Capacitors: Study Guide (Physics 2 10.6)

Two plates, equal and opposite charge

A parallel-plate capacitor is two flat conducting plates facing each other, a small distance d apart, with an insulator (or empty space) between them. Connect the plates to a battery and electrons move off one plate and onto the other. The result is +Q on one plate and −Q on the other: equal sizes, opposite signs. The total charge stays zero.

Separating the charges creates a potential difference ΔV between the plates. The more charge you separate, the larger ΔV becomes, in direct proportion. The ratio is the capacitance:

C = Q / ΔV

Q is the size of the charge on one plate. The unit is the farad: 1 F = 1 C/V. A farad is very large; real capacitors are usually measured in microfarads (µF, 10⁻⁶ F), nanofarads (nF, 10⁻⁹ F) or picofarads (pF, 10⁻¹² F).

A large C means the capacitor holds a lot of charge for each volt across it.

What sets the capacitance

Capacitance depends only on the capacitor itself: its shape, its size and the material between the plates. It does not depend on how much charge is on it or what voltage is applied. Double Q and ΔV doubles too, so their ratio stays the same.

For parallel plates:

C = κε₀A / d

• A is the area of one plate (m²).
• d is the distance between the plates (m).
• ε₀ = 8.85 × 10⁻¹² C²/(N·m²) is the permittivity of free space.
• κ (kappa) is the dielectric constant of the material between the plates. κ = 1 for a vacuum, air is very close to 1, and insulating materials have κ > 1. κ has no unit.

Why these trends make sense:

• Larger A: the same charge spreads over more area, so the field between the plates is weaker and ΔV is smaller. Smaller ΔV for the same Q means larger C. So C ∝ A.
• Smaller d: with the same field, ΔV = Ed is smaller over a shorter distance. Again, larger C. So C ∝ 1/d.

You can test C ∝ 1/d in the lab. Mount two flat plates on insulating spacers of different thicknesses, measure C with a capacitance meter for each spacing, and plot C against 1/d. A straight line confirms the relationship, and its slope estimates κε₀A.

The field between the plates

Away from the edges, the field between the plates is uniform: the same size and direction everywhere. It points from the positive plate to the negative plate, at right angles to the plates. Near the edges the field bulges outward; in this course you ignore these edge effects unless told otherwise.

Because the field is uniform, the average field equals the actual field:

E = ΔV / d

With empty space between the plates, the field also depends directly on the charge per unit area:

E = Q / (ε₀A)

The two formulas agree. Put Q = CΔV = (ε₀A/d)ΔV into the second one and you get ΔV/d.

The equipotentials between the plates are flat surfaces parallel to the plates, equally spaced for equal steps of V. That is the pattern of a uniform field.

Charged particles between the plates

A charge q between the plates feels a constant force F = qE, so it has a constant acceleration a = qE/m. That is exactly the situation of a thrown ball near Earth’s surface, with qE/m playing the part of g. If the particle enters moving parallel to the plates:

• along the plates: no force, so constant velocity, x = v₀t;
• across the plates: constant acceleration from rest, y = ½at².

The path is a parabola. A positive charge curves toward the negative plate; a negative charge curves toward the positive plate. For electrons and protons the electric force is so much larger than the weight that you can ignore gravity.

Energy stored in a capacitor

Charging a capacitor takes work. Each small bit of charge must be pushed from one plate to the other against the potential difference that already exists. The first bit is easy (ΔV ≈ 0); the last bit is pushed through the full ΔV. The energy stored equals that work.

On a graph of Q against ΔV (Figure 2), the line through the origin has slope C, and the energy stored is the area under the line, a triangle:

U_C = ½QΔV = ½C(ΔV)² = Q²/(2C)

The ½ appears because the average potential difference during charging is half the final value.

Dielectrics

A dielectric is an insulator placed between the plates. Its charges cannot flow, but the plates’ field pulls them slightly apart: the positive charges shift a little toward the negative plate and the negative charges toward the positive plate. This creates an induced field inside the dielectric that points opposite to the field from the plates. The net field is smaller.

• For a fixed Q, a smaller field means a smaller ΔV = Ed. Since C = Q/ΔV, the capacitance increases by the factor κ when the dielectric fills the gap.
• If the capacitor stays connected to a battery, ΔV cannot change. The larger C then means the battery pushes more charge onto the plates: Q = κC₀ΔV.

So before any “what changes?” question, decide which quantity is held fixed:

Situation	Held fixed	Why
Charged, then disconnected (isolated)	Q	the charge has nowhere to go
Kept connected to a battery	ΔV	the battery sets the potential difference

Worked example 1: a parallel-plate capacitor

Question. Two plates, each 0.20 m × 0.15 m, are 1.5 mm apart with air between them (κ = 1). The capacitor is connected to a 12 V battery. Find the capacitance, the charge on each plate, the field between the plates and the energy stored.

1. Area of one plate: A = 0.20 m × 0.15 m = 0.030 m². Separation d = 1.5 × 10⁻³ m.
2. Capacitance: C = ε₀A/d = (8.85 × 10⁻¹²)(0.030) ÷ (1.5 × 10⁻³) = 1.77 × 10⁻¹⁰ F (177 pF).
3. Charge: Q = CΔV = (1.77 × 10⁻¹⁰ F)(12 V) = 2.12 × 10⁻⁹ C. The other plate holds −2.12 × 10⁻⁹ C.
4. Field: E = ΔV/d = 12 V ÷ 1.5 × 10⁻³ m = 8.0 × 10³ V/m, from the + plate to the − plate.
5. Energy: U_C = ½C(ΔV)² = 0.5 × (1.77 × 10⁻¹⁰)(12)² = 1.27 × 10⁻⁸ J.

Check. The field from the charge gives the same result: E = Q/(ε₀A) = 2.124 × 10⁻⁹ ÷ (8.85 × 10⁻¹² × 0.030) = 8000 V/m. And ½QΔV = 0.5 × 2.124 × 10⁻⁹ × 12 = 1.27 × 10⁻⁸ J, which agrees. A capacitor with plates about half the size of a sheet of printer paper holds only a couple of nanocoulombs at 12 V, which is why practical capacitors use very thin dielectric layers rolled or stacked into a small volume.

Worked example 2: isolated or connected?

Question. Start with the charged capacitor from Worked example 1 (177 pF, 12 V). Predict what happens to C, Q, ΔV, E and U_C in each case: (a) it is disconnected and the plate separation is doubled; (b) it stays connected and the separation is doubled; (c) it is disconnected and a dielectric with κ = 3.0 is slid in to fill the gap.

1. (a) Isolated, so Q is fixed at 2.12 nC. C = ε₀A/d halves to 88.5 pF. ΔV = Q/C doubles to 24 V. E = ΔV/d = 24 V ÷ 3.0 mm = 8.0 × 10³ V/m, unchanged (E = Q/(ε₀A) depends only on Q and A). U_C = ½QΔV doubles to 2.55 × 10⁻⁸ J.
2. (b) Connected, so ΔV is fixed at 12 V. C halves to 88.5 pF. Q = CΔV halves to 1.06 nC. E = 12 V ÷ 3.0 mm = 4.0 × 10³ V/m, halved. U_C = ½C(ΔV)² halves to 6.37 × 10⁻⁹ J.
3. (c) Isolated, so Q stays at 2.12 nC. C triples to 531 pF. ΔV = Q/C falls to 4.0 V. The field in the dielectric is 4.0 V ÷ 1.5 mm = 2.7 × 10³ V/m, one third of before: the induced field (5.3 × 10³ V/m, opposite) cancels two thirds of the plates’ field. U_C = Q²/(2C) falls to one third, 4.25 × 10⁻⁹ J.

Case	C	Q	ΔV	E	U_C
(a) isolated, d doubled	½	same	×2	same	×2
(b) connected, d doubled	½	½	same	½	½
(c) isolated, κ = 3.0 added	×3	same	⅓	⅓	⅓

Interpretation. In (a) the stored energy rises: you must do work to pull the oppositely charged plates apart, and that work is stored. In (c) the energy falls: the field pulls the dielectric into the gap and does work on it, so the capacitor loses energy.

Worked example 3: an electron between the plates

Question. Two plates 2.0 cm apart and 4.0 cm long have a potential difference of 50 V. An electron enters along the midline, moving parallel to the plates at 2.0 × 10⁷ m/s (Figure 1). How far is it deflected by the time it leaves the plates? Which way?

1. Field: E = ΔV/d = 50 V ÷ 0.020 m = 2.5 × 10³ V/m, from the + plate to the − plate.
2. Force: F = eE = (1.60 × 10⁻¹⁹ C)(2500 V/m) = 4.0 × 10⁻¹⁶ N, toward the positive plate (the electron is negative, so the force is opposite to E).
3. Acceleration: a = F/m = 4.0 × 10⁻¹⁶ ÷ 9.11 × 10⁻³¹ = 4.39 × 10¹⁴ m/s².
4. Time between the plates: t = L/v₀ = 0.040 m ÷ 2.0 × 10⁷ m/s = 2.0 × 10⁻⁹ s.
5. Deflection: y = ½at² = 0.5 × (4.39 × 10¹⁴)(2.0 × 10⁻⁹)² = 8.8 × 10⁻⁴ m, about 0.88 mm toward the positive plate.

Check. 0.88 mm is less than the 10 mm from the midline to the plate, so the electron gets through. Gravity would give an acceleration of 9.8 m/s², about 2 × 10⁻¹⁴ of the electric one, so ignoring it is safe.

Common misconceptions

• “C depends on Q or ΔV.” C = Q/ΔV is how you measure C, not what sets it. Only A, d and κ set it.
• Adding the charges on both plates. Q is the charge on one plate. The capacitor’s total charge is zero.
• Forgetting the ½ in the stored energy. QΔV would assume all the charge was moved through the full ΔV.
• Not deciding what is fixed. Most errors in “what happens if…” questions come from assuming Q is fixed when the battery is still connected, or the reverse.
• “The field is stronger near the plates.” Away from the edges it is the same everywhere between the plates.
• “A dielectric is a conductor.” A conductor filling the gap would join the plates and discharge them. A dielectric is an insulator whose charges only shift slightly.
• Bending the electron the wrong way. A negative charge is pushed opposite to the field, toward the positive plate.

Where this leads

Topic 10.7 uses ΔU_E = qΔV with conservation of energy to find the speed a charge gains crossing a potential difference such as the gap in a capacitor: see the conservation of electric energy study guide. Capacitors return in Unit 11, in circuits that charge and discharge them. Test yourself with the practice questions, then use the revision notes and the topic checklist. For potential and equipotentials, look back at the electric potential study guide.