Electric Potential: Potential Difference, Energy and Volts

By Engr. Henry Akpan

Energy in an Electric Field

Lift a stone and it gains gravitational potential energy, which it gives back when it falls. Electric charges behave the same way in an electric field: push a positive charge towards another positive charge and you store energy, which is released when the charge is let go. Electric potential describes this stored energy per unit charge, and the difference in electric potential between two points is the familiar "voltage" of circuits. This tutorial covers electric potential for WAEC, NECO, JAMB and first-year university physics (PHY 102): the definition, potential due to a point charge, potential of several charges, potential difference and work, electric potential energy, equipotential surfaces, the link between field and potential, and the electronvolt, with worked examples. It follows on from the tutorial on electric fields and Coulomb's law.

What Is Electric Potential?

The electric potential V at a point is the work done per unit positive charge in bringing a small positive test charge from infinity to that point:

V = W ÷ q

Its unit is the volt (V), where 1 V = 1 J/C. Potential is a scalar: it has size and sign but no direction. That makes potentials much easier to add than fields.

By convention, the potential at infinity is zero. In circuits, the earth is also taken as zero potential.

Potential Due to a Point Charge

At a distance r from a point charge Q, the electric potential is:

V = kQ ÷ r = Q ÷ (4πε₀r)

where k = 1/(4πε₀) ≈ 9 × 10⁹ N m² C⁻². Compare this with the field strength E = kQ/r²: the potential falls off as 1/r, while the field falls off as 1/r².

  • Near a positive charge, V is positive, and it increases as you get closer.
  • Near a negative charge, V is negative, and it becomes more negative as you get closer.

Positive charges naturally move from high potential to low potential, just as a ball rolls downhill. Electrons, being negative, move from low potential to high potential.

Worked example 1

Find the electric potential at a point 0.3 m from a charge of +2 μC. How much work is needed to bring a +3 μC charge from infinity to that point?

V = kQ/r = (9 × 10⁹ × 2 × 10⁻⁶) ÷ 0.3 = 6.0 × 10⁴ V.

W = qV = 3 × 10⁻⁶ × 6.0 × 10⁴ = 0.18 J.

Potential Due to Several Charges

Because potential is a scalar, the total potential at a point is simply the algebraic sum of the potentials due to each charge. Include the signs, but there are no angles or components to worry about.

V = kQ1/r1 + kQ2/r2 + kQ3/r3 + …

Worked example 2

Charges of +4 nC and −2 nC are placed 0.4 m apart at points A and B. Find (a) the potential at the midpoint of AB, and (b) the points on line AB where the potential is zero.

(a) Each charge is 0.2 m from the midpoint, so V = 9 × 10⁹ × (4 × 10⁻⁹ − 2 × 10⁻⁹) ÷ 0.2 = 90 V.

(b) Let the point be a distance x from A. Between A and B, the zero-potential condition is 4/x = 2/(0.4 − x), so 1.6 − 4x = 2x, giving x ≈ 0.27 m from A. Beyond B, the condition is 4/x = 2/(x − 0.4), so 4x − 1.6 = 2x, giving x = 0.8 m from A (0.4 m beyond B).

Notice that zero potential does not mean zero field. At these points the field is not zero, but the positive and negative potentials cancel.

Potential Difference and Work

The potential difference (p.d.) between two points A and B is the work done per unit charge in moving a charge from one point to the other:

V_AB = V_A − V_B = W ÷ q, so W = qV

The p.d. is what a voltmeter measures. A 12 V battery does 12 J of work on every coulomb of charge it drives round a circuit.

The work done depends only on the start and end points, not on the path taken. The electric field is said to be conservative, like the gravitational field.

Electric Potential Energy

A charge q at a point of potential V has electric potential energy U = qV. For two point charges Q1 and Q2 a distance r apart:

U = kQ1Q2 ÷ r

  • If the charges have the same sign, U is positive: work had to be done to push them together, and they will fly apart if released.
  • If the charges have opposite signs, U is negative: work must be done to pull them apart.

Worked example 3

Find the potential energy of charges of +4 μC and −2 μC placed 0.3 m apart.

U = 9 × 10⁹ × (4 × 10⁻⁶) × (−2 × 10⁻⁶) ÷ 0.3 = −0.24 J. So 0.24 J of work is needed to separate them completely.

Equipotential Surfaces

An equipotential surface is a surface on which every point has the same potential. Their key properties are:

  • No work is done in moving a charge along an equipotential surface, because the potential difference is zero.
  • Equipotentials are always perpendicular to the field lines.
  • Around a point charge, equipotentials are concentric spheres (circles on a diagram).
  • Between two parallel plates, the equipotentials are flat planes parallel to the plates, equally spaced.
  • Where equipotentials are close together, the field is strong.
  • The surface of a conductor in equilibrium is an equipotential, so the whole conductor is at one potential.

A contour map is a good picture: contour lines join points of equal height, and water flows perpendicular to them, fastest where they are crowded.

The Link Between Field and Potential

The electric field strength equals the potential gradient, with a minus sign:

E = −dV/dr

The minus sign says that the field points in the direction of decreasing potential. In a uniform field, such as between two parallel plates with p.d. V and separation d, this becomes:

E = V ÷ d

This shows why the field can be measured in volts per metre (V/m) as well as newtons per coulomb (N/C). The two units are equivalent.

Worked example 4

Two parallel plates 4 mm apart are connected to a 200 V supply. Find the field strength between them.

E = V/d = 200 ÷ 0.004 = 5.0 × 10⁴ V/m.

Accelerating Charges and the Electronvolt

When a charge q moves freely through a potential difference V, the work done on it becomes kinetic energy:

qV = ½mv²

This is how electron guns in cathode-ray tubes, X-ray tubes and particle accelerators work.

Worked example 5

An electron starts from rest and is accelerated through 2000 V. Find its kinetic energy and speed. (e = 1.6 × 10⁻¹⁹ C, m = 9.1 × 10⁻³¹ kg)

KE = eV = 1.6 × 10⁻¹⁹ × 2000 = 3.2 × 10⁻¹⁶ J.

v = √(2 × 3.2 × 10⁻¹⁶ ÷ 9.1 × 10⁻³¹) ≈ 2.65 × 10⁷ m/s, nearly a tenth of the speed of light.

The electronvolt

Joules are inconveniently large for atomic particles, so physicists use the electronvolt (eV): the energy gained by an electron accelerated through a potential difference of 1 V.

1 eV = 1.6 × 10⁻¹⁹ J

The electron in Worked example 5 has an energy of exactly 2000 eV (2 keV). The electronvolt is used throughout the tutorials on the photoelectric effect and atomic spectra.

Electric Potential: Practice Questions

  1. Find the potential 0.2 m from a charge of −5 nC.
  2. How much work is done in moving a charge of 4 C through a potential difference of 12 V?
  3. Charges of +3 μC are placed at two corners of an equilateral triangle of side 0.3 m. Find the potential at the third corner.
  4. The p.d. between two plates 2 cm apart is 600 V. Find the field strength.
  5. Express 5 MeV in joules.

Answers: 1. V = 9 × 10⁹ × (−5 × 10⁻⁹) ÷ 0.2 = −225 V. 2. W = qV = 48 J. 3. V = 2 × (9 × 10⁹ × 3 × 10⁻⁶ ÷ 0.3) = 1.8 × 10⁵ V (potentials add as scalars). 4. E = 600 ÷ 0.02 = 3.0 × 10⁴ V/m. 5. 5 × 10⁶ × 1.6 × 10⁻¹⁹ = 8.0 × 10⁻¹³ J.

Conclusion

Electric potential is the work done per unit charge in bringing a positive charge from infinity, measured in volts. For a point charge V = kQ/r, and for several charges you add the potentials as signed scalars. The work to move a charge between two points is W = qV, the potential energy of two charges is kQ1Q2/r, and the field strength is the negative potential gradient, which becomes E = V/d between parallel plates. Equipotentials are perpendicular to field lines, and the electronvolt is the natural energy unit for particles accelerated through a potential difference.

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