Difficulty: Intermediate-Advanced | Prerequisites: Electric potential energy, Coulomb's law, basic integral calculus
Electric potential (voltage) is the bridge between the force-based view of electrostatics (fields, Coulomb's law) and the energy-based view (PE, work). It tells you the potential energy per unit charge at a point in space. Calculating V from a known electric field requires integrating E along a path, which is the central skill tested in this section. If you are comfortable with PE between point charges and have working knowledge of definite integrals, you are ready.
Electric potential at a point equals the negative line integral of the electric field from a reference point to that location. For non-uniform fields (where E changes with position), you must integrate rather than simply multiply E by distance. The result is path-independent because the electric force is conservative.
Electric potential (V)
The electric potential energy per unit positive test charge at a point in space. Measured in volts (V), where 1 V = 1 J/C.
Think of it as the "electrical height" at a point. Just as water flows downhill, positive charges move from high V to low V.
Potential difference (ΔV or voltage)
The difference in electric potential between two points: ΔV = V_B - V_A. This is what voltmeters measure.
In simple terms, it tells you how much energy per coulomb a charge gains or loses moving between two points.
Line integral of E
The integral ∫E·dl along a path from A to B. It sums up the component of the electric field along the direction of travel, weighted by the infinitesimal displacement dl.
Think of it as adding up tiny "pushes" from the field along your path, step by step.
Non-uniform electric field
An electric field whose magnitude or direction (or both) changes from point to point. E(x) = 2000x + 6000x² from the homework is a classic example.
In simple terms, the field is not the same everywhere, so you cannot just do E times d. You need calculus.
Conservative field
A field where the work done moving a charge from A to B depends only on the endpoints, not on the path taken. All electrostatic fields are conservative.
In simple terms, it does not matter whether you take the scenic route or the direct one. The voltage difference is the same.
The fundamental relationship is:
V_B - V_A = -∫(from A to B) E·dl
For a field that varies only along one axis (say x), this simplifies to:
ΔV = -∫(from x_A to x_B) E_x dx
The negative sign means that moving in the direction of the electric field decreases the potential. Moving against the field increases it.
From the homework, E(x) = 2000x + 6000x². To find the potential difference between two points:
ΔV = -∫ (2000x + 6000x²) dx
ΔV = -[1000x² + 2000x³] evaluated between the limits
This is straightforward power-rule integration. The challenge is keeping track of signs and limits.
Another homework problem integrates E = -2kλ/x (the field from a line charge) from x = 4.3 to x = 7:
ΔV = -∫ (-2kλ/x) dx = 2kλ ∫ (1/x) dx = 2kλ [ln(x)] evaluated from 4.3 to 7
This produces a logarithmic potential, characteristic of line charges.
For a point charge: V = kQ/r (no integration needed, this is the closed-form result).
For continuous distributions, you either integrate contributions from each dq, or integrate E along a path.
The electric field points from high potential to low potential. The magnitude of E is largest where V changes most rapidly (steep "slope" in the potential landscape).
Because the electrostatic field is conservative, the integral ∫E·dl from A to B gives the same answer regardless of the path.
This means ΔV between two points is uniquely defined. You can pick whichever path makes the integral easiest (usually along a coordinate axis or along a field line).
Potential difference (general):
V_B - V_A = -∫(A to B) E·dl
For E along the x-axis only:
ΔV = -∫(x_A to x_B) E_x(x) dx
Point charge potential:
V = kQ / r (referenced to V = 0 at infinity)
Line charge potential (from integration):
ΔV = (λ / 2πε₀) ln(r_A / r_B)
where λ is the linear charge density. Note the logarithmic dependence on distance, characteristic of cylindrical symmetry.
Relationship between E and V:
E_x = -dV/dx
The electric field is the negative gradient of the potential. In one dimension, this is just the negative derivative.
This integration technique is how engineers calculate voltage drops across non-uniform materials and how physicists model the potential inside particle accelerators where fields vary with position. Every time you measure a voltage with a multimeter, you are measuring the result of exactly this integral.
Students often forget the negative sign in ΔV = -∫E·dl. This is not optional decoration. Omitting it flips the sign of every answer.
Treating a non-uniform field as uniform and using ΔV = -Ed. This shortcut only works when E is constant. If E depends on position, you must integrate.
Confusing the direction of integration with the direction of the field. The limits of integration define your path (A to B). The field has its own direction. The dot product E·dl sorts out the relationship.
Mixing up V (potential, a scalar) with E (field, a vector). You cannot add potentials like vectors, and you cannot treat fields like scalars.
⚠️ Integration of E to find V is a core exam skill. Expect at least one problem requiring you to evaluate a definite integral of a given E(x).
⚠️ The negative sign in ΔV = -∫E·dl is tested both computationally and conceptually. "Does potential increase or decrease in the direction of E?" (It decreases.)
⚠️ Line charge problems yield logarithmic potentials. If you see ln in an answer, that is expected, not a sign you have gone wrong.
True or False: Moving in the direction of the electric field increases the electric potential.
False. Potential decreases in the direction of E.
Fill in the blank: For a uniform electric field, ΔV = ______.
-Ed (where d is the displacement in the direction of E).
True or False: The potential difference between two points depends on the path you take.
False. Electrostatic fields are conservative, so ΔV depends only on the endpoints.
If E(x) = 3000x², what is ∫(from 0 to 2) E(x) dx?
3000 x [x³/3] from 0 to 2 = 1000(8) = 8000. So ΔV = -8000 V.
Q: The electric field in a region is given by E(x) = 2000x + 6000x² N/C. Find the potential difference V(0) - V(0.1) (i.e., from x = 0.1 m back to x = 0).
A: V(0) - V(0.1) = -∫(from 0.1 to 0) E dx = +∫(from 0 to 0.1) E dx = ∫(0 to 0.1)(2000x + 6000x²) dx = [1000x² + 2000x³] from 0 to 0.1 = 1000(0.01) + 2000(0.001) = 10 + 2 = 12 V.
Q: A line charge produces a field E = -6.48 x 10⁻¹⁰ / x along the x-axis. Find the potential difference between x = 4.3 m and x = 7 m.
A: ΔV = -∫(4.3 to 7)(-6.48x10⁻¹⁰ / x) dx = 6.48x10⁻¹⁰ [ln x] from 4.3 to 7 = 6.48x10⁻¹⁰ (ln 7 - ln 4.3) = 6.48x10⁻¹⁰ x 0.488 ≈ 3.16 x 10⁻¹⁰ V.
Q: If V = 100 V at point A and V = 60 V at point B, how much work does the electric field do on a +2 μC charge moving from A to B?
A: W = qΔV = q(V_A - V_B) = (2x10⁻⁶)(100 - 60) = 8x10⁻⁵ J. The field does positive work because the charge moves from higher to lower potential.
This material connects directly to Gauss's law: once you find E using a Gaussian surface, you integrate it to find V. It also leads into equipotential surfaces and the gradient relationship E = -∇V, which you will use extensively in later chapters.
The integration skills here carry over to calculating capacitance, where you integrate E between the plates to find the voltage, then use C = Q/V.
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