal}}(0.400\,\mathrm m)&=(180-240)\,\mathrm{eV}=-60\,\mathrm{eV}\\x_{\mathrm{turn}}&=0.300\,\mathrm m\end{aligned}$

Result: The target is inaccessible under these conditions. The particle turns at $0.300\,\mathrm m$ ; formal negative kinetic energy is not a physical state.

Energy partition up to the turning point

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The plotted domain ends at the accessible turning point.

K+\Delta U=180\,\mathrm{eV}

stays constant.

\Delta U

is the increase relative to the start, not an absolute potential-energy reference.[1]

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Five sign and energy checks

Calculate · Foundation

A charge +3.00 μC crosses $\Delta V=-20.0\,\mathrm V$ . Find $\Delta U$ and field work.

Show hint

Field work is negative potential-energy change.

Learning goal: Calculate signed energy change and field work.

Show solution and calculation

Approach: $\Delta U=q\Delta V,\quad W_{\mathrm{el}}=-\Delta U$ .
$\Delta U=-60.0\,\mathrm{\mu J}$ and $W_{\mathrm{el}}=+60.0\,\mathrm{\mu J}$ .
With only electric work, K increases by 60.0 µJ.

\Delta U=-60.0\,\mathrm{\mu J}

Rearrange · Application

A fixed charge $q=-2.00\,\mathrm{nC}$ has $\Delta U=+6.00\,\mathrm{\mu J}$ . Find $\Delta V$ .

Show hint

Divide by negative charge.

Learning goal: Infer potential difference from negative charge and energy change.

Show solution and calculation

Approach: $\Delta V=\frac{\Delta U}{q}$ .
$\Delta V=\frac{6.00\cdot10^{-6}\,\mathrm J}{-2.00\cdot10^{-9}\,\mathrm C}=-3000\,\mathrm V$ .
A potential drop increases negative-charge potential energy.

\Delta V=-3.00\times10^3\,\mathrm V

Transfer · Application

An electron initially has 100 eV kinetic energy and must cross $\Delta V=-75.0\,\mathrm V$ without other work. Is the endpoint energetically reachable, and what K remains?

Show hint

$q=-e$ gives positive $\Delta U$ for a potential drop.

Learning goal: Check final electron energy and energetic accessibility.

Show solution and calculation

Approach: $K_f=K_i-q\Delta V$ .
$\Delta U=+75.0\,\mathrm{eV}$ , so $K_f=(100-75)\,\mathrm{eV}=25.0\,\mathrm{eV}$ .
Positive remaining energy does not rule out the endpoint based on its energy.
It alone does not establish accessibility: a larger potential-energy barrier en route could turn the probe back earlier, and trajectory geometry must also allow passage.
On a monotonic decelerating path, the potential drop at the turning point would have magnitude 100 V.

K_f=25.0\,\mathrm{eV}

Interpret · Application

Two small probes +2 μC and −2 μC are each moved quasistatically from $V_A=10\,\mathrm V$ to $V_B=40\,\mathrm V$ in the same unchanged electrostatic field. No dissipative forces occur. Compare ΔU, field work and external work.

Show hint

Both have the same ΔV but opposite charge signs.

Learning goal: Compare signed field work and energy changes for opposite charges.

Show solution and calculation

Approach: $\Delta U=q(V_B-V_A)$ .
$\Delta U=q\Delta V$ gives +60 μJ for the positive probe and −60 μJ for the negative one.
Field work is $-\Delta U$ ; quasistatic external work equals $+\Delta U$ .

\begin{aligned}\Delta U_+&=(2\,\mathrm{\mu C})(30\,\mathrm V)=60\,\mathrm{\mu J}\\\Delta U_-&=-60\,\mathrm{\mu J}\\W_{\mathrm{el}}&=-\Delta U,\qquad W_{\mathrm{ext}}=\Delta U\end{aligned}

[1]

Error analysis · Foundation

For $q=+3\,\mathrm{\mu C}$ , $V_A=5\,\mathrm V$ and $V_B=-15\,\mathrm V$ . A solution adds 100 V to both and claims field work has changed. Check using both calculations.

Show hint

The same offset cancels in $V_B-V_A$ .

Learning goal: Separate a constant potential offset from potential difference.

Show solution and calculation

Approach: $(V_B+V_0)-(V_A+V_0)=V_B-V_A$ . ΔV stays −20 V before and after the shift, leaving $\Delta U=-60\,\mathrm{\mu J}$ and field work +60 μJ.
Individual $qV_A$ and $qV_B$ values each increase by 300 μJ, but their difference stays fixed.

Check: a spatially constant offset also leaves the gradient and electric field unchanged. A chosen zero cannot change force or measured work difference.

\begin{aligned}\Delta V_{\mathrm{shift}}&=(V_B+100\,\mathrm V)-(V_A+100\,\mathrm V)=\Delta V\\W_{\mathrm{el}}&=-q\Delta V=60\,\mathrm{\mu J}\end{aligned}

[1][2]

When a scalar potential difference is insufficient

In brief

Sources for this explanation

Quanta · Updated October 2, 2026

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An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

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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Change history

October 2, 2026Highlighted signed potential-energy change and energy units in question headings; added paragraph connections to field strength and kinetic reachability checks.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Clarified introduction, model connections and solution steps; added an original third example, quantitative diagram and concise summary.

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Quanta. Change in electric potential energy. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/elektrische-potentialenergie-spannung

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Electric potential-energy change from voltage difference

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Worked example

Applications

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A potential reference shift does not change energy differences

How do you convert between joules and electronvolts?

Negative probe and accelerated model ion

A negative test charge rises in potential

A doubly positive model ion starts at rest

A charge cannot always reach a potential endpoint

Example: a proposed endpoint cannot be reached

A positive model particle slows down

Energy partition up to the turning point

Five sign and energy checks

A charge +3.00 μC crosses ΔV=−20.0 V\Delta V=-20.0\,\mathrm VΔV=−20.0V. Find ΔU\Delta UΔU and field work.

A fixed charge q=−2.00 nCq=-2.00\,\mathrm{nC}q=−2.00nC has ΔU=+6.00 μJ\Delta U=+6.00\,\mathrm{\mu J}ΔU=+6.00μJ. Find ΔV\Delta VΔV.

An electron initially has 100 eV kinetic energy and must cross ΔV=−75.0 V\Delta V=-75.0\,\mathrm VΔV=−75.0V without other work. Is the endpoint energetically reachable, and what K remains?

Two small probes +2 μC and −2 μC are each moved quasistatically from VA=10 VV_A=10\,\mathrm VVA​=10V to VB=40 VV_B=40\,\mathrm VVB​=40V in the same unchanged electrostatic field. No dissipative forces occur. Compare ΔU, field work and external work.

For q=+3 μCq=+3\,\mathrm{\mu C}q=+3μC, VA=5 VV_A=5\,\mathrm VVA​=5V and VB=−15 VV_B=-15\,\mathrm VVB​=−15V. A solution adds 100 V to both and claims field work has changed. Check using both calculations.

When a scalar potential difference is insufficient

In brief

Sources for this explanation

Frequently asked questions about Change in electric potential energy

Subject references and further reading

More formulas: Electricity and magnetism

Retain Change in electric potential energy for exams

Electric potential-energy change from voltage difference

Variables & units

Electric potential energy calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electric potential and energy

Worked example

Applications

How do you calculate with Change in electric potential energy?

The same potential difference affects charges differently

How do you calculate signed change in electric potential energy?

Potential follows from electric work

A potential reference shift does not change energy differences

How do you convert between joules and electronvolts?

Negative probe and accelerated model ion

A negative test charge rises in potential

A doubly positive model ion starts at rest

A charge cannot always reach a potential endpoint

Example: a proposed endpoint cannot be reached

A positive model particle slows down

Energy partition up to the turning point

Five sign and energy checks

A charge +3.00 μC crosses ΔV=−20.0 V\Delta V=-20.0\,\mathrm VΔV=−20.0V. Find ΔU\Delta UΔU and field work.

A fixed charge q=−2.00 nCq=-2.00\,\mathrm{nC}q=−2.00nC has ΔU=+6.00 μJ\Delta U=+6.00\,\mathrm{\mu J}ΔU=+6.00μJ. Find ΔV\Delta VΔV.

An electron initially has 100 eV kinetic energy and must cross ΔV=−75.0 V\Delta V=-75.0\,\mathrm VΔV=−75.0V without other work. Is the endpoint energetically reachable, and what K remains?

Two small probes +2 μC and −2 μC are each moved quasistatically from VA=10 VV_A=10\,\mathrm VVA​=10V to VB=40 VV_B=40\,\mathrm VVB​=40V in the same unchanged electrostatic field. No dissipative forces occur. Compare ΔU, field work and external work.

For q=+3 μCq=+3\,\mathrm{\mu C}q=+3μC, VA=5 VV_A=5\,\mathrm VVA​=5V and VB=−15 VV_B=-15\,\mathrm VVB​=−15V. A solution adds 100 V to both and claims field work has changed. Check using both calculations.

When a scalar potential difference is insufficient

In brief

Sources for this explanation

Frequently asked questions about Change in electric potential energy

Subject references and further reading

A charge +3.00 μC crosses $\Delta V=-20.0\,\mathrm V$ . Find $\Delta U$ and field work.

A fixed charge $q=-2.00\,\mathrm{nC}$ has $\Delta U=+6.00\,\mathrm{\mu J}$ . Find $\Delta V$ .

An electron initially has 100 eV kinetic energy and must cross $\Delta V=-75.0\,\mathrm V$ without other work. Is the endpoint energetically reachable, and what K remains?

Two small probes +2 μC and −2 μC are each moved quasistatically from $V_A=10\,\mathrm V$ to $V_B=40\,\mathrm V$ in the same unchanged electrostatic field. No dissipative forces occur. Compare ΔU, field work and external work.

For $q=+3\,\mathrm{\mu C}$ , $V_A=5\,\mathrm V$ and $V_B=-15\,\mathrm V$ . A solution adds 100 V to both and claims field work has changed. Check using both calculations.

A charge +3.00 μC crosses $\Delta V=-20.0\,\mathrm V$ . Find $\Delta U$ and field work.

A fixed charge $q=-2.00\,\mathrm{nC}$ has $\Delta U=+6.00\,\mathrm{\mu J}$ . Find $\Delta V$ .

An electron initially has 100 eV kinetic energy and must cross $\Delta V=-75.0\,\mathrm V$ without other work. Is the endpoint energetically reachable, and what K remains?

Two small probes +2 μC and −2 μC are each moved quasistatically from $V_A=10\,\mathrm V$ to $V_B=40\,\mathrm V$ in the same unchanged electrostatic field. No dissipative forces occur. Compare ΔU, field work and external work.

For $q=+3\,\mathrm{\mu C}$ , $V_A=5\,\mathrm V$ and $V_B=-15\,\mathrm V$ . A solution adds 100 V to both and claims field work has changed. Check using both calculations.