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Conservation of Electric Energy

Track qΔV, kinetic-energy change, and external work with the correct charge sign.

AP Physics 2 Topic 10.7 · College Board CED printed p. 54 / PDF p. 61 · See this topic in the course outline

Once you know the starting and ending potentials, energy can predict speed without finding the force at every point along the path.

For another explanation, read The Electric Potential and Potential Energy. Use the corrected energy discussion to track qΔV, kinetic energy, and external work.

Before you start

Track the object-field system

A charge q moved from A to B changes the electric potential energy of its object-field system by ΔUE=q(VB−VA)=qΔV\Delta U_E=q(V_B-V_A)=q\Delta V. Electric work is WE=−qΔVW_E=-q\Delta V. A positive charge lowers U by moving to lower V; a negative charge lowers U by moving to higher V. The sign of charge is essential.

If only the conservative electric force does work, ΔK=−ΔUE=−qΔV\Delta K=-\Delta U_E=-q\Delta V. Hence 12mvB2=12mvA2−q(VB−VA)\tfrac12 mv_B^2=\tfrac12 mv_A^2-q(V_B-V_A). If an external agent also does work, ΔK+ΔUE=Wext\Delta K+\Delta U_E=W_{\mathrm{ext}} for the chosen system. A slow controlled move can increase U while keeping K unchanged; a free accelerating move generally exchanges U and K.

On an energy-bar representation, any decrease in U must be matched by an increase in K if no external work or dissipation occurs. An attempted solution that predicts negative final K means the charge cannot reach that endpoint with the given initial energy. Potential difference predicts an energy change independent of path, though travel time and trajectory can depend on the field and initial velocity.

Energy alone gives a bound on accessible potential. A turning point where an uphill velocity component vanishes can still retain transverse kinetic energy. To set the entire K to zero at an actual turn, state one-dimensional uphill motion with no transverse component and a potential that extends to the required energy level.

An electron-volt (eV) is an energy unit: it is the energy corresponding to charge magnitude e crossing a potential difference of 1 V. Thus 1 eV = (1.60 × 10⁻¹⁹ C)(1 V) = 1.60 × 10⁻¹⁹ J. A volt is energy per charge; an electron-volt is energy. Use the charge sign separately to decide whether kinetic or potential energy increases.

An electron gains energy by rising in potential

An electron starts from rest at V = 0 V and reaches V = +150 V. Find its kinetic energy and speed. Use me = 9.11 × 10⁻³¹ kg, ignore other forces, and use the nonrelativistic model.

  1. q = −e and ΔV = +150 V, so ΔU = −2.40 × 10⁻¹⁷ J.

    Why this step? A negative charge loses potential energy when it moves to higher potential.

  2. Kf = −ΔU = 2.40 × 10⁻¹⁷ J = 150 eV.

    Why this step? Electric work converts the decrease in object-field potential energy into kinetic energy.

  3. v=2K/me=7.26×106 m/sv=\sqrt{2K/m_e}=7.26\times10^6\ \mathrm{m/s}.

    Why this step? The starting kinetic energy is zero; the speed is much smaller than the speed of light, consistent with the nonrelativistic assumption.

Answer and check

The electron gains 150 eV and reaches about 7.26 × 10⁶ m/s. The field points toward decreasing V, so its force on the electron points in the opposite direction.

Make the model respond

Focused Physics 2 model

Capacitor: connected or isolated

Predict first. Before adding dielectric, decide whether the battery fixes voltage or isolation fixes charge.

C = κε₀A/d; Q = CV; E = V/d; U = ½CV² = Q²/(2C)

Capacitor: connected or isolated: schematicTwo oppositely charged parallel plates with uniform field arrows from the positive to the negative plate. Plate width changes with the square root of area; the gap changes with separation. The dielectric fills the entire gap. Geometry is schematic and arrows show direction, not field magnitude.Battery connected: V = 12 V stays fixed+−κ = 1; A = 100 cm²; d = 1 mm
Two oppositely charged parallel plates with uniform field arrows from the positive to the negative plate. Plate width changes with the square root of area; the gap changes with separation. The dielectric fills the entire gap. Geometry is schematic and arrows show direction, not field magnitude.

On a narrow screen, scroll within a figure to read its labels. The table and explanation provide the same physics in text.

Capacitance88.54 pF
Voltage12 V
Stored energy6.375 nJ

The battery maintains 12 V and supplies or removes charge as C changes. Q = 1.063 × 10^−9 C and U = 6.375 × 10^−9 J.

Read the model data and comparisons
Selected-state values. SI units are shown explicitly; display controls may use named SI prefixes.
QuantityValueUnitMeaning or limit
Area0.01m²
Separation0.001m
Relative permittivity1
Capacitance8.854 × 10^−11F
Charge magnitude on either plate1.063 × 10^−9C
Voltage across plates12V
Field magnitude between plates12,000V/m
Stored field energy6.375 × 10^−9J
Isolated reference charge Q₀1.063 × 10^−9CPrepared at 12 V, A = 100 cm², d = 1 mm, κ = 1; then disconnected

Increasing area or permittivity raises capacitance; increasing separation lowers it. With a connected ideal battery, fixed voltage means charge and stored energy increase with C. With the isolated reference charge held fixed, voltage and stored energy decrease as C increases. The battery exchanges energy in the connected case, so changes in capacitor energy alone are not the work done by an external agent.

Assumptions and limits. Ideal parallel plates; edge fringing neglected. Linear, uniform dielectric fully fills the gap without changing d. Battery condition means a 12 V ideal source remains attached. Isolated condition means the charge prepared at the explicitly stated reference geometry is retained; no leakage. Energy is the capacitor field energy, not total battery-plus-capacitor energy.

Try it from memory

Write your answer and reason before opening the worked answer. Your self-assessment is a study decision; it is not an automatically graded score.

Question 1

A +2.0 μC charge moves from 12 V to 5 V under only electric forces. Find ΔU and ΔK.

Worked answer and reason

ΔU = −14 μJ and ΔK = +14 μJ.

The final-minus-initial voltage is −7 V, and conservation reverses the energy-change sign.

Question 2

A proton with 20 eV of kinetic energy heads toward a region 30 V higher in potential. Can it arrive there if only electric forces act?

Worked answer and reason

No. Reaching that potential would require a 30 eV increase in U, but only 20 eV is available. A 20 V rise is an energy upper bound for the potential the proton can reach. In one-dimensional uphill motion with no transverse velocity, it turns at that rise if the potential continues increasing far enough. A general trajectory need not attain that bound.

Nonnegative total kinetic energy sets an accessibility bound. Turning of one velocity component need not make total kinetic energy zero when transverse motion remains.

JavaScript enables saved drafts and self-assessment. You can always work on paper.

Catch the tempting mistake

Every charged particle speeds up toward lower potential.
Positive charges do; negative charges can gain kinetic energy toward higher potential.
Potential difference by itself sets speed.
Charge, mass, initial kinetic energy, and any other work also matter.

Put it to work

Give the physical reason as well as the calculation. The worked solutions are on a separate page.

Scientific calculator workspace

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Scientific calculator

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For the teacher
  • Ask students to draw energy bars and state whether external work is included.
  • Use a turning-point question to expose negative-K arithmetic as a physical impossibility.
  • Ask whether all velocity components vanish at the proposed turning point. Without one-dimensional motion, K₀/q gives an energy accessibility bound rather than a guaranteed turning-potential rise.

Unit planning and inquiry guidance →

Source objectives and scope
  • 10.7.A — Describe changes in energy in a system due to a difference in electric potential between two locations.

Suggested skills are teaching guidance. The course applies all science practices across its objectives.

Science references for this teaching
Glossary and deeper reading · 10 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

  • Kinetic energy

    Kinetic energy is energy associated with motion relative to a chosen reference frame. For a nonrelativistic translating object it is one half of its mass times speed squared. For example, doubling a cart’s speed quadruples its translational kinetic energy when its mass stays fixed.

    Read more: Kinetic energy in detail →

  • Mechanical work

    Mechanical work is energy transferred by a force through displacement of its point of application. For a constant force on a particle, it depends on the force component along the particle’s displacement, and it can be positive, negative, or zero. For example, a magnetic force perpendicular to a particle’s displacement does zero work on that particle.

    Read more: Mechanical work in detail →

  • Potential difference

    Potential difference is the change in electric potential between two locations. It represents electric potential-energy change per unit charge and is measured in volts. For example, the potential difference across a resistor describes the energy transferred per coulomb passing through it.

    Read more: Potential difference in detail →

  • Electric charge

    Electric charge is a conserved property that determines electric interactions. It can be positive or negative and is measured in coulombs; ordinary net charges arise from an imbalance of charged particles. For example, transferring electrons between initially neutral objects leaves one negatively charged and the other positively charged.

    Read more: Electric charge in detail →

  • Electric potential energy

    Electric potential energy is energy associated with the configuration of interacting charges. Its value depends on the chosen reference, and a charge pair is counted once when calculating a system’s total pair energy. For example, bringing two positive point charges closer requires positive external work in a slow controlled approach.

    Read more: Electric potential energy in detail →

  • Electric force

    Electric force is the interaction of an electric field with a charged object. For point charges at rest, like signs repel and opposite signs attract, with direction along their separation. For example, an electron and proton exert equal-magnitude opposite forces on one another despite having different masses.

    Read more: Electric force in detail →

  • Velocity

    Velocity describes how position changes with time and includes direction. Instantaneous velocity describes one moment, while average velocity is displacement divided by elapsed time. For example, a car turning a corner can change velocity while maintaining the same speed.

    Read more: Velocity in detail →

  • Energy level

    An energy level is an allowed energy of a quantum system. Bound systems have discrete levels, and a transition between them requires an energy transfer equal to the level difference. For example, an atom emitting a photon drops to a lower energy level and transfers the difference to that photon.

    Read more: Energy level in detail →

  • Electronvolt

    An electronvolt is an energy unit equal to the elementary-charge magnitude multiplied by one volt. It is approximately 1.60 × 10⁻¹⁹ joule and is not a unit of voltage. For example, atomic transition energies are often more conveniently expressed in electronvolts than in very small fractions of a joule.

    Read more: Electronvolt in detail →

  • Potential energy

    Potential energy describes energy associated with the configuration of interacting parts of a system. Its numerical zero is a chosen reference, while differences in potential energy determine physical energy changes. For example, separating Earth and a ball changes their gravitational potential energy even though the energy belongs to their interaction.

    Read more: Potential energy in detail →

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