Capacitors: Study Guide (Physics C: E&M 10.3)

Course note. This guide is for the calculus-based Physics C: Electricity and Magnetism course, Topic 10.3. The algebra-based Physics 2 course also meets parallel-plate capacitors; this course adds Gauss’s-law derivations, spherical and cylindrical capacitors, and energy by integration.

Constants used throughout: ε₀ = 8.85 × 10⁻¹² C²/(N·m²), 1/(4πε₀) = 8.99 × 10⁹ N·m²/C². Unless a question says otherwise, the space between the conductors is air or vacuum (dielectric constant κ = 1). Topic 10.4 deals with other materials.

What a capacitor is

A capacitor is two conductors separated by an insulating gap. When you connect it to a battery, the battery moves charge from one conductor to the other. One conductor ends up with +Q and the other with −Q. The net charge stays zero. When we say “the charge on a capacitor”, we mean Q, the size of the charge on either conductor.

The separated charges create a field in the gap and so a potential difference ΔV. Double Q and both the field and ΔV double. The fixed ratio is the capacitance:

C = Q / ΔV

The unit is the farad: 1 F = 1 C/V. A farad is very large, so real capacitors are usually rated in μF, nF or pF.

C depends only on the capacitor itself: the shape, size and spacing of the conductors and the material between them. Charging it more changes Q and ΔV in proportion, never C.

The parallel-plate capacitor

The simplest design is two flat plates of area A, a distance d apart, with d much smaller than the width of the plates. Find the field with Gauss’s law and superposition.

1. In Topic 8.6 you found that one large thin sheet with charge density σ makes a field σ/(2ε₀) on each side, pointing away from positive charge.
2. The + plate has density +σ = Q/A; the − plate has −σ.
3. Between the plates, both fields point from + to −. They add: E = σ/(2ε₀) + σ/(2ε₀) = σ/ε₀.
4. Outside the plates, the two fields point in opposite directions and cancel: E ≈ 0.

So between the plates:

E = σ/ε₀ = Q/(ε₀A)

The field is uniform in size and direction, except near the edges, where the lines bulge outward (“fringing”). E is proportional to σ: put twice the charge on the same plates and E doubles.

You can reach the same result in one step. Take a Gaussian pillbox with one end inside the metal of the + plate (where E = 0) and the other end in the gap. Only the gap end has flux: EA′ = σA′/ε₀, so E = σ/ε₀.

Capacitance. The field is uniform, so the potential difference is ΔV = Ed = Qd/(ε₀A). Then:

C = Q/ΔV = ε₀A/d

(With a material of dielectric constant κ filling the gap, C = κε₀A/d; see Topic 10.4.) Larger plates hold more charge for the same ΔV; a wider gap needs a larger ΔV for the same field, so C falls.

A method for any capacitor

The parallel-plate result follows a recipe that works for every shape in this course:

1. Put +Q on one conductor and −Q on the other.
2. Use Gauss’s law to find E in the gap.
3. Integrate along a path from one conductor to the other: ΔV = ∫E·dr (take the size).
4. Divide: C = Q/ΔV. Q should cancel. If it does not, check your work.

Concentric spherical capacitor. An inner sphere of radius a holds +Q; a thin outer shell of radius b holds −Q. A spherical Gaussian surface with a < r < b encloses +Q, so E = Q/(4πε₀r²). Then:

ΔV = ∫ₐᵇ Q/(4πε₀r²) dr = (Q/4πε₀)(1/a − 1/b) = (Q/4πε₀)(b − a)/(ab)

C = 4πε₀ab/(b − a)

Two checks:

• Let b → ∞. Then C → 4πε₀a, the capacitance of an isolated sphere.
• Let the gap be thin, b − a = d ≪ a. Then ab ≈ a², and C ≈ ε₀(4πa²)/d. That is ε₀A/d with A the area of the sphere: up close, the gap looks like parallel plates.

The coaxial cylindrical capacitor follows the same recipe; it is Worked example 2.

A charged particle between the plates

The field between the plates is uniform, so a charged particle there feels a constant force F = qE. Its acceleration a = qE/m is constant. The motion is just like a projectile near Earth’s surface, with qE/m in place of g:

• along the plates: no force, so constant velocity;
• across the plates: constant acceleration, so displacement ½at².

The path is a parabola. For electrons and protons, gravity is negligible next to qE (Worked example 3).

Energy stored in a capacitor

To charge a capacitor, something (usually a battery) must move charge from one conductor to the other against the field. The work done is stored as electric potential energy.

Suppose the capacitor already holds charge q, so the potential difference is q/C. Moving a further small charge dq takes work dW = (q/C) dq. Charging from 0 to Q:

U = ∫₀^Q (q/C) dq = Q²/(2C)

Using Q = CΔV, the same result can be written three ways:

U = Q²/(2C) = ½QΔV = ½C(ΔV)²

On a graph of ΔV against q, the line is straight with slope 1/C. Each thin strip of width dq under the line has area (q/C) dq, the work for that bit of charge. So the area under the line, a triangle, is the stored energy, ½QΔV.

Why ½QΔV and not QΔV? Early charge crossed a small ΔV; only the last bit crossed the full ΔV.

Fixed charge or fixed potential difference?

When the geometry changes, first ask what stays constant:

• Isolated capacitor (disconnected): Q is fixed, because the charge has nowhere to go.
• Capacitor connected to a battery: ΔV is fixed by the battery.

Double the plate separation d	C	Q	ΔV	E	U
Isolated (Q fixed)	×½	same	×2	same	×2
Connected (ΔV fixed)	×½	×½	same	×½	×½

Isolated: E = Q/(ε₀A) does not involve d, and U = Q²/(2C) doubles as C halves (you do work pulling the plates apart). Connected: U = ½C(ΔV)² halves.

Worked example 1: a parallel-plate capacitor, then pulling the plates apart

Question. Two plates, each of area 0.050 m², are 1.0 mm apart in air and connected to a 6.0 V battery. (a) Find C, Q, σ, E and U. (b) The battery is disconnected, then the plates are pulled apart to 3.0 mm. Find the new ΔV, E and U, and the work done by the person pulling.

(a)

1. C = ε₀A/d = (8.85 × 10⁻¹²)(0.050) ÷ (1.0 × 10⁻³) = 4.43 × 10⁻¹⁰ F = 443 pF.
2. Q = CΔV = (4.425 × 10⁻¹⁰ F)(6.0 V) = 2.66 × 10⁻⁹ C.
3. σ = Q/A = 5.31 × 10⁻⁸ C/m².
4. E = ΔV/d = 6.0 V ÷ 1.0 × 10⁻³ m = 6.0 × 10³ V/m. Check: σ/ε₀ = 5.31 × 10⁻⁸ ÷ 8.85 × 10⁻¹² = 6.0 × 10³ N/C. (1 V/m = 1 N/C.)
5. U = ½C(ΔV)² = ½(4.425 × 10⁻¹⁰)(6.0)² = 7.97 × 10⁻⁹ J.

(b) The capacitor is isolated, so Q stays 2.66 × 10⁻⁹ C.

1. C′ = ε₀A/(3.0 × 10⁻³ m) = 1.48 × 10⁻¹⁰ F, one third of C.
2. ΔV′ = Q/C′ = 18 V, three times larger.
3. E′ = Q/(ε₀A) is unchanged: 6.0 × 10³ V/m (check: 18 V ÷ 3.0 mm).
4. U′ = Q²/(2C′) = 2.39 × 10⁻⁸ J, three times larger.
5. Work done by the person = U′ − U = 1.59 × 10⁻⁸ J.

Interpretation. The plates attract, so separating them takes work, and that work is stored. ΔV rises although no charge was added.

Worked example 2: a coaxial cylindrical capacitor

Question. A coaxial cable has an inner wire of radius a = 0.40 mm and a thin outer conducting tube of radius b = 1.5 mm, with air between them. Its length is L = 2.0 m. (a) Derive C for a length L. (b) Evaluate C. (c) The cable is charged to 100 V. Find Q and the field at the surface of the inner wire.

(a)

1. Put +Q on the inner wire and −Q on the tube. The charge per length is λ = Q/L.
2. Gaussian surface: a coaxial cylinder of radius r (a < r < b) and length L. E is radial, so the end caps have zero flux: E(2πrL) = Q/ε₀, giving E = Q/(2πε₀Lr).
3. ΔV = ∫ₐᵇ Q/(2πε₀Lr) dr = [Q/(2πε₀L)] ln(b/a).
4. C = Q/ΔV = 2πε₀L / ln(b/a). Q cancels, as it should.

(b) ln(1.5/0.40) = ln 3.75 = 1.322. C = 2π(8.85 × 10⁻¹²)(2.0) ÷ 1.322 = 8.41 × 10⁻¹¹ F = 84 pF (42 pF per metre).

(c) Q = CΔV = 8.41 × 10⁻⁹ C, so λ = 4.21 × 10⁻⁹ C/m. At r = a: E = λ/(2πε₀a) = 1.89 × 10⁵ V/m.

Check. Rearranging, E(a) = ΔV/[a ln(b/a)] = 100 ÷ (4.0 × 10⁻⁴ × 1.322) = 1.89 × 10⁵ V/m. The field is strongest next to the thin wire (E ∝ 1/r). C grows with L, as expected.

Worked example 3: an electron passing between plates

Question. An electron enters the gap between two horizontal plates midway between them, moving at 1.0 × 10⁷ m/s parallel to the plates. The plates are 5.0 cm long and 2.0 cm apart, with ΔV = 40 V. Does it leave the plates, and at what angle?

1. E = ΔV/d = 40 ÷ 0.020 = 2.0 × 10³ V/m, uniform.
2. a = eE/mₑ = (1.60 × 10⁻¹⁹)(2.0 × 10³) ÷ (9.11 × 10⁻³¹) = 3.51 × 10¹⁴ m/s², towards the positive plate (the electron is negative).
3. Time between the plates: t = 0.050 m ÷ 1.0 × 10⁷ m/s = 5.0 × 10⁻⁹ s.
4. Sideways displacement: y = ½at² = ½(3.51 × 10¹⁴)(5.0 × 10⁻⁹)² = 4.4 × 10⁻³ m = 4.4 mm. That is less than the 10 mm to a plate, so the electron gets out.
5. Sideways velocity at exit: v_y = at = 1.76 × 10⁶ m/s. Angle: tan θ = 1.76 × 10⁶ ÷ 1.0 × 10⁷, so θ = 10° from its original direction.

Check. The electric force, 3.2 × 10⁻¹⁶ N, is more than 10¹³ times the electron’s weight (about 8.9 × 10⁻³⁰ N), so ignoring gravity is safe. A slower electron spends longer between the plates; y ∝ 1/v², so at half the speed it would deflect four times as far and hit the plate.

Common misconceptions

• “The capacitor’s charge is 2Q” or “its net charge is Q.” Each conductor holds Q in size; the net charge is zero.
• “More charge means more capacitance.” C is set by geometry and material. Extra charge raises ΔV in proportion, so Q/ΔV is unchanged.
• Using σ/(2ε₀) between the plates. That is one isolated sheet. Two plates add to σ/ε₀.
• “E between the plates depends on d.” With Q fixed, E = Q/(ε₀A) does not involve d. Only when ΔV is fixed does E = ΔV/d change with d.
• Forgetting to ask what is constant. Isolated means Q fixed; connected to a battery means ΔV fixed.
• U = QΔV. That is the energy for charge Q crossing a fixed ΔV. Charging a capacitor gives ½QΔV.
• Using the parallel-plate formula for spheres or cylinders. Use the spherical or coaxial result unless the gap is thin compared with the radius.

Where this leads

Next, in Topic 10.4, Dielectrics, you fill the gap with an insulating material and see C rise by the factor κ. In Unit 11 you will combine capacitors in series and parallel and watch them charge and discharge in RC circuits. Practise now with the practice questions, then use the revision notes and the topic checklist.