AP Physics C: Electricity & Magnetism • Electric Potential

Conservation of Electric Energy

How the interplay of kinetic and electric potential energy governs charged-particle motion in electrostatic fields.

A magnet with electric energy and an atomic field.
Inside this lesson 9 sections 5 questions 8 figures

SECTION 1

Historical Context & Motivation

The idea that energy is neither created nor destroyed—merely transformed—stands as one of the most powerful unifying principles in all of physics. In mechanics, students learn to trade kinetic energy and gravitational potential energy using conservation laws. The extension of this framework to electrostatics—where charges accelerate, decelerate, and exchange energy through electric potential energy—was a conceptual breakthrough that unified electricity with the broader energy paradigm of classical physics.

1785 • Coulomb's Torsion Balance

Charles-Augustin de Coulomb quantified the inverse-square force law between point charges, providing the foundation from which electric potential energy could later be derived.

1800 • Volta's Pile

Alessandro Volta constructed the first chemical battery, demonstrating that a sustained potential difference could drive charge through a circuit and convert chemical energy into electrical energy.

1847 • Helmholtz & Energy Conservation

Hermann von Helmholtz published a rigorous formulation of the conservation of energy encompassing mechanical, thermal, and electrical phenomena, cementing the principle as universal.

1929 • Van de Graaff Generator

Robert Van de Graaff built high-voltage electrostatic accelerators that converted electric potential energy into kinetic energy of charged particles, directly applying energy conservation in experimental nuclear physics.

The central question this lesson addresses is deceptively simple: when a charged particle moves through an electric field, how do we track the energy transformations quantitatively? Since the electrostatic force is conservative—meaning the work it does depends only on initial and final positions, not on the path taken—we can define a scalar potential energy function and apply conservation of energy with the same rigor used in gravitational problems. This insight dramatically simplifies the analysis of charged-particle dynamics and underpins everything from capacitor design to particle accelerator physics.

SECTION 2

Core Principles & Definitions

Conservation of electric energy rests on several interlocking ideas. At its core, the electrostatic force is conservative, which guarantees the existence of a well-defined electric potential energy function U. The electric potential V is the potential energy per unit charge, linking the scalar field description to the energy of any specific charge placed in it. When only conservative forces act, the total mechanical energy—kinetic plus electric potential—remains constant throughout the motion.

1

Conservative Force

The electrostatic (Coulomb) force does path-independent work. The work around any closed loop is zero: ∮ E · dl = 0. This is the prerequisite for defining a potential energy.
2

Electric Potential Energy U

For a system of point charges, U equals the work required by an external agent to assemble the configuration from infinity. For two charges: U = kq₁q₂/r. The sign reflects attraction (U < 0) or repulsion (U > 0).
3

Electric Potential V

Defined as V = U/q, the potential is a property of the field itself, independent of the test charge. It is measured in volts (1 V = 1 J/C). The potential difference ΔV between two points determines the work done per unit charge.
4

Energy Conservation Statement

When no non-conservative forces (friction, applied pushes) act on a charge, K₁ + U₁ = K₂ + U₂. Equivalently, ΔK + ΔU = 0. Kinetic energy gained equals potential energy lost.
5

Work-Energy Theorem Link

The work done by the electric field on charge q moving through potential difference ΔV is W = −qΔV = qV₁ − qV₂. This work equals the change in kinetic energy, connecting field quantities directly to particle dynamics.

SECTION 3

Visual Explanation — Energy Bar Charts

A powerful way to visualize conservation of electric energy is through energy bar charts (sometimes called LOL diagrams). These charts display the kinetic energy K and electric potential energy U of a charged particle at two positions, showing how the total remains constant. The diagram below illustrates a positive charge released from rest near a positive source charge: as U decreases (the charge moves to larger separation or lower potential), K increases by exactly the same amount.

Energy Conservation: Positive Charge Repelled by Positive SourcePosition A (near source)Position B (far from source)K≈0U = largeK = gainedU = smallE_totalE_totalΔU → ΔKKinetic Energy KElectric Potential Energy U
Energy bar chart showing a positive charge released from rest near a positive source. At position A the energy is almost entirely potential energy U (tall violet bar). At position B the charge has gained kinetic energy K (tall cyan bar) while U has decreased. The total height (E_total) is the same at both positions.

Notice how the total bar height is identical on both sides—this is the graphical statement of energy conservation. For a negative charge in the same scenario, the signs of U would differ, but the conservation principle remains the same. The bar chart representation is particularly useful on AP exams for qualitative/quantitative translation problems, where you must reason about energy without fully solving equations.

SECTION 4

Mathematical Framework

The mathematical formulation of energy conservation in electrostatics connects several quantities: kinetic energy, electric potential energy, electric potential, and the work-energy theorem. We develop the key equations below, starting from the definition of work done by the electric field and arriving at the conservation statement.

WORK BY ELECTRIC FIELD

WE=∫ABqE⋅dl=−ΔU=−(UB−UA)W_E = ∫ᴬᴮ q E · dl = −ΔU = −(U_B − U_A)

The work done by the electrostatic force on charge q moving from A to B equals the negative change in potential energy. The integral is path-independent because E is conservative.

POTENTIAL ENERGY & POTENTIAL

U=qV⟹ΔU=qΔV=q(VB−VA)U = qV ⟹ ΔU = qΔV = q(V_B − V_A)

The electric potential energy of charge q at a point equals q times the electric potential V at that point. For two point charges: U = kq₁q₂/r.

CONSERVATION OF ENERGY (NO NON-CONSERVATIVE WORK)

KA+UA=KB+UB⟹½mvA2+qVA=½mvB2+qVBK_A + U_A = K_B + U_B ⟹ ½mv²_A + qV_A = ½mv²_B + qV_B

When the electric force is the only force doing work, the total mechanical energy is conserved. This is the master equation for charged-particle kinematics problems in electrostatics.

GENERAL FORM (WITH NON-CONSERVATIVE WORK)

Wnc=ΔK+ΔU=(KB−KA)+q(VB−VA)W_nc = ΔK + ΔU = (K_B − K_A) + q(V_B − V_A)

If non-conservative forces (e.g., friction, applied force) also act, their net work Wnc equals the change in total mechanical energy. When Wnc = 0, the conservation form is recovered.

ℹ
Sign Convention Reminder

SECTION 5

Energy Diagrams for Point-Charge Systems

For a system of two point charges, the electric potential energy varies as U(r) = kq₁q₂/r. Plotting this function alongside a constant total energy line yields an energy diagram that reveals turning points, allowed regions of motion, and equilibrium behavior—analogous to the U(x) diagrams used in mechanics. The kinetic energy at any separation r is the vertical gap between the total energy line and the U(r) curve.

Energy Diagram: Two Like Charges (q₁, q₂ > 0)0Separation r →EnergyU(r) = kq₁q₂/rE_totalForbidden(K would be < 0)Turning point r₀r₀K = E − U
Energy diagram for two like positive charges. The violet curve shows U(r) = kq₁q₂/r. The dashed amber line is the constant total energy E. At the turning point r₀, K = 0 and the charge momentarily stops. The cyan arrow shows the kinetic energy at any separation as the gap between E and U(r). The red-shaded region is forbidden because K cannot be negative.

Several features of this diagram merit attention. First, U → 0 as r → ∞, so at large separations K approaches E — the charge reaches its maximum speed far from the source. Second, the turning point r₀ is the distance of closest approach when a charge is launched inward with known energy. Setting E = kq₁q₂/r₀ yields r₀ = kq₁q₂/E. Third, for opposite charges (q₁q₂ < 0), U is negative and the curve dips below the axis; bound states and escape conditions can then be analyzed in direct analogy with gravitational orbits.

SECTION 6

Worked Example — Proton Accelerated Through a Potential Difference

A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.60 × 10⁻¹⁹ C) is released from rest and accelerated through a potential difference of ΔV = −500 V (i.e., it moves from a region at 500 V to a region at 0 V). Find the proton's final speed.

Worked example

  1. 1

    Step 1 — Identify Given Values

    m = 1.67 × 10⁻²⁷ kg, q = +1.60 × 10⁻¹⁹ C, vAv_A = 0 (released from rest), VAV_A = 500 V, VBV_B = 0 V, so ΔV = VBV_B − VAV_A = −500 V.
  2. 2

    Step 2 — Write Conservation of Energy

    With no non-conservative forces: KAK_A + UAU_A = KBK_B + UBU_B. Substituting U = qV and KAK_A = 0:
    0 + qV_A = ½mv²_B + qV_B
  3. 3

    Step 3 — Solve for vBv_B

    Rearranging: ½mv²_B = q(VAV_A − VBV_B) = qΔV_drop, where ΔV_drop = VAV_A − VBV_B = 500 V. Then vBv_B = √(2qΔV_drop / m).
  4. 4

    Step 4 — Substitute Numbers

    vBv_B = √(2 × 1.60 × 10⁻¹⁹ × 500 / 1.67 × 10⁻²⁷) = √(1.60 × 10⁻¹⁶ / 1.67 × 10⁻²⁷) = √(9.58 × 10¹⁰)
    vBv_B ≈ 3.10 × 10⁵ m/s
  5. 5

    Step 5 — Sanity Check

    This is about 0.1% of the speed of light, so the non-relativistic treatment is valid. The kinetic energy gained is qΔV_drop = 500 eV, consistent with the definition of the electron-volt.

SECTION 7

Gravitational vs. Electric Energy Conservation

Students often find it helpful to compare electric energy conservation with its gravitational analog—the structure is nearly identical, and the parallels reinforce the universality of energy methods. Below is a systematic comparison highlighting both the shared framework and the key differences.

FeatureGravitationalElectrostatic
Source of fieldMass MCharge Q
Force lawF = GMm/r²F = kQq/r²
Potential energyU = −GMm/r (always attractive)U = kQq/r (sign depends on charges)
Potential (energy per unit)Φ = −GM/r (J/kg)V = kQ/r (J/C = V)
Sign of UAlways negative (attraction)Positive (like charges) or negative (unlike charges)
Conservation equation½mv² − GMm/r = const½mv² + kQq/r = const

Side-by-side comparison of gravitational and electrostatic energy conservation

SECTION 8

Connection to Circuits, Capacitors & Beyond

The conservation of electric energy does not stop at point charges in vacuum. In circuit theory, a battery maintains a constant potential difference (EMF), and energy conservation for a charge traversing the loop becomes Kirchhoff's loop rule: the sum of all potential differences around a closed circuit is zero. In capacitor problems, the energy stored (U = ½CV² = Q²/2C) represents the work done to separate charge against the electric field inside the device. Mastering energy conservation for point charges thus provides the conceptual substrate for the entire second half of the AP Physics C: E&M course.

ConceptThis LessonAdvanced Extension
Energy sourceStatic Coulomb fieldEMF (batteries, generators with non-conservative forces)
StorageU = kq₁q₂/r (point charges)U = ½CV² (capacitors), u = ½ε₀E² (field energy density)
DissipationNone (purely conservative)P = I²R (resistive heating in circuits)
Conservation lawK + U = constKirchhoff's loop rule: Σ ΔV = 0 around closed loop

From point-charge energy conservation to circuit energy analysis

Looking further ahead, the energy stored in the electric field itself—expressed as an energy density u = ½ε₀E²—generalizes conservation of energy to distributed electromagnetic systems. This perspective becomes essential when studying electromagnetic waves, where energy is carried through space by oscillating E and B fields without any charges present. The point-charge energy conservation you master here is the conceptual stepping stone to Poynting's theorem and the full energy budget of Maxwell's equations.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A positive charge is released from rest in a uniform electric field. As it accelerates in the direction of the field, which statement is true about its electric potential energy U and the electric potential V at its location?
  • A. Both U and V increase.
  • B. U increases and V decreases.
  • C. Both U and V decrease.
  • D. U decreases and V increases.
Show answer (C)
A positive charge accelerates from high V to low V (the direction of E), so V decreases. Since U = qV and q > 0, U decreases as well. The lost potential energy becomes kinetic energy. Choice A is wrong because a freely accelerating positive charge moves toward lower potential. Choice B is wrong because U = qV with q > 0 means U and V change in the same direction. Choice D is wrong because V cannot increase along the direction of E; V decreases in the direction of the electric field.

Question 2: BASIC CALCULATION

An electron (m = 9.11 × 10⁻³¹ kg, q = −1.60 × 10⁻¹⁹ C) is accelerated from rest through a potential difference ΔV = VBV_B − VAV_A = +120 V. What is the electron's final speed?
  • A. 6.49 × 10⁶ m/s
  • B. 3.25 × 10⁶ m/s
  • C. 6.49 × 10⁵ m/s
  • D. 4.21 × 10¹³ m/s
Show answer (A)
Using ½mv² = |q|×|ΔV| = (1.60 × 10⁻¹⁹)(120) = 1.92 × 10⁻¹⁷ J. Then v = √(2 × 1.92 × 10⁻¹⁷ / 9.11 × 10⁻³¹) = √(4.22 × 10¹³) ≈ 6.49 × 10⁶ m/s. Note: the electron accelerates toward higher V because q < 0, so ΔU = qΔV = (−)(+) < 0, meaning U decreases and K increases. Choice B omits the factor of 2 under the square root. Choice C is off by a factor of 10 due to a power-of-ten arithmetic error. Choice D exceeds the speed of light and results from forgetting the square root.

Question 3: INTERMEDIATE

Two protons are initially held at rest separated by 2.0 × 10⁻¹⁰ m and then released simultaneously. What is the speed of each proton when they are very far apart? (mpm_p = 1.67 × 10⁻²⁷ kg, k = 8.99 × 10⁹ N·m²/C²)
  • A. 1.87 × 10⁴ m/s
  • B. 2.63 × 10⁴ m/s
  • C. 3.72 × 10⁴ m/s
  • D. 5.26 × 10⁴ m/s
Show answer (B)
Initially: K = 0, UiU_i = ke²/r = (8.99 × 10⁹)(1.60 × 10⁻¹⁹)²/(2.0 × 10⁻¹⁰) = 1.15 × 10⁻¹⁸ J. At r → ∞: UfU_f = 0, and by symmetry each proton has the same speed v, so KfK_f = 2 × ½m_pv² = m_pv². Setting UiU_i = m_pv²: v = √(1.15 × 10⁻¹⁸ / 1.67 × 10⁻²⁷) = √(6.89 × 10⁸) ≈ 2.63 × 10⁴ m/s. Choice A incorrectly assigns all kinetic energy to one proton (divides by 2m instead of m). Choice C uses ½mv² = UiU_i for each proton separately (double-counting energy). Choice D is twice the correct answer, likely from omitting the factor of ½ in kinetic energy.

Question 4: APPLIED

A small sphere of mass m = 5.0 × 10⁻³ kg and charge q = +3.0 × 10⁻⁶ C is launched vertically upward with initial speed v₀ = 8.0 m/s from ground level in a region where a uniform downward electric field E = 2000 N/C exists in addition to gravity (g = 9.8 m/s²). (a) Derive an expression for the maximum height h in terms of m, q, v₀, E, and g. (b) Calculate the numerical value of h. (c) Determine the speed of the sphere when it returns to its launch height. (d) If the electric field were reversed (pointing upward), explain qualitatively whether the sphere would reach a greater or lesser maximum height, and identify the condition on E for which the sphere would never return to the ground.
Show answer
(a) Both gravity (mg, downward) and the electric force (qE, downward on +q in a downward field) act on the sphere. Using energy conservation from launch (height 0) to maximum height h (v = 0): ½mv₀² = mgh + qEh. Solving: h = v₀² / [2(g + qE/m)]. (b) qE/m = (3.0 × 10⁻⁶)(2000)/(5.0 × 10⁻³) = 1.2 m/s². So g + qE/m = 9.8 + 1.2 = 11.0 m/s². h = (8.0)² / [2(11.0)] = 64 / 22.0 ≈ 2.91 m. (c) Since both gravity and the electric force are conservative, the total mechanical energy is the same at launch height on the way back down. The sphere returns at v = v₀ = 8.0 m/s. (d) If E points upward, the electric force on +q is upward, opposing gravity. The effective downward acceleration is g − qE/m, which is smaller, so h increases. If qE ≥ mg (i.e., E ≥ mg/q = (5.0 × 10⁻³)(9.8)/(3.0 × 10⁻⁶) ≈ 16,333 N/C), the net force is upward and the sphere accelerates away from the ground indefinitely, never returning.

Question 5: CRITICAL THINKING

An alpha particle (q = +2e, m = 6.64 × 10⁻²⁷ kg) is fired directly toward a gold nucleus (Q = +79e) with initial kinetic energy K₀ = 5.0 MeV from very far away. (a) Derive an expression for the distance of closest approach d. (b) Calculate d numerically. (c) At the distance of closest approach, is the system's total energy kinetic, potential, or a mix? Justify your answer. (d) Explain why the conservation-of-energy approach is valid even though the gold nucleus is not fixed but can recoil.
Show answer
(a) At large r, U ≈ 0 and K = K₀. At closest approach, v = 0 (momentarily at rest), so K = 0 and U = kQq/d. By conservation: K₀ = kQq/d, giving d = kQq/K₀. (b) Qq = (79)(2)(1.60 × 10⁻¹⁹)² = 4.045 × 10⁻³⁶ C². K₀ = 5.0 MeV = 8.0 × 10⁻¹³ J. d = (8.99 × 10⁹)(4.045 × 10⁻³⁶)/(8.0 × 10⁻¹³) = 3.637 × 10⁻²⁶ / 8.0 × 10⁻¹³ ≈ 4.55 × 10⁻¹⁴ m. (c) At closest approach the energy is entirely electric potential energy. The alpha particle is momentarily at rest (K = 0), so U = E_total. (d) The gold nucleus (A = 197) is about 49 times more massive than the alpha particle (A = 4). By momentum conservation, the recoil speed of the gold nucleus is ~m_α/m_Au times the alpha's initial speed, so its kinetic energy is ~m_α/m_Au ≈ 2% of K₀. This small correction is often neglected. For an exact treatment, one replaces m with the reduced mass μ = m_α m_Au/(m_α + m_Au) and works in the center-of-mass frame—the conservation-of-energy framework remains fully valid.

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