Electric potential and potential gradient
Potential difference as a line integral of E, absolute potential, E = −∇V, equipotentials, the dipole and the work done moving charges.
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Why it matters
Potential is a scalar, so it is far easier to add up than the vector field E; engineers almost always find V first and then get E by differentiation. Voltage ratings, insulation stress (the potential gradient in kV/mm), electrode design in sensors and electrostatic deflection in CROs are all questions about potential and its gradient.
Key ideas
Work and potential difference. Moving a charge Q against a field E needs work W = −Q ∫ E·dl from the start A to the end B. The potential difference is the work per unit charge: V_BA = V_B − V_A = −∫_A^B E·dl (volts = J/C). The minus sign matters: moving along E lowers the potential.
Absolute potential. Choosing the reference V = 0 at infinity (for finite charge distributions), the potential at a point is the work per unit charge to bring a positive test charge from infinity to that point. For an infinite line or sheet the field does not fall off fast enough, so the reference must be a finite point instead.
Conservative field. The electrostatic field is conservative: ∮ E·dl = 0 around any closed path, equivalently ∇ × E = 0. So the work between two points does not depend on the path, and potential is single-valued. (This fails for time-varying magnetic fields — see Faraday's law.)
Potential gradient. The gradient ∇V is a vector whose direction is that of the fastest increase of V and whose magnitude is the rate of that increase (V/m). The field is E = −∇V, so E points from high to low potential, along the steepest descent. Breakdown strength of insulation is quoted as a maximum potential gradient (for example, about 3 kV/mm for dry air at normal conditions).
Equipotential surfaces. Surfaces on which V is constant. No work is done moving along them, so E is everywhere perpendicular to them. A conductor in electrostatic equilibrium is an equipotential volume.
Superposition. Potentials of several charges add as scalars (with signs), which is why finding V and then E = −∇V is usually quickest.
Electric dipole. Two equal and opposite charges ±Q separated by d. Far away (r ≫ d) the potential falls as 1/r² and the field as 1/r³, with dipole moment p = Qd pointing from −Q to +Q.
Potential energy. A charge q at potential V has potential energy U = qV. A positive charge moving along E loses potential energy; a negative charge moving along E gains it.
Formulas
V_AB = V_A − V_B = −∫_B^A E·dl— potential difference (V); the integral runs from B to A.V = Q / (4πεr)— absolute potential of a point charge, V = 0 at infinity; r in m.V = Σ Qk / (4πε|r − rk|)— several point charges (scalar sum with signs).V = ∫ ρL dl / (4πεR)— line charge (similarly with ρS dS or ρv dv).E = −∇V— field from potential (V/m); in Cartesian E = −(∂V/∂x a_x + ∂V/∂y a_y + ∂V/∂z a_z).V_AB = (ρL / 2πε) ln(ρB / ρA)— between radii ρA and ρB of an infinite line charge.V = Qd cos θ / (4πεr²)andE = Qd (2cos θ a_r + sin θ a_θ) / (4πεr³)— dipole far field, r ≫ d.W = Q·V_BA— work done by an external agent to move Q from A to B (J).∮ E·dl = 0,∇ × E = 0— conservative (static) field.
Worked examples
Example 1 (standard). A 5 nC point charge is at the origin in free space. Find the potential at A (r = 1 m) and B (r = 3 m), V_AB, and the work needed to move a 2 μC charge from B to A.
V = kQ/rwith k = 8.988 × 10⁹: V_A = 8.988 × 10⁹ × 5 × 10⁻⁹ / 1 = 44.94 V; V_B = 44.94/3 = 14.98 V.- V_AB = V_A − V_B = 29.96 V (A is at the higher potential).
- W = q·V_AB = 2 × 10⁻⁶ × 29.96 = 59.9 μJ, done by the external agent (the positive charge is pushed towards the positive source).
Example 2 (GATE level). In free space V = 2x²y − 5z volts. Find V, E and |E| at P(1, 2, −1) m.
- V(P) = 2(1)²(2) − 5(−1) = 4 + 5 = 9 V.
∇V = 4xy a_x + 2x² a_y − 5 a_z. At P: ∇V = 8 a_x + 2 a_y − 5 a_z V/m.E = −∇V= −8 a_x − 2 a_y + 5 a_z V/m.- |E| = √(64 + 4 + 25) = √93 = 9.64 V/m.
- Extra: ∇²V = 4y ≠ 0, so (by Poisson's equation) this potential needs a volume charge ρv = −ε₀·4y; at P, ρv = −8ε₀ = −70.8 pC/m³.
Example 3 (path independence). In a uniform field E = 10 a_x V/m, find V_B − V_A for A(0, 0, 0) and B(2, 3, 0) m.
- V_B − V_A = −∫ E·dl from A to B = −∫ 10 dx from 0 to 2 = −20 V. The y-displacement contributes nothing because it is perpendicular to E. B is 20 V lower than A, whatever path is taken.
Common mistakes
- Dropping the minus sign in V = −∫E·dl or E = −∇V — the most common sign error in the subject.
- Saying the potential gradient points the way potential decreases. ∇V points uphill; E points downhill.
- Adding potentials as vectors, or adding fields as scalars. V adds with signs; E adds as vectors.
- Taking V = 0 at infinity for an infinite line or sheet charge (gives infinite potential); use a finite reference.
- Confusing V_AB (= V_A − V_B) with V_BA, and the work done by the field (−qΔV) with the work done by an external agent (+qΔV).
- Assuming E = 0 wherever V = 0 (between +Q and −Q the midpoint has V = 0 but E ≠ 0), or V = 0 wherever E = 0.
For GATE IN
Expect E from a given V (gradient at a point), potential difference between two points for a given E (path integrals that test path independence), potential of point-charge arrangements (square, triangle) and dipoles, work done moving a charge, and identifying equipotentials. NAT items often need the gradient in cylindrical or spherical form, so practise all three.
Quick check
- If V = 50x volts, what is E?
- Along an equipotential surface, how much work is done moving a charge?
- What is V at the midpoint of +Q and −Q separated by d?
- What is the work done moving a 2 C charge through a potential difference of 12 V?
- In which direction does ∇V point?
Answers: 1. −50 a_x V/m. 2. Zero. 3. Zero (though E is not). 4. 24 J. 5. Towards increasing V (opposite to E).
Interview questions
All Electricity and Magnetism interview questionsTry answering each one aloud before you open it.
1.What is electric potential, and how is it different from electric potential energy?Concept
Electric potential is the amount of electric potential energy per unit charge at a point in an electric field. It is a scalar quantity measured in volts (V). Electric potential energy, on the other hand, is the energy that a charge possesses due to its position in an electric field. It is measured in joules (J). While electric potential is a property of the electric field itself, electric potential energy depends on the charge placed in the field.
2.Explain the concept of potential gradient and its significance.Concept
The potential gradient ∇V is a vector whose direction is that in which the potential increases fastest and whose magnitude is that rate of increase, in V/m. The electric field is its negative, E = −∇V, so E points the way the potential falls fastest. Engineers use it as the electric stress on insulation: dielectric strength is a maximum allowable potential gradient (for example about 3 kV/mm for dry air).
3.How is the electric field related to the electric potential gradient?Concept
The electric field is the negative gradient of the electric potential. Mathematically, this is expressed as E = -dV/dx, where E is the electric field, and dV/dx is the potential gradient. This relationship shows that the electric field points in the direction of the steepest decrease in potential and its magnitude is proportional to the rate of change of potential with distance.
4.Why is the concept of electric potential important in electrical engineering?Application
Electric potential is important in electrical engineering because it helps in understanding and analyzing electric circuits and fields. It allows engineers to calculate the work done by or against electric forces, design circuits with specific voltage requirements, and ensure safety by managing potential differences. It also aids in the analysis of energy conversion and storage in devices like capacitors and batteries.
5.What happens to the electric potential energy of a charge when it moves in the direction of the electric field?Application
Moving along E means moving to lower potential, and the potential energy is U = qV. For a positive charge U therefore decreases: the field does positive work and, if nothing else acts, that energy appears as kinetic energy. For a negative charge U increases, so an external agent must do work to move it along the field.
6.Why are equipotential surfaces perpendicular to electric field lines?Application
Equipotential surfaces are perpendicular to electric field lines because there is no change in potential along an equipotential surface. If they were not perpendicular, there would be a component of the electric field along the surface, causing a potential difference, which contradicts the definition of an equipotential surface. Thus, the electric field lines, which indicate the direction of the greatest potential decrease, must be perpendicular to these surfaces.
7.What is the potential difference between two points if the electric field is uniform and has a magnitude of 5 V/m over a distance of 10 meters?Numerical
The potential difference (ΔV) between two points in a uniform electric field can be calculated using the formula ΔV = E × d, where E is the electric field strength and d is the distance. Here, E = 5 V/m and d = 10 m. Therefore, ΔV = 5 V/m × 10 m = 50 V.
8.Calculate the work done in moving a 2 C charge between two points with a potential difference of 12 V.Numerical
The work done (W) in moving a charge (q) through a potential difference (ΔV) is given by W = q × ΔV. Here, q = 2 C and ΔV = 12 V. Therefore, W = 2 C × 12 V = 24 J. The work done is 24 joules.
9.How does the potential gradient affect the motion of a charged particle in an electric field?Application
The force on the particle is F = qE = −q∇V, with magnitude proportional to the potential gradient. A positive charge is pushed down the potential slope (along E) and a negative charge up it (against E). The kinetic energy gained between two points depends only on the potential difference: ½mv² = |q|·|ΔV| for a particle starting from rest, as in the electron gun of a CRO.
10.Explain why a conductor in electrostatic equilibrium has a constant electric potential throughout its volume.Concept
In electrostatic equilibrium, the charges within a conductor have redistributed themselves such that there is no net movement of charge. This means that the electric field inside the conductor is zero. Since the electric field is the negative gradient of the electric potential, a zero electric field implies that there is no change in potential within the conductor. Therefore, the electric potential is constant throughout the volume of the conductor.
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