Source: University Physics: Elec & Mag, UIUC
Tags: electric potential, voltage, equipotential, gradient, capacitance, parallel plate capacitor, cylindrical capacitor, dielectric, energy stored in capacitor, PHY 212
Difficulty: Intermediate Prerequisites: Gauss's law and electric potential energy study notes, basic integration, gradient (partial derivatives).
Electric potential (V) is the per-unit-charge version of electric potential energy. Where potential energy tells you about a specific charge in a field, potential tells you about the field itself, at every point in space. It is a scalar, which makes it much easier to work with than the vector electric field, and you can always recover E from V by taking the gradient. Capacitance then brings everything together: you have two conductors, you find E between them (often via Gauss's law), integrate E to get V, and the ratio Q/V gives you the capacitance, a quantity that depends only on geometry. Dielectrics modify that geometry-dependent number. This block of material ties the entire first half of the course into a single chain: charge → E → V → C.
Electric potential is potential energy per unit charge: ΔV = −∫ E · dl. It is a scalar, and equipotential surfaces are always perpendicular to electric field lines. Capacitance is the ratio C = Q / ΔV for a pair of conductors; it depends only on geometry (and on any dielectric present). Energy stored in a capacitor is U = ½CV².
Electric Potential (V)
Potential energy per unit charge: ΔV_a→b = ΔU_a→b / q = −∫ₐᵇ E · dl. Think of it as the "voltage landscape" of space. A positive charge rolls downhill from high V to low V, just as a ball rolls downhill in a gravitational field.
Equipotential Surface
A surface on which every point has the same electric potential. In simple terms, no work is done moving a charge along an equipotential, and E is always perpendicular to it.
Gradient Relationship (E from V)
E = −∇V. In Cartesian coordinates: Eₓ = −∂V/∂x, Eᵧ = −∂V/∂y, E_z = −∂V/∂z. This converts the scalar V back into the vector field E.
Capacitance (C)
The ratio of charge stored to potential difference: C = Q / ΔV, measured in farads (F). Think of it as how much charge a pair of conductors can hold per volt of potential difference between them. It depends only on the shape, size, and spacing of the conductors (and on any dielectric material between them).
Dielectric
An insulating material placed between the plates of a capacitor. It increases capacitance by a factor κ (the dielectric constant): C₁ = κC.
Dielectric Constant (κ)
A dimensionless number (always ≥ 1) that characterises how much a dielectric material reduces the electric field inside it and thereby increases capacitance.
Energy Density (u)
Energy stored per unit volume in an electric field: u = ½ε₀E². This tells you that wherever there is an electric field, there is stored energy.
ΔV_a→b = −∫ₐᵇ E · dl.
V is a scalar (but it carries a sign).
For a point charge: V(r) = kQ / r (taking V = 0 at infinity).
A positive charge creates a positive potential that decreases with distance; a negative charge creates a negative potential.
E = −∇V.
E points in the direction of the steepest decrease in V.
E is proportional to how rapidly V changes with position: stronger gradients mean stronger fields.
Perpendicular to electric field lines everywhere.
Closer spacing of equipotentials corresponds to a stronger electric field (just like contour lines on a topographic map).
The surface of a conductor at equilibrium is an equipotential.
No work is done when a charge moves along an equipotential.
Setup: point charge q at the centre of a spherical shell (inner radius a₃, outer radius a₄) carrying charge Q.
r > a₄: V = k(Q + q) / r (shell and point charge look like a single charge from far away).
a₃ < r < a₄ (inside the conductor): V is constant (conductor is an equipotential). Break the path integral into segments where E is known.
r < a₃: V = kq / r + constant contribution from the shell.
The graph of V(r) is continuous, with a flat region through the conductor where E = 0.
Outside (r > a): E = kQ / r², so V(r) = kQ / r. Identical to a point charge.
Inside (r < a): Use Gauss's law with ρ = Q / (4πa³/3) to find E, then integrate.
E_inside = kQr / a³.
V(r) = kQ(3a² − r²) / (2a³) for r < a.
V is continuous at r = a and has a parabolic shape inside the sphere.
ΔV = −∫ E · dl = 0, so V is constant throughout that region.
Zero field does not mean zero potential; it means the potential is the same everywhere in that region.
C = Q / ΔV. Units: farads (F). 1 F = 1 C/V.
Capacitance is determined entirely by geometry (shape, area, separation) and by any dielectric present. It does not depend on Q or V individually.
The algorithm for finding C:
Assume some charge Q on the plates.
Use Gauss's law (or other methods) to find E between the plates.
Integrate E to find V = −∫ E · dl.
Take the ratio C = Q / V. Q cancels, leaving only geometry.
Two parallel conducting plates, area A, separation d, with charge +Q on one and −Q on the other.
E between plates (from Gauss's law, planar symmetry): E = σ / ε₀ = Q / (A · ε₀).
V = E · d = Q · d / (A · ε₀).
C = Q / V = ε₀A / d.
Capacitance increases with larger plate area and smaller separation.
Two coaxial cylinders of length L (L ≫ radii), inner radius a₂, outer radius a₃.
E between the cylinders (cylindrical symmetry): E = Q / (2πε₀ · L · r).
V = (Q / 2πε₀L) · ln(a₃ / a₂).
C = 2πε₀L / ln(a₃ / a₂).
A metal conductor inserted between parallel plates reduces the effective separation (from d to d − t, where t is the slab thickness).
E = 0 inside the metal slab, so only the remaining gap contributes to V.
New capacitance: C = ε₀A / (d − t) > ε₀A / d. Capacitance increases.
Wires are conductors, so they are equipotentials.
In a simple circuit with a battery of voltage V: Q = CV.
Parallel combination: C_eq = C₁ + C₂ + ... (same voltage across each, charges add).
Series combination: 1/C_eq = 1/C₁ + 1/C₂ + ... (same charge on each, voltages add).
Three equivalent forms: U = ½QV = ½Q²/C = ½CV².
Energy density in the electric field: u = ½ε₀E².
The energy is stored in the electric field between the plates, not on the plates themselves.
Inserting a dielectric (constant κ) between the plates multiplies the capacitance: C₁ = κC.
Two scenarios to distinguish:
Connected to a battery (constant V): V stays the same. C increases by κ, so Q increases by κ (Q₁ = κQ).
Isolated (constant Q): Q stays the same. C increases by κ, so V decreases by κ (V₁ = V/κ).
In both cases C₁ = κC. The difference is which quantity is held fixed.
A dielectric of constant κ fills a fraction of the plate area (say ¼ of the area covered by dielectric, ¾ uncovered).
Treat as two capacitors in parallel:
C₁ = (3/4) · ε₀A / d (air portion).
C₂ = κ · (1/4) · ε₀A / d (dielectric portion).
C_total = C₁ + C₂ = C · (3/4 + κ/4), where C = ε₀A / d is the original capacitance.
Quantity | Formula |
|---|---|
Electric potential (point charge) | V = kQ / r |
Potential difference | ΔV = −∫ E · dl |
E from V | E = −∇V |
Parallel-plate capacitance | C = ε₀A / d |
Cylindrical capacitance | C = 2πε₀L / ln(a_outer / a_inner) |
Capacitors in parallel | C_eq = ΣCᵢ |
Capacitors in series | 1/C_eq = Σ(1/Cᵢ) |
Energy stored | U = ½QV = ½Q²/C = ½CV² |
Energy density | u = ½ε₀E² |
Dielectric effect | C₁ = κC |
Capacitors are everywhere in electronics: they smooth voltage in power supplies, store energy for camera flashes, and tune radio circuits to specific frequencies. The dielectric in a capacitor is not just filler; choosing the right material (ceramic, polymer, electrolyte) determines the capacitor's voltage rating, size, and reliability. The energy-density formula u = ½ε₀E² also explains why high-voltage power lines are spaced far apart: the energy stored in strong electric fields can cause dielectric breakdown of air.
Students often think capacitance depends on the charge on the plates or the voltage applied. It does not. C is a property of the geometry (and the dielectric). Q and V can change; their ratio stays the same.
Confusing the two dielectric scenarios. When connected to a battery, V is fixed and Q changes. When isolated, Q is fixed and V changes. Mixing these up inverts the answer.
Forgetting that inserting a conductor (metal) between plates is different from inserting a dielectric. A conductor forces E = 0 inside the slab; a dielectric merely reduces E by a factor of κ.
Assuming E = 0 means V = 0. Zero field in a region means V is constant there, not necessarily zero.
⚠️ The relationship E = −∇V (or equivalently ΔV = −∫ E · dl) appears in nearly every potential problem. Know it both ways.
⚠️ Capacitance derivations (parallel plate, cylindrical) are classic exam problems. The algorithm is always the same: assume Q, find E, integrate for V, divide Q/V.
⚠️ Dielectric problems almost always test whether you can identify the correct constraint: constant V (battery connected) vs. constant Q (isolated). State this clearly in your working.
⚠️ Energy stored in a capacitor has three equivalent expressions. Pick the one that uses the quantity held constant in the problem.
⚠️ Equipotentials and field lines being perpendicular is a common conceptual question.
Q: True or false – Electric potential is a vector quantity.
A: False. It is a scalar.
Q: Fill in the blank – The capacitance of a parallel-plate capacitor is C = ___.
A: ε₀A / d.
Q: True or false – Inserting a dielectric into an isolated capacitor increases the voltage across it.
A: False. It decreases the voltage (V₁ = V/κ), because Q stays constant and C increases.
Q: If E = 0 throughout a region, what can you say about V in that region?
A: V is constant (the same everywhere in that region).
Q: True or false – Capacitors in series have the same charge on each capacitor.
A: True.
Q: What is the electric potential at a distance of 0.3 m from a point charge of +2 μC?
A: V = kQ/r = (8.99 × 10⁹)(2 × 10⁻⁶) / 0.3 ≈ 5.99 × 10⁴ V ≈ 60 kV.
Q: A parallel-plate capacitor has plates of area 0.01 m² separated by 0.002 m. What is its capacitance?
A: C = ε₀A/d = (8.85 × 10⁻¹²)(0.01) / 0.002 = 4.43 × 10⁻¹¹ F ≈ 44.3 pF.
Q: A 10 μF capacitor is charged to 100 V. How much energy is stored?
A: U = ½CV² = ½(10 × 10⁻⁶)(100)² = 0.05 J = 50 mJ.
Q: A dielectric with κ = 3 is inserted into a capacitor that remains connected to a 12 V battery. If the original capacitance was 5 μF, what are the new capacitance and charge?
A: C₁ = κC = 3 × 5 = 15 μF. Since V is constant (battery connected), Q₁ = C₁V = 15 × 10⁻⁶ × 12 = 180 μC.
Q: Two capacitors, 4 μF and 6 μF, are connected in series. What is the equivalent capacitance?
A: 1/C_eq = 1/4 + 1/6 = 5/12, so C_eq = 12/5 = 2.4 μF.
Electric potential is the bridge between fields and circuits. Once you have V, you can define current (rate of charge flow driven by potential difference), which launches the second half of the course. Capacitance connects back to Gauss's law (you need it to find E) and forward to RC circuits, where the time constant τ = RC governs charging and discharging behaviour. Dielectrics reappear in the study of polarisation and bound charges in later electrostatics topics.
Electric potential, voltage, potential difference, equipotential, gradient, del operator, capacitance, capacitor, parallel plate, cylindrical capacitor, dielectric, dielectric constant, kappa, energy stored in capacitor, energy density, series capacitors, parallel capacitors, farad, PHY 212 midterm 1, UIUC physics, university physics electricity and magnetism