Unit 9 · Topic 9.2

9.2 Electric Potential

Electric potential is electric potential energy per unit charge, measured in volts. Because it's a scalar, potentials from several charges simply add, and the field can be found from how fast the potential changes with position.

Key terms

electric potential
potential difference (voltage)
volt
scalar superposition
equipotential lines
field–potential relationship

Potential is energy per charge

The electric potential at a point is $V = \dfrac{U_E}{q}$ : the potential energy a charge q would have there, divided by q. Its unit is the volt, 1 V = 1 J/C. Potential is set by the source charges, so it exists at a point whether or not a test charge is there.

Only differences in potential are physically meaningful. The potential difference between two points, $\Delta V = V_B - V_A$ , is often called the voltage. Moving charge q across it changes the system's potential energy by $\Delta U_E = q\Delta V$ .

Charges don't have to be spread around by hand to make a potential difference. A battery uses chemical reactions to push positive and negative charge to opposite terminals, which keeps a steady potential difference between them, such as 1.5 V for an AA cell.

Point charges and superposition

For a point charge, taking V = 0 at infinity: $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ . Keep the sign of q: potential is positive near a positive charge and negative near a negative one.

With several charges, add the potentials as plain numbers: $V = \displaystyle\sum\frac{kq_i}{r_i}$ . There are no components, which makes potential much easier than field. For continuous charge, integrate: $V = \displaystyle\int\frac{k\,dq}{r}$ .

Distribution	Potential
Ring, on its axis at distance x	$V = \dfrac{kQ}{\sqrt{x^2 + R^2}}$
Semicircular arc, at its center	$V = \dfrac{kQ}{R}$
Rod of length L, on its line, distance d from the near end	$V = k\lambda\ln\dfrac{d + L}{d}$

From field to potential

$\Delta V = -\int_A^B\vec{E}\cdot d\vec{r}$

Moving along the field, potential drops. In a uniform field, moving a distance d along the field gives $\Delta V = -Ed$ , so field strength can be written in V/m, which is the same as N/C. Outside a charged sphere, integrating $kQ/r^2$ inward from infinity gives $V = kQ/r$ . For an infinite line, V can't be zero at infinity, so you only work with differences: $V_b - V_a = -2k\lambda\ln(r_b/r_a)$ .

From potential to field

Going the other way, the field component in any direction is the negative rate of change of potential: $E_x = -\dfrac{dV}{dx}$ . On a graph of V against x, the field is the negative of the slope. Where V is flat, the field is zero; where V drops steeply, the field is strong and points toward lower V.

V = 0 at a point doesn't mean E = 0 there, and E = 0 doesn't mean V = 0. Field depends on how V changes, not on its value.

Equipotentials

An equipotential line (or surface) connects points at the same potential. Field lines always cross equipotentials at right angles, and point from higher to lower potential. Moving a charge along an equipotential takes no work. Around a point charge the equipotentials are concentric circles that get farther apart as you move out; in a uniform field they are evenly spaced parallel lines. Where equipotentials (drawn at equal steps of V) are close together, the field is strong.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Zero potential but not zero field (classic trap)
A square has sides of 0.20 m. Charges of +3.0 nC sit at the two top corners and −3.0 nC at the two bottom corners. Find the potential and the field at the center.
Show the solutionHide the solution
1. Step 1: Each corner is $r = \sqrt{0.10^2 + 0.10^2} \approx 0.141$ m from the center.
2. Step 2: Potential: $V = \dfrac{k}{r}(3.0 + 3.0 - 3.0 - 3.0)\times10^{-9} = 0$ .
3. Step 3: Field: each charge gives $\dfrac{kq}{r^2} = \dfrac{(9.0\times10^{9})(3.0\times10^{-9})}{0.020} = 1350$ N/C. Fields from the top (+) charges point away from them, down and to the side; fields from the bottom (−) charges point toward them, also down and to the side.
4. Step 4: The sideways parts cancel in pairs; the downward parts add: $E = 4(1350)\cos 45^\circ \approx 3.8\times10^{3}$ N/C, pointing down, toward the negative charges.
5. Step 5: The trap is assuming V = 0 means E = 0. They measure different things.
Answer: V = 0; E ≈ 3.8 × 10³ N/C toward the side with the negative charges
Example 2Calculator allowed
Field from a potential function
Along the x-axis, the potential is $V(x) = 200 - 40x^2$ , with V in volts and x in meters. Find the field at x = 0.50 m.
Show the solutionHide the solution
1. Step 1: $E_x = -\dfrac{dV}{dx} = -(-80x) = 80x$ .
2. Step 2: At x = 0.50 m, E = 40 V/m, in the +x direction.
3. Step 3: Check: V decreases as x increases past 0, and the field points toward lower potential, in +x. It agrees.
Answer: 40 V/m (40 N/C) in the +x direction
Example 3Calculator allowed
Ring: potential first, then field
A ring of radius R carries total charge Q spread uniformly. Find the potential on its axis a distance x from the center, then use it to find the field there.
Show the solutionHide the solution
1. Step 1: Every piece dq is the same distance $\sqrt{x^2 + R^2}$ from the point, and potential has no direction, so nothing cancels: $V = \displaystyle\int\frac{k\,dq}{\sqrt{x^2+R^2}} = \frac{kQ}{\sqrt{x^2+R^2}}$ .
2. Step 2: Differentiate: $E_x = -\dfrac{dV}{dx} = -kQ\cdot\left(-\dfrac{1}{2}\right)(x^2+R^2)^{-3/2}(2x) = \dfrac{kQx}{(x^2+R^2)^{3/2}}$ .
3. Step 3: This matches the field found by direct integration in 8.4, with much less work, because the scalar integral needed no components.
Answer: $V = \dfrac{kQ}{\sqrt{x^2+R^2}}$ ; $E_x = \dfrac{kQx}{(x^2+R^2)^{3/2}}$

Common mistakes

Adding potentials as vectors, or using the magnitude of each charge. Potential is a scalar: keep the signs and add the numbers.
Mixing up the formulas. Field goes as kq/r² and has a direction; potential goes as kq/r and doesn't.
Forgetting the minus sign in E = −dV/dx, and so pointing the field toward higher potential.
Assuming the field is zero where the potential is zero (or the reverse).

On the exam

Expect graph questions: given V against x, sketch E against x (the negative slope), or find where the field is zero (where the V graph is flat).
Equipotential maps are common in multiple choice: find the field direction, rank the field strength by line spacing, or find the work to move a charge between two lines.
For continuous charges, set up $\int k\,dq/r$ with limits. It's usually simpler than integrating the field.

Connected topics

Videos

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Check yourself

4 questions on 9.2 Electric Potential. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

What is the electric potential 0.50 m from a −4.0 nC point charge, taking V = 0 infinitely far away?

Question 2 of 4Calculator allowed

Charges +q, +q, −q and −q sit at the corners of a square, with the two positive charges on the top side and the two negative charges on the bottom side. At the center of the square, which is true?

Question 3 of 4Calculator allowed

A ring of radius 0.30 m carries 4.0 nC spread evenly around it. What is the potential on the ring's axis 0.40 m from its center?

Question 4 of 4Calculator allowed

Along the x-axis the electric potential is $V(x) = 4x^2 - 3x$ , with V in volts and x in meters. What is the electric field at x = 2.0 m?

0 of 4 answered