Unit 21: Potential, Potential Difference and Potential Energy

Example 3.1 – Point Charge

Consider a +2 µC charge at the origin. The equipotential sphere of radius r = 0.10 m corresponds to a potential:

V = kq/r = (8.99 × 10^{9})(2 × 10^{-6})/0.10 ≈ 1.80 × 10^{5} V.

All points on that sphere share this value.

Example 3.2 – Electric Dipole

For a dipole consisting of charges +q and –q separated by distance d, the equipotential lines in the plane containing the dipole are more complex. Near the midpoint, they resemble distorted circles that become elongated along the axis of the dipole. Far away, the dipole’s potential falls off as 1/r², and equipotentials approach those of a point charge of zero net charge (i.e., they become more spherical).

Sketching these lines helps visualize why the electric field lines (which are perpendicular) emerge from the positive charge and terminate on the negative charge.

4. Potential Gradient and Its Relation to Electric Field

Definition of Potential Gradient

The potential gradient describes how rapidly the electric potential changes with position. In one dimension along the x‑axis it is defined as:

\frac{dV}{dx}

Its SI unit is volt per metre (V/m). In three dimensions the gradient is a vector:

\(\nabla V = \left(\frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z}\right)\)

Relation to Electric Field

The electric field is the negative gradient of the potential:

\(\vec{E} = - \nabla V\)

In radial coordinates for a spherically symmetric potential V(r):

E_{r} = -\frac{dV}{dr}

The minus sign indicates that the electric field points in the direction of decreasing potential (from high to low V).

Units Consistency

Since V is measured in volts and distance in metres, V/m is equivalent to newton per coulomb (N/C), confirming that both expressions for E have the same unit.

Example 4.1 – Point Charge

For a point charge q, the potential is V(r) = kq/r. Differentiate:

\frac{dV}{dr} = -\frac{kq}{r^{2}}

Hence the radial electric field magnitude is:

E = -\frac{dV}{dr} = \frac{kq}{r^{2}}

which matches Coulomb’s law for the field of a point charge.

Example 4.2 – Uniform Field

In a uniform electric field E₀ directed along the +x axis, the potential varies linearly: V(x) = –E₀ x + C (where C is a constant). The gradient is:

\frac{dV}{dx} = -E₀

Thus E = -dV/dx = E₀, as expected.

5. Worked Examples and Problem‑Solving Strategies

To solidify understanding, we solve a few representative problems that combine the concepts discussed.

Problem 5.1 – Potential of a Charge Array

Three charges are placed at the corners of an equilateral triangle of side 0.15 m: q₁ = +4 µC at the top, q₂ = –3 µC at the bottom‑left, q₃ = +2 µC at the bottom‑right. Find the electric potential at the centroid of the triangle.

Solution:

1. Determine the distance from each corner to the centroid. For an equilateral triangle, the centroid is at a distance r = \frac{a}{\sqrt{3}} from each vertex, where a = side length.
2. r = 0.15 / √3 ≈ 0.0866 m.
3. Compute the potential due to each charge using V = kq/r and add algebraically:
• V₁ = (8.99 × 10⁹)(4 × 10⁻⁶)/0.0866 ≈ 4.15 × 10⁵ V
• V₂ = (8.99 × 10⁹)(–3 × 10⁻⁶)/0.0866 ≈ –3.11 × 10⁵ V
• V₃ = (8.99 × 10⁹)(2 × 10⁻⁶)/0.0866 ≈ 2.07 × 10⁵ V
4. Sum: V_total ≈ (4.15 – 3.11 + 2.07) × 10⁵ V ≈ 3.11 × 10⁵ V = 311 kV.

The positive net potential indicates that the positive charges dominate at the centroid.

Problem 5.2 – Work to Assemble a Charge Configuration

Calculate the work required to bring three charges, q₁ = +1 µC, q₂ = +2 µC, q₃ = +3 µC, from infinity to the vertices of an equilateral triangle of side 0.10 m.

Solution: The total work equals the sum of the potential energies of each pair (since bringing the first charge costs no work).

Pair distances are all 0.10 m.

• U₁₂ = k q₁ q₂ / r = (8.99 × 10⁹)(1 × 10⁻⁶)(2 × 10⁻⁶)/0.10 ≈ 0.1798 J
• U₁₃ = k q₁ q₃ / r = (8.99 × 10⁹)(1 × 10⁻⁶)(3 × 10⁻⁶)/0.10 ≈ 0.2697 J
• U₂₃ = k q₂ q₃ / r = (8.99 × 10⁹)(2 × 10⁻⁶)(3 × 10⁻⁶)/0.10 ≈ 0.5394 J

Total work W = U₁₂ + U₁₃ + U₂₃ ≈ 0.1798 + 0.2697 + 0.5394 = 0.9889 J ≈ 0.99 J.

Problem 5.3 – Speed of an Accelerated Electron

An electron initially at rest is accelerated through a potential difference of 5 kV. Find its final speed.

Solution: Use energy conservation: loss in electric potential energy equals gain in kinetic energy.

ΔU = qΔV = (–e)(5 000 V) = –(1.602 × 10⁻¹⁹)(5 000) J = –8.01 × 10⁻¹⁶ J.

The magnitude of energy gained is 8.01 × 10⁻¹⁶ J.

Set kinetic energy K = ½ m v² equal to this:

½ (9.11 × 10⁻³¹ kg) v² = 8.01 × 10⁻¹⁶ J → v² = (2 × 8.01 × 10⁻¹⁶) / (9.11 × 10⁻³¹) ≈ 1.76 × 10¹⁵ → v ≈ 1.33 × 10⁸ m/s.

This is about 44 % of the speed of light; relativistic corrections would be needed for higher accuracies, but the non‑relativistic result illustrates the principle.

Problem‑Solving Tips

• Always identify whether you need potential (scalar) or field (vector). Use superposition for potentials directly; for fields, add vectorially.
• When calculating work or energy, remember that moving a charge against the field increases its potential energy, while moving with the field decreases it.
• Equipotential surfaces are invaluable for visualizing fields: draw them first, then sketch field lines perpendicular to them.
• Check units: potential in volts, energy in joules (or eV), electric field in V/m or N/C.
• For problems involving multiple steps (e.g., assemble a configuration), break the process into sequential steps and sum the contributions.

Summary

In this chapter we have defined electric potential difference as work per unit charge and derived the potential of a point charge. We introduced electric potential energy for charge pairs and for a charge in a potential, highlighted the electron‑volt as a convenient energy unit, and explored equipotential surfaces and their geometric and physical properties. The potential gradient was shown to be the negative of the electric field, linking the scalar potential to the vector field. Numerous examples and problem‑solving strategies illustrated how these concepts are applied in typical electrostatic scenarios. Mastery of these topics provides a strong foundation for further study of capacitors, dielectrics, and electric circuits.