Electrostatic Potential and Capacitance

The electrostatic potential V at a point in an electric field is defined as the work done in bringing a unit positive test charge from infinity to that point, without any change in its kinetic energy. Since the electrostatic force is conservative, this work is independent of the path taken, so potential is a well-defined scalar function of position: V = W_∞→P/q₀, measured in volts (1 V = 1 J/C).

For an isolated point charge Q, the potential at a distance r is obtained by integrating the field along a radial path from infinity: V(r) = kQ/r, where k = 1/(4πε₀) = 9×10⁹ N·m²/C². Unlike the electric field, potential has no direction — it is simply a number (positive, negative, or zero) at each point, and it falls off as 1/r rather than 1/r².

When several point charges q₁, q₂, q₃, … are present, the superposition principle applies to potential just as it does to field, but the sum is a simple algebraic (scalar) sum rather than a vector sum: V = k(q₁/r₁ + q₂/r₂ + q₃/r₃ + …), where each rᵢ is the distance from charge qᵢ to the point where V is being calculated. This scalar addition makes potential calculations for multi-charge systems considerably simpler than field calculations.

An electric dipole — two equal and opposite charges +q and −q separated by a small distance 2a — has a dipole moment p = q×2a directed from −q to +q. The potential at a general point P at distance r from the centre of the dipole, making angle θ with the dipole axis, is V = kp cosθ/r² (valid for r ≫ a). Two special cases follow directly: on the axial line (θ = 0°), V = kp/r²; on the equatorial line (θ = 90°), cosθ = 0 so V = 0 everywhere, even though the electric field there is not zero. Dipole potential falls off as 1/r², faster than the 1/r of a single point charge, because the potentials of the two opposite charges partially cancel at large distances.

Worked example: A point charge Q = 1×10⁻⁷ C is placed at the origin. Find the potential at r = 9 cm.
V = kQ/r = (9×10⁹ × 1×10⁻⁷) / 0.09 = 900/0.09 = 1×10⁴ V = 10 kV.

Worked example (dipole): A dipole has moment p = 4×10⁻⁹ C·m. Find the potential at a point 10 cm from its centre, (a) on the axial line, (b) on the equatorial line.
(a) V_axial = kp/r² = (9×10⁹ × 4×10⁻⁹)/(0.1)² = 36/0.01 = 3600 V.
(b) V_equatorial = 0 V, since θ = 90° and cos 90° = 0.

An equipotential surface is a surface on which the electric potential has the same value at every point. Since no work is done moving a charge between two points at the same potential, moving a test charge anywhere along an equipotential surface requires zero work — this is the defining practical property of such surfaces.

Equipotential surfaces have several general properties that follow directly from the physics: (i) the electric field is always perpendicular to an equipotential surface at every point — if it had a component along the surface, that component would do work moving a charge along the (constant-potential) surface, which is impossible; (ii) two equipotential surfaces can never intersect, since a point cannot have two different potentials at once; (iii) equipotential surfaces are spaced closer together where the field is strong and farther apart where the field is weak, because a large potential change over a short distance implies a large field.

For a single point charge, the equipotential surfaces are concentric spheres centred on the charge, since V = kQ/r depends only on r. For a uniform electric field (such as between the plates of a parallel-plate capacitor), the equipotential surfaces are flat planes perpendicular to the field lines.

The field and potential are linked quantitatively through E = −dV/dr: the component of electric field in any direction equals the negative rate of change of potential in that direction. This is why field lines always point from higher to lower potential, in the direction of the steepest potential decrease.

Worked example: Two equipotential planes in a uniform field are separated by 4 mm along the field direction, with a potential difference of 40 V between them. Find the electric field.
E = ΔV/Δr = 40 V / (4×10⁻³ m) = 1×10⁴ V/m = 10 kV/m, directed from the higher-potential plane to the lower-potential plane.

The electrostatic potential energy of a system of charges is the total work an external agent must do to assemble the charges, bringing each one in from infinity, against the electrostatic forces of the charges already in place. For two point charges q₁ and q₂ separated by r₁₂, this work is U = kq₁q₂/r₁₂. The energy is positive for like charges (work must be done against repulsion) and negative for unlike charges (the field does the work, releasing energy as they come together).

For three or more charges, the total potential energy is the sum of the potential energies of every distinct pair, since each pair interacts independently: for three charges q₁, q₂, q₃, U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₁q₃/r₁₃).

When a charge q is placed at a point where the potential (due to some external sources) is V, its potential energy is simply U = qV. For an electric dipole of moment p placed in a uniform external field E, making angle θ with the field, the potential energy is U = −pE cosθ = −p·E. This is minimum (most stable, U = −pE) when the dipole is aligned with the field (θ = 0°) and maximum (least stable, U = +pE) when anti-aligned (θ = 180°). The field also exerts a torque τ = pE sinθ on the dipole, which tries to rotate it into alignment with E.

Worked example (system of charges): Two point charges q₁ = 2 μC and q₂ = 3 μC are 0.6 m apart in vacuum. Find their electrostatic potential energy.
U = kq₁q₂/r = (9×10⁹ × 2×10⁻⁶ × 3×10⁻⁶)/0.6 = (9×10⁹ × 6×10⁻¹²)/0.6 = 0.054/0.6 = 0.09 J.

Worked example (dipole in a field): A dipole of moment p = 2×10⁻⁸ C·m sits in a uniform field E = 5×10⁴ N/C at θ = 30° to the field. Find the torque, and the work needed to rotate it to θ = 90°.
Torque: τ = pE sinθ = 2×10⁻⁸ × 5×10⁴ × sin30° = 1×10⁻³ × 0.5 = 5×10⁻⁴ N·m.
U(30°) = −pE cos30° = −1×10⁻³ × 0.866 = −8.66×10⁻⁴ J.
U(90°) = −pE cos90° = 0 J.
Work done = U(90°) − U(30°) = 0 − (−8.66×10⁻⁴) = 8.66×10⁻⁴ J.

Inside a conductor in electrostatic equilibrium, the electric field is always zero — if it were not, free electrons would keep moving until the internal field vanished. Because E = −dV/dr and E = 0 throughout the conductor's interior, the potential is the same everywhere inside a conductor and on its surface; the whole conductor, including its surface, is an equipotential region. Since the field inside is zero, any excess charge given to a conductor resides entirely on its outer surface, and just outside that surface the field must be perpendicular to the surface (a tangential component would drive surface currents, which cannot persist in equilibrium).

Gauss's law applied to a thin pillbox straddling the surface gives the field just outside a charged conductor's surface as E = σ/ε₀, where σ is the local surface charge density. Because the charge distributes itself to keep the surface an equipotential, σ (and hence E) is generally not uniform — it is largest where the surface curvature is sharpest, which is why charge concentrates at pointed conductors and why lightning conductors are made with sharp tips.

A cavity carved inside a conductor, with no charge inside the cavity, has zero field everywhere within it, however the conductor is charged or however strong the external field around it is. This is the principle of electrostatic shielding: a conducting enclosure protects whatever is inside it from external electric fields (used, for example, to shield sensitive electronic equipment).

A dielectric is an insulator that has no free charges to move, but its molecules can still respond to an external field. Non-polar molecules (like O₂, which have no permanent dipole moment) develop an induced dipole moment when a field is applied; polar molecules (like H₂O, with a permanent dipole moment) are torqued into partial alignment with the field. Either way, the material becomes polarised, developing a net dipole moment per unit volume called the polarisation P = ε₀(K−1)E, where K is the dielectric constant of the material. This polarisation creates a field inside the dielectric that opposes the external field, so the net field inside is reduced: E_inside = E₀/K, where E₀ is the field that would exist there in vacuum.

Worked example: A charged conducting sphere has a uniform surface charge density σ = 8.85×10⁻⁷ C/m². Find the electric field just outside its surface.
E = σ/ε₀ = (8.85×10⁻⁷)/(8.85×10⁻¹²) = 1×10⁵ N/C.

A capacitor is a system of two conductors separated by an insulator, used to store electric charge and energy. When a potential difference V is applied between the two conductors, they acquire equal and opposite charges +Q and −Q, and the ratio Q/V is a constant for a given capacitor called its capacitance, C = Q/V. Capacitance depends only on the geometry of the conductors (their shape, size, separation) and the medium between them — not on Q or V themselves. The SI unit of capacitance is the farad (F); since a farad is a very large unit for practical devices, capacitances are usually quoted in microfarads (μF, 10⁻⁶ F) or picofarads (pF, 10⁻¹² F).

The most common capacitor is the parallel plate capacitor: two large, flat, parallel conducting plates of area A separated by a small distance d. Treating each plate as an infinite sheet of charge, the field between the plates is E = σ/ε₀ = Q/(ε₀A) (uniform, and effectively zero outside the plates because the two sheets' fields cancel there). Since the field is uniform, V = Ed = Qd/(ε₀A), so the capacitance in vacuum (or air) is C₀ = ε₀A/d. This shows capacitance increases with plate area and decreases with plate separation.

If the space between the plates is completely filled with a dielectric of dielectric constant K, the field inside is reduced to E₀/K (from the previous section), so for the same charge, the potential difference V drops by the same factor K, and the capacitance increases: C = Kε₀A/d = KC₀. This is precisely why capacitors are built with a dielectric — for the same size, they can store far more charge at the same voltage than an air-gap capacitor.

Worked example: A parallel plate capacitor has plates of area A = 6×10⁻³ m² separated by d = 3 mm in air.
(a) Find its capacitance.
C₀ = ε₀A/d = (8.85×10⁻¹² × 6×10⁻³)/(3×10⁻³) = 8.85×10⁻¹² × 2 = 1.77×10⁻¹¹ F = 17.7 pF.
(b) If a dielectric slab of K = 6 completely fills the gap, find the new capacitance.
C = KC₀ = 6 × 17.7 pF = 106.2 pF.
(c) If connected to a 100 V battery (air-filled case), find the charge stored.
Q = C₀V = 1.77×10⁻¹¹ × 100 = 1.77×10⁻⁹ C = 1.77 nC.

Capacitors can be combined to obtain an effective capacitance different from any single unit. In a series combination, capacitors are connected end to end so the same charge Q flows onto each one, but the total voltage splits across them; the reciprocals of capacitance add: 1/C_s = 1/C₁ + 1/C₂ + 1/C₃ + …, so the equivalent capacitance is always smaller than the smallest individual capacitor. In a parallel combination, capacitors share the same voltage across each one, but the charges (and hence the effective capacitance) add directly: C_p = C₁ + C₂ + C₃ + …, giving an equivalent capacitance larger than any individual one.

Charging a capacitor means doing work to move charge from one plate to the other against the growing electric field, and this work is stored as electrostatic potential energy. If a small charge dq is moved when the plates already carry charge q (at that instant the potential difference is q/C), the work done is dq·(q/C). Integrating from 0 to the final charge Q gives the total energy stored: U = Q²/(2C). Using Q = CV, this can equally be written as U = ½QV = ½CV². This same energy, spread through the volume between the plates (Ad for a parallel plate capacitor), corresponds to an energy density u = ½ε₀E² stored in the electric field itself — a result that holds generally, not just for capacitors, and supports the view that the field itself carries energy.

A practical application of these ideas at very high voltage is the Van de Graaff generator, used to accelerate charged particles for nuclear and particle physics experiments. Charge is sprayed onto a moving insulating belt by a corona discharge from a sharp electrode at the base, carried up inside a large hollow conducting sphere, and transferred to the sphere's inner surface by another pointed electrode. Because charge given to the inside of a hollow conductor always migrates entirely to its outer surface (regardless of how much charge is already there), this process can be repeated continuously, building up an extremely high potential — several million volts — on the outer shell, limited in practice only by the electrical breakdown of the surrounding air (often mitigated by housing the generator in high-pressure gas).

Worked example (combination): C₁ = 2 μF and C₂ = 3 μF are connected in series, and this series combination is connected in parallel with C₃ = 4 μF. Find the total capacitance.
Series part: 1/C_s = 1/2 + 1/3 = 5/6, so C_s = 6/5 = 1.2 μF.
Total (parallel with C₃): C = C_s + C₃ = 1.2 + 4 = 5.2 μF.

Worked example (energy): A 4 μF capacitor is charged to 100 V. Find the charge stored and the energy stored.
Q = CV = 4×10⁻⁶ × 100 = 4×10⁻⁴ C.
U = ½CV² = ½ × 4×10⁻⁶ × (100)² = ½ × 4×10⁻⁶ × 10⁴ = ½ × 0.04 = 0.02 J.