Electric Potential: Study Guide (Physics 2 10.5)

Energy per coulomb

In Topic 10.4 you found the electric potential energy U_E of a system of charges. That number depends on every charge in the system, including the one you just placed. It is useful to separate the two parts: what the fixed charges do to a point in space, and how much charge you put there.

The electric potential V at a point is the electric potential energy per unit charge that a small test charge q would have at that point:

V = U_E / q

The unit is the volt: 1 V = 1 J/C. V is a property of the point, set by the charges around it. It does not depend on the test charge. Put a charge q at a point of potential V and the system’s potential energy is U_E = qV.

Two things to keep straight:

• V is a scalar. It has a size and a sign, but no direction.
• In this course V is zero infinitely far from all charges. So V tells you the work per coulomb needed to bring a positive test charge slowly in from very far away.

Potential of point charges

For a single point charge q, the potential at distance r is

V = kq / r, with k = 9.0 × 10⁹ N·m²/C²

The sign of q goes into the formula. Near a positive charge V is positive; near a negative charge V is negative. For example, 0.30 m from a +2.0 nC charge, V = (9.0 × 10⁹)(2.0 × 10⁻⁹) ÷ 0.30 = 60 V.

V falls off as 1/r, more slowly than the field, which falls off as 1/r². If you double the distance, V halves but E drops to a quarter.

For several charges, use scalar superposition: find the potential due to each charge on its own, then add the signed numbers.

V = Σ kqᵢ / rᵢ

This is much easier than adding fields. There are no components and no angles: only distances and signs. The course asks you to do this for four or fewer point charges, or for more charges when the arrangement is highly symmetric.

Potential difference

What matters in most problems is the change in potential between two points. The potential difference from point A to point B is

ΔV = V_B − V_A = ΔU_E / q

So when a charge q moves from A to B, the electric potential energy of the system changes by

ΔU_E = q ΔV

Watch the signs. A positive charge that moves to a lower potential (ΔV < 0) loses potential energy, because qΔV < 0. A negative charge that moves to a higher potential also loses potential energy: now q < 0 and ΔV > 0, so again qΔV < 0. Put the signs of both q and ΔV into the product every time, rather than relying on a rule you half remember.

When ΔU_E < 0, the field does positive work on the charge, and the charge speeds up if nothing else acts on it. Positive charges, left alone, move toward lower potential. Negative charges move toward higher potential. (Topic 10.7 turns this into an energy calculation.)

Batteries

Potential differences do not only come from fixed charges. Inside a battery, chemical reactions push positive and negative charges apart and keep them apart. That separation holds one terminal at a higher potential than the other. A 1.5 V cell gives each coulomb that passes through it from the negative to the positive terminal 1.5 J of electric potential energy, so 2.0 C gains 3.0 J.

Conductors in contact

Electrons move freely inside a conductor. If two parts of a conductor were at different potentials, electrons would drift toward the higher potential until the difference vanished. So in electrostatic equilibrium the whole conductor, including its surface, is at one potential.

The same applies when you join two conductors, for example with a wire. Electrons flow from one to the other until both surfaces are at the same potential. Charge is conserved during this, but the charge is usually not shared equally. For two well-separated conducting spheres joined by a thin wire, each surface has V = kQ/R, so the sphere with the larger radius ends up with the larger charge.

Potential and field

Imagine moving a positive test charge q a short distance Δr straight along the field. The field does work qEΔr on it, so the potential energy falls by that amount, and the potential falls by EΔr. Turned around, the average field between two points is

|E| = |ΔV / Δr|

where Δr is the distance between the points measured along the field. Two consequences:

• The unit V/m is the same as N/C. (Check: 1 V/m = 1 J/(C·m) = 1 N·m/(C·m) = 1 N/C.)
• The field points in the direction of decreasing potential. If you know which way V falls fastest, you know which way E points.

Equipotential lines

An equipotential line (an isoline of potential) joins points that have the same V. A map of these lines is like a contour map of hills: closely spaced lines mean a steep slope, which here means a strong field.

• Equipotential lines are always perpendicular to the field vectors. If the field had a component along an isoline, moving along it would change V, so it would not be an isoline.
• So there is no field component along an equipotential, and the field does no work on a charge that moves along one.
• You can build either map from the other. From a field map, draw lines that cross every field vector at a right angle. From an isoline map, draw field vectors at right angles to the lines, pointing toward lower V, longer where the lines are closer together.

Predicting motion from a map

A map of isolines tells you how a charged particle released at rest will start to move. A positive charge accelerates along the field, toward lower V and at right angles to the isolines. A negative charge accelerates the opposite way, toward higher V. Where the lines are crowded, the field is strong and the acceleration is large.

Worked example 1: potential from three point charges

Question. Three point charges sit at three corners of a rectangle 4.0 cm wide and 3.0 cm tall (Figure 2): q₁ = +4.0 nC at the bottom left, q₂ = −2.0 nC at the bottom right and q₃ = +3.0 nC at the top left. Find the electric potential at P, the empty top-right corner. Then find the work an external force must do to bring a +1.5 nC charge slowly from very far away to P.

1. Distances to P: from q₂, r₂ = 3.0 cm; from q₃, r₃ = 4.0 cm; from q₁, along the diagonal, r₁ = √(4.0² + 3.0²) = 5.0 cm.
2. Potential from each charge, with its sign, in metres:
   • V₁ = (9.0 × 10⁹)(+4.0 × 10⁻⁹) ÷ 0.050 = +720 V
   • V₂ = (9.0 × 10⁹)(−2.0 × 10⁻⁹) ÷ 0.030 = −600 V
   • V₃ = (9.0 × 10⁹)(+3.0 × 10⁻⁹) ÷ 0.040 = +675 V
3. Add the scalars: V_P = 720 − 600 + 675 = +795 V (8.0 × 10² V to 2 significant figures).
4. Work to bring the test charge in from infinity, where V = 0: W = qΔV = (1.5 × 10⁻⁹ C)(795 V − 0) = 1.19 × 10⁻⁶ J.

Interpretation and check. No angles were needed, even though the charges lie in two directions from P. That is the advantage of a scalar. The negative charge is closest to P, but the two positive charges together win, so V_P is positive. The work is positive: you must push a positive charge “uphill” to a point of positive potential. If you brought an electron to P instead, the potential energy would change by (−1.60 × 10⁻¹⁹ C)(795 V) = −1.27 × 10⁻¹⁶ J; the field would pull it in.

Worked example 2: reading an equipotential map

Question. Use Figure 1 (charge +0.50 nC). (a) Find the average field between the 90 V and 75 V isolines, and between the 45 V and 30 V isolines. State the direction. (b) A proton moves from A (on the 90 V line) to B (on the 30 V line). Find the change in electric potential energy and the work done on the proton by the field. (c) The proton then moves from B to C along the 30 V line. How much work does the field do? (d) An electron is released at rest near A. Which way does it start to move?

1. (a) The 90 V and 75 V lines are 6.0 − 5.0 = 1.0 cm apart, so |E| = 15 V ÷ 0.010 m = 1.5 × 10³ V/m. The 45 V and 30 V lines are 15 − 10 = 5.0 cm apart, so |E| = 15 V ÷ 0.050 m = 3.0 × 10² V/m. In both places E points radially outward, from higher V to lower V.
2. (b) ΔV = V_B − V_A = 30 V − 90 V = −60 V. For the proton, ΔU_E = qΔV = (1.60 × 10⁻¹⁹ C)(−60 V) = −9.6 × 10⁻¹⁸ J. The work done by the field is the negative of the change in potential energy: +9.6 × 10⁻¹⁸ J. The result does not depend on the path from A to B.
3. (c) B and C are on the same isoline, so ΔV = 0 and the field does zero work. The field is radial and the path along the circle is everywhere perpendicular to it.
4. (d) An electron has negative charge, so it moves toward higher potential: inward, toward the positive charge.

Check. The exact field of this charge is kq/r² = 1800 V/m at 5.0 cm and 1250 V/m at 6.0 cm. The average, 1500 V/m, lies between them, as it should. Further out, crowded lines have become sparse: the field is five times weaker between 10 cm and 15 cm.

Common misconceptions

• Adding potentials as vectors. V is a scalar. Add the signed values directly. Do not split them into x and y parts.
• Dropping the sign of the charge. A negative charge makes a negative contribution to V. Leaving out the sign is the most common error in superposition problems.
• “V = 0 means E = 0” (or the reverse). Halfway between +q and −q, V = 0 but E is not. Inside a charged conducting shell, E = 0 but V is not zero; it equals the surface potential.
• Confusing V and U_E. V is per coulomb and belongs to a point. U_E = qV belongs to the system and depends on the charge you place there.
• “Conductors in contact share charge equally.” They reach equal potential. A sphere of radius 0.10 m at 900 V holds kQ/R → Q = 1.0 × 10⁻⁸ C; a sphere twice as wide at the same potential holds twice the charge.
• Field pointing toward higher potential. E always points toward lower V. A positive charge goes “downhill” in V; a negative charge goes “uphill”.
• Using the straight-line distance when it is not along the field. In |E| = |ΔV/Δr|, Δr is measured along the field (across the isolines), and the result is an average.

Where this leads

Topic 10.6 applies these ideas to two parallel plates with equal and opposite charges, where the field is uniform and ΔV = Ed exactly: the capacitors study guide. Topic 10.7 then uses ΔU_E = qΔV with conservation of energy to find speeds. Test yourself with the practice questions, then use the revision notes and the topic checklist to consolidate. For the energy of a system of charges, look back at the electric potential energy study guide.