Grade by Grade
D. Gravitational, Electric, and Magnetic Fields · Gravitational, Electric and Magnetic Fields

Electric Potential Energy and Electric Potential

D2.1 D2.3 D2.5About 75 minutes
Learning goals
  • I can calculate the electric potential energy of two or more point charges, Ee = kq₁q₂/r, and interpret its sign.
  • I can define electric potential (V = Ee/q) and potential difference, and calculate V = kQ/r for point charges.
  • I can use conservation of energy with W = qΔV, including the electron volt.
  • I can describe equipotential surfaces and explain Rutherford's scattering experiment using energy.

Words to know

electric potential energy (Ee)
The energy stored in a system of charges because of their positions: Ee = kq₁q₂/r for two point charges, with Ee = 0 at infinite separation.
electric potential (V)
Electric potential energy per unit positive charge at a point: V = Ee/q. Unit: volt (V) = joule per coulomb (J/C). For a point charge, V = kQ/r.
potential difference (ΔV)
The change in electric potential between two points; the work done per unit charge moving a charge between them: W = qΔV. Also called voltage.
electron volt (eV)
The energy gained by one elementary charge moving through a potential difference of 1 V: 1 eV = 1.60 × 10⁻¹⁹ J.
equipotential surface
A surface on which every point has the same electric potential. It is always perpendicular to the field lines.
scalar
A quantity with size but no direction. Electric potential and potential energy are scalars; field and force are vectors.

Energy without arrows

Forces and fields are vectors, and adding vectors takes work. Energy is a scalar: you just add numbers (with signs). In this lesson you will meet electric potential energy and electric potential, the quantity behind every "volt" printed on a battery. Using energy often turns a hard force problem into a few lines of algebra.

Remember from Grade 11 (Electricity and Magnetism): potential difference (voltage) is energy per unit charge, ΔV = E/Q, measured in volts. Now you will see where that idea comes from.

Electric potential energy of two charges

By the same reasoning that gave Eg = −GMm/r, with the zero at infinite separation:

Ee = k q₁ q₂ / r

Here you do keep the signs of the charges, because the sign of Ee tells you something:

  • Like charges: Ee > 0. They repel, so you had to push them together. Release them and they fly apart, turning Ee into kinetic energy.
  • Unlike charges: Ee < 0. They attract, so they are bound, like a planet and the Sun. You must add energy to pull them apart.

Compare: Eg = −GMm/r is always negative because gravity always attracts.

For three or more charges, add Ee for every pair. Three charges have three pairs; four charges have six.

Worked example

Example 1: Charges flying apart

Two small spheres, each of mass 1.0 g and charge +2.0 μC, are held 0.10 m apart and released. How fast is each moving when they are far apart? (Ignore gravity and friction.)

  1. Initial Ee = kq₁q₂/r = (8.99 × 10⁹)(2.0 × 10⁻⁶)²/0.10 = 0.360 J. Final Ee ≈ 0 (far apart).
  2. Conservation of energy: 0.360 J = ½mv² + ½mv² = mv² (by symmetry they have equal speeds).
  3. v = √(0.360/1.0 × 10⁻³) = √360 = 19 m/s.

Each sphere ends up moving at about 19 m/s, in opposite directions (momentum is conserved too).

Electric potential

Just as the field is force per unit charge, electric potential is potential energy per unit charge:

V = Ee/q (unit: the volt, 1 V = 1 J/C)

For a point charge Q, dividing Ee = kQq/r by q:

V = kQ/r

  • V is a scalar. Keep the sign of Q: positive charges make positive potential around them, negative charges make negative potential.
  • For several charges, the potential at a point is the algebraic sum V = kQ₁/r₁ + kQ₂/r₂ + ... No vectors!
  • Like g and E, V describes the place, not the charge you put there.

Only differences in potential matter physically. When a charge q moves through a potential difference ΔV, the change in its potential energy is

ΔEe = qΔV

and the work needed (with no change in kinetic energy) is W = qΔV. If only electric forces act, ΔEk = −qΔV: positive charges speed up moving toward lower potential, and negative charges speed up moving toward higher potential.

rV0positive Q: V = +kQ/rnegative Q: V = −k|Q|/r(a + charge here has positive Ee: pushed away)(a + charge here has negative Ee: held in)
Electric potential around a positive charge (top curve) and a negative charge (bottom curve). Both approach zero far away and follow V = kQ/r.
Worked example

Example 2: Potential and work

(a) Find the potential at 0.30 m and at 0.60 m from a +5.0 μC charge. (b) How much work does it take to move a +2.0 μC charge from 0.60 m in to 0.30 m?

  1. (a) V₁ = kQ/r = (8.99 × 10⁹)(5.0 × 10⁻⁶)/0.30 = 1.50 × 10⁵ V. V₂ = (8.99 × 10⁹)(5.0 × 10⁻⁶)/0.60 = 7.49 × 10⁴ V.
  2. (b) ΔV = V_final − V_initial = 1.50 × 10⁵ − 7.49 × 10⁴ = 7.49 × 10⁴ V.
  3. W = qΔV = (2.0 × 10⁻⁶)(7.49 × 10⁴) = 0.15 J.

(a) 1.5 × 10⁵ V and 7.5 × 10⁴ V. (b) 0.15 J of work must be done to push the like charge closer.

Worked example

Example 3: Zero potential, but not zero field

A +4.0 μC charge is at x = 0 and a −2.0 μC charge is at x = 0.30 m. Find the potential at x = 0.60 m.

  1. V = kQ₁/r₁ + kQ₂/r₂ = (8.99 × 10⁹)(4.0 × 10⁻⁶)/0.60 + (8.99 × 10⁹)(−2.0 × 10⁻⁶)/0.30.
  2. V = 5.99 × 10⁴ V − 5.99 × 10⁴ V = 0.

V = 0. But the field there is not zero: 1.0 × 10⁵ N/C from the +4.0 μC charge pointing +x, and 2.0 × 10⁵ N/C toward the −2.0 μC charge pointing −x, a net 1.0 × 10⁵ N/C in the −x direction. Potential and field are different quantities; one can be zero while the other is not.

The electron volt

In atomic and particle physics, joules are awkwardly large. The electron volt is the energy one elementary charge gains moving through 1 V:

1 eV = (1.60 × 10⁻¹⁹ C)(1 V) = 1.60 × 10⁻¹⁹ J

An electron accelerated through 100 V gains 100 eV. Chemical bonds involve a few eV; alpha particles from radioactive decay have a few MeV (10⁶ eV); the Large Hadron Collider gives protons several TeV (10¹² eV).

Equipotential surfaces

An equipotential is a surface where V is the same everywhere. Around a point charge, V = kQ/r depends only on r, so the equipotentials are spheres centred on the charge (circles in a 2D drawing).

  • Equipotentials are always perpendicular to field lines.
  • Moving a charge along an equipotential takes no work (ΔV = 0).
  • The field points from high to low potential.
  • The surface of a conductor in equilibrium is an equipotential (otherwise charges would move).

They are like contour lines on a topographic map: the field is the "downhill" direction, and the field is strongest where equipotentials are closest together.

300 V200 V100 VDashed circles: equipotentialsSolid arrows: field linesField lines cross equipotentials at 90°.The field points toward lower potential.No work is done moving a charge alongan equipotential.Spacing grows: V ∝ 1/r, so equal stepsof V are farther apart far away.
Equipotentials (dashed) and field lines (arrows) for a positive point charge. The equipotentials are circles, crossed at right angles by the field lines.

Rutherford's experiment: energy at the atomic scale

In 1909, Hans Geiger and Ernest Marsden, working with Ernest Rutherford, fired alpha particles (charge +2e) at a very thin gold foil. Most went straight through, but about 1 in 8000 bounced back at large angles. Rutherford, who had done earlier Nobel-winning work on radioactivity at McGill University in Montréal, said it was as if a shell fired at tissue paper had come back and hit you. In 1911 he explained the result: the atom's positive charge and almost all of its mass are concentrated in a tiny nucleus.

Energy tells us how close an alpha gets. In a head-on approach, the alpha slows until all its kinetic energy has become electric potential energy: Ek = k(2e)(Ze)/r_min. This distance is an upper limit on the size of the nucleus.

Worked example

Example 4: Closest approach

A 5.0 MeV alpha particle heads straight for a gold nucleus (Z = 79). Find its distance of closest approach.

  1. Ek = 5.0 × 10⁶ eV × 1.60 × 10⁻¹⁹ J/eV = 8.0 × 10⁻¹³ J.
  2. At closest approach, Ek = k(2e)(79e)/r: r = k(2)(79)e²/Ek.
  3. r = (8.99 × 10⁹)(158)(1.60 × 10⁻¹⁹)²/(8.0 × 10⁻¹³) = 4.5 × 10⁻¹⁴ m.

About 4.5 × 10⁻¹⁴ m. The atom is about 10⁻¹⁰ m across, so the nucleus is at least thousands of times smaller than the atom.

gold nucleus (+79e)alpha particles (+2e) from the leftCloser aim → bigger deflection. A head-on alpha stops, then bounces straight back.
Rutherford scattering (not to scale). Alpha particles aimed closer to the nucleus are repelled more strongly and deflect through larger angles.
Just for fun

CERN likes to point out that 1 TeV (10¹² eV) is about the kinetic energy of a flying mosquito. The Large Hadron Collider's trick is not the amount of energy but squeezing several mosquitoes' worth into a single proton.

Hands-on activity

Potential maps with a simulation (or a pencil)

Simulation: open PhET "Charges and Fields" (free, phet.colorado.edu). Turn on "Values" and "Grid".

  1. Place one +1 nC charge. Use the voltage tool to find points where V = 18 V, 9 V and 4.5 V. Measure their distances from the charge. (Expect 0.5 m, 1.0 m and 2.0 m, since V = kQ/r.) Use the tool's "plot" button to draw each equipotential.
  2. Make a dipole (+1 nC and −1 nC 1 m apart). Plot the V = 0 equipotential. What shape is it?
  3. Drag the E-field sensor along an equipotential. Check that the field arrow always crosses the equipotential at a right angle.

No computer? Draw a + charge and a − charge 10 cm apart on paper. Use a ruler to find points where V = k(1/r₁ − 1/r₂) has the same value (for example, points where r₁ = r₂, then points where r₂ = 2r₁) and sketch the equipotentials.

What to look for: around one charge, equipotentials are circles with the predicted radii. For the dipole, the V = 0 line is the perpendicular bisector: every point on it is equally far from +1 nC and −1 nC, so the potentials cancel. Points where r₂ = 2r₁ form a circle around the positive charge (on the paper version, check a few).

Your turn

1

Find the electric potential energy of a +3.0 μC charge and a −5.0 μC charge 0.20 m apart. What does the sign mean?

2

Find the electric potential 0.0180 m from a +2.0 nC charge.

3

Convert 2.0 × 10⁻¹⁵ J to electron volts, and 1.0 keV to joules.

4

An electron starts at rest and moves through a potential difference of 100 V. Find its kinetic energy (eV and J) and speed.

5

What is the electric potential at the electron's distance from the proton in a hydrogen atom (5.29 × 10⁻¹¹ m)? What is the electron's potential energy in eV?

6

Why can V be zero at a point where E is not zero? Give an example.

More practice: warm-up

1

What is the unit of electric potential, in base form?

2

Is electric potential a vector or a scalar?

3

Two like charges are pushed closer together. Does Ee increase or decrease?

4

How many joules is 1 eV?

5

What angle do field lines make with equipotential surfaces?

More practice: on your own

1

A proton is released from rest at a point where V = 500 V and moves to a point where V = 0. Find its final speed.

2

Charges of +1.0 μC sit at two corners of a square of side 0.20 m and −1.0 μC at a third corner, opposite the empty corner. Find V at the empty corner.

3

How much work is needed to bring a +1.0 μC charge from far away to that empty corner?

4

Find the total electric potential energy of three charges, +2.0 μC, +2.0 μC and −2.0 μC, at the corners of an equilateral triangle with sides 0.10 m.

5

A charge moves along an equipotential surface. How much work does the electric field do? Explain.

Challenge

1

In Rutherford's experiment, why did most alpha particles pass straight through the foil, and what does the closest-approach distance tell us?

2

A 5.0 MeV alpha particle (mass 6.64 × 10⁻²⁷ kg) is fired head-on at a gold nucleus. What is its initial speed? If its energy were doubled, how would the closest-approach distance change?

3

Explain why the electron in a hydrogen atom has negative potential energy, and what it means that 13.6 eV are needed to ionize it, even though Ee = −27.2 eV.

Check your understanding

Quiz

1.

The electric potential energy of a positive and a negative charge is

2.

Electric potential is measured in volts, where 1 V equals

3.

The potential 2.0 m from a point charge is 600 V. At 6.0 m it is

4.

An electron is accelerated from rest through a potential difference of 5000 V. It gains

5.

Which statement about equipotential surfaces is true?

6.

Rutherford's gold-foil results showed that

For parents and teachers

Big idea (D2.3): energy gives a second, simpler way to handle electric interactions. Ee = kq₁q₂/r (signs kept), V = kQ/r, ΔEe = qΔV. Potentials add as plain numbers, not vectors.

How to help: Ask your teen to explain the difference between field (N/C, a vector) and potential (V, a scalar), and between potential and potential energy. A good check question: "Can the potential be zero where the field is not?" (Yes, at the midpoint of a dipole.)

Common difficulties:

  • Using 1/r² in potential (V = kQ/r has r, not r²).
  • Dropping the signs of charges in Ee and V. Unlike for forces, signs are kept here.
  • Mixing up eV (an energy) with V (a potential).
  • Forgetting to count every pair when finding the energy of three or more charges.