Difficulty: Intermediate | Prerequisites: Gauss's Law and electric fields (see Part 1 notes), basic integration.
Electric potential gives you a scalar way to describe the same physics that the electric field describes as a vector. Because potential is a scalar, superposition is simpler: you just add numbers, not vectors. This makes it the go-to tool for problems with multiple point charges. Capacitance builds on potential by quantifying how much charge a pair of conductors can store per volt of potential difference. Inserting a dielectric between the plates changes the capacitance and stored energy in ways that depend on whether the battery stays connected. These ideas underpin every circuit component that stores energy, from camera flashes to the bypass capacitors on a circuit board.
Electric potential is the potential energy per unit charge. Equipotential surfaces are perpendicular to field lines, and no work is done moving a charge along one. Capacitance is C = Q/V, and inserting a dielectric multiplies C by κ. Whether the stored energy goes up or down depends on whether voltage or charge is held constant.
Electric potential (V)
The electric potential energy per unit charge at a point in space: V = U/q. In simple terms, it tells you how much energy a unit positive charge would have at that location. Measured in volts (1 V = 1 J/C).
Potential difference (ΔV)
The change in potential between two points: ΔV = V_b − V_a = −∫ E · dl. This is what voltmeters measure and what drives current through a circuit.
Equipotential surface
A surface on which every point has the same electric potential. In simple terms, a charge sitting anywhere on this surface has the same potential energy, so moving it along the surface costs no work.
Capacitance (C)
The ratio of stored charge to potential difference for a pair of conductors: C = Q / ΔV. Measured in farads (F). Think of it as how much charge the device can hold per volt you apply.
Dielectric
An insulating material inserted between capacitor plates. It reduces the effective electric field inside, which increases the capacitance by a factor κ (the dielectric constant).
Dielectric constant (κ)
A dimensionless number (always ≥ 1) that describes how much a dielectric material reduces the electric field. For vacuum, κ = 1. For common materials, κ ranges from about 2 (teflon) to 80 (water).
Spherical capacitor
Two concentric conducting spherical shells of radii a and b (a < b), one carrying +Q and the other −Q. Its capacitance is C = 4πε₀ab / (b − a).
For a single point charge Q at the origin: V(r) = kQ / r (with V(∞) = 0).
For multiple charges, add the potentials as scalars: V = Σ kQᵢ / rᵢ.
Potential can be positive or negative depending on the sign of the charge.
For charges +Q and −2Q separated by distance d, set V = kQ/r₁ + k(−2Q)/r₂ = 0.
This gives 1/r₁ = 2/r₂, so r₂ = 2r₁.
The point where V = 0 is between the charges, closer to the smaller-magnitude charge (+Q), at a distance d/3 from +Q.
There is also a second point to the left of +Q (outside the pair), but the exam question specifies "between."
Electric field lines are always perpendicular to equipotential surfaces. This follows directly from E = −∇V.
The work done moving a charge along an equipotential surface is zero, because there is no component of E along the surface.
Equipotential surfaces can be closed (e.g. spheres around a point charge). The claim that they "can never be closed loops" is false.
Parallel-plate capacitor: C = ε₀A / d, where A is plate area and d is plate separation.
Spherical capacitor (radii a and b, charges +Q and −Q):
E(r) for r < a: 0 (inside the inner conductor).
E(r) for a < r < b: kQ / r² (only +Q enclosed).
E(r) for r > b: 0 (net enclosed charge is zero).
Potential: V(r) for r ≥ b is 0. For a ≤ r ≤ b: V(r) = (Q / 4πε₀)(1/r − 1/b).
Potential difference: ΔV = V(a) − V(b) = Q(b − a) / (4πε₀ab).
Capacitance: C = 4πε₀ab / (b − a).
Inserting a dielectric with constant κ > 1 multiplies the capacitance: C' = κC.
Energy stored in a capacitor: U = ½CV² = Q²/(2C).
Battery stays connected (V constant): C increases → U = ½CV² increases by factor κ. The battery supplies extra charge.
Battery disconnected (Q constant): C increases → U = Q²/(2C) decreases by factor κ. Energy goes into pulling the dielectric slab in.
This distinction (constant V vs. constant Q) is a favourite exam trap.
Quantity | Expression |
|---|---|
Potential from point charge | V = kQ / r |
Relationship E and V | E = −∇V, or ΔV = −∫ E · dl |
Parallel-plate capacitance | C = ε₀A / d |
Spherical capacitance | C = 4πε₀ab / (b − a) |
Cylindrical capacitance (length L) | C = 2πε₀L / ln(b/a) |
With dielectric | C' = κC |
Energy stored | U = ½CV² = ½QV = Q²/(2C) |
Capacitors with dielectrics are everywhere: the ceramic capacitors on a phone's circuit board use high-κ materials to store more charge in a tiny space. Defibrillators store energy in a large capacitor and release it in a single pulse to restart a heart rhythm.
Students often confuse potential (a scalar, in volts) with electric field (a vector, in N/C or V/m). A region can have nonzero potential but zero field (e.g. inside a charged conductor).
A common error is thinking that inserting a dielectric always increases stored energy. It depends on whether the battery is connected. With V constant, energy goes up. With Q constant, energy goes down.
Students sometimes think equipotential surfaces must be flat planes. They can be any shape, including closed surfaces like spheres.
When finding V = 0 between two charges of different magnitude, students often place the zero point at the midpoint. The zero is closer to the smaller-magnitude charge, not at the centre.
⚠️ The battery-connected vs. battery-disconnected dielectric question is a classic multiple-choice item. Know which quantity (V or Q) stays constant and use the correct energy formula.
⚠️ The spherical capacitor derivation (E in three regions, V by integration, then C = Q/ΔV) is a 10-mark long-answer question. Practise it end to end.
⚠️ Equipotential surfaces being perpendicular to field lines is a standard true/false or multiple-choice question.
⚠️ For the two-charge V = 0 problem, remember that potential is a scalar sum and the zero point is between the charges, closer to the smaller one.
True or false: The electric field is always perpendicular to equipotential surfaces.
Fill in the blank: The work done moving a charge along an equipotential surface is _______.
True or false: Inserting a dielectric into a capacitor connected to a battery decreases the stored energy.
Fill in the blank: The capacitance of a spherical capacitor with inner radius a and outer radius b is C = _______.
True or false: Electric potential is a vector quantity.
Answers: 1. True. 2. Zero. 3. False (it increases). 4. 4πε₀ab / (b − a). 5. False (it is a scalar).
Q: Two point charges +Q and −2Q are separated by distance d. Where along the line joining them (other than infinity) is V = 0?
A: Between the charges, closer to +Q. Setting kQ/r₁ = 2kQ/r₂ gives r₂ = 2r₁, so the point is at d/3 from +Q.
Q: Which statement is true about equipotential surfaces? (a) E-field lines are parallel to them, (b) work done along them is maximum, (c) E-field lines are perpendicular to them, (d) they can never be closed.
A: (c). By definition, E = −∇V, so the field points in the direction of steepest potential decrease, which is perpendicular to surfaces of constant V.
Q: A parallel-plate capacitor is connected to a battery at voltage V. A dielectric (κ > 1) is inserted. What happens to the stored energy?
A: It increases. V is constant (battery connected), C increases to κC, so U = ½CV² increases by a factor of κ.
Q: For two concentric spherical shells (radii a and b, charges +Q and −Q), find the electric field in the region a < r < b.
A: E = Q / (4πε₀r²) = kQ / r², directed radially outward. Only the inner shell's charge is enclosed by a Gaussian sphere at radius r.
Q: Calculate the capacitance of the spherical capacitor described above.
A: C = 4πε₀ab / (b − a). This follows from C = Q / ΔV where ΔV = Q(b − a)/(4πε₀ab).
Potential difference is what drives current in circuits (see the DC/AC circuits notes). The capacitance formulas derived here feed directly into RC circuit time constants (τ = RC) and energy storage in RLC circuits. The relationship E = −∇V connects this material to the Gauss's Law notes (Part 1) and reappears when you study Faraday's Law and electromagnetic induction.
electric potential, voltage, potential difference, equipotential surface, capacitance, capacitor, parallel-plate capacitor, spherical capacitor, cylindrical capacitor, dielectric, dielectric constant, κ, stored energy, energy density, Coulomb potential, superposition of potentials, PHYS 212, University Physics electricity and magnetism