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2.6 - Capacitor

What a Capacitor Is

A capacitor is a device that stores electric charge and energy.
A parallel-plate capacitor is two parallel conducting plates separated by an insulator or a vacuum. Its key features: - The plates hold equal and opposite charges, with +Q on one plate and −Q on the other. - It creates a uniform electric field between the plates. - It stores electrical energy.

Capacitance

Capacitance (CC) measures how much charge a capacitor can store per unit voltage.
C=QΔVC = \frac{Q}{\Delta V}
where: - CC = capacitance (farads, F) - QQ = magnitude of charge on each plate (C) - ΔV\Delta V = potential difference between the plates (V)
The unit is the farad: 1 F = 1 coulomb/volt (C/V). You’ll usually see smaller units like μF (10−610^{-6} F), nF (10−910^{-9} F), and pF (10−1210^{-12} F).
What it means: - A large CC stores more charge for a given voltage. - A small CC stores less charge for a given voltage.
Importantly, capacitance depends only on the physical setup (the geometry and materials), not on the charge or the voltage.

Capacitance of a Parallel-Plate Capacitor

C=κϵ0AdC = \frac{\kappa\epsilon_0 A}{d}
where: - CC = capacitance (F) - κ\kappa = dielectric constant (dimensionless, κ≥1\kappa \geq 1) - ϵ0=8.85×10−12\epsilon_0 = 8.85 \times 10^{-12} C²/(N·m²) (permittivity of free space) - AA = area of one plate (m²) - dd = separation between the plates (m)
The dependencies are: - C∝AC \propto A: a larger plate area means more capacitance, since there’s more room for charge. - C∝1dC \propto \frac{1}{d}: a bigger gap means less capacitance, since the interaction is weaker. - C∝κC \propto \kappa: a higher dielectric constant means more capacitance.
For a vacuum or air, κ=1\kappa = 1, so C=ϵ0AdC = \frac{\epsilon_0 A}{d}.
Testing this in a lab: change one variable at a time and measure CC with a capacitance meter (or charge the plates to a known ΔV\Delta V, measure QQ, and use C=Q/ΔVC = Q/\Delta V). - To test the dd dependence, keep the plates and the material the same and measure CC at several gaps. A graph of CC vs. 1/d1/d should be a straight line through the origin with slope κϵ0A\kappa\epsilon_0 A. - To test the AA dependence, keep dd the same and use plates of different sizes. A graph of CC vs. AA should be a straight line through the origin with slope κϵ0/d\kappa\epsilon_0/d. - Graphing it this way (as a straight line) lets you find ϵ0\epsilon_0 or κ\kappa from the slope.

Electric Field Between the Plates

The field between parallel plates is uniform, ignoring edge effects:
E=Qκϵ0AE = \frac{Q}{\kappa\epsilon_0 A}
or, equivalently:
E=ΔVdE = \frac{\Delta V}{d}
where: - EE = field magnitude (N/C or V/m) - QQ = charge on each plate (C) - ΔV\Delta V = potential difference (V) - dd = plate separation (m)
Its properties: - It’s uniform, meaning constant magnitude and direction between the plates. - It points from the positive plate to the negative plate, perpendicular to the plates. - Edge effects are ignored. The field isn’t uniform near the edges, but that’s negligible when dd is much smaller than the plate dimensions.

Particle Motion Between the Plates

A charged particle in the uniform field between the plates feels:
F⃗=qE⃗=constant\vec{F} = q\vec{E} = \text{constant}a⃗=qE⃗m=constant\vec{a} = \frac{q\vec{E}}{m} = \text{constant}
This works just like projectile motion: - The uniform field acts like uniform gravity. - The acceleration is constant, like g=9.8g = 9.8 m/s² pointing down. - The kinematics equations apply.
Using the horizontal-launch analogy: - The direction parallel to the plates (“horizontal”) has uniform velocity. - The direction perpendicular to the plates (“vertical”) has constant acceleration from the field. - The path is parabolic.
For example, an electron that enters between the plates moving horizontally will deflect toward the positive plate along a parabolic path.

Energy Stored in a Capacitor

The stored electric potential energy is:
UC=12QΔVU_C = \frac{1}{2}Q\Delta V
Using C=Q/ΔVC = Q/\Delta V, you can rewrite it a couple of ways:
UC=12QΔV=12C(ΔV)2=Q22CU_C = \frac{1}{2}Q\Delta V = \frac{1}{2}C(\Delta V)^2 = \frac{Q^2}{2C}
All three forms are equivalent, so use whichever fits what you’re given.
where: - UCU_C = energy stored (J) - QQ = charge on the capacitor (C) - ΔV\Delta V = voltage across the capacitor (V) - CC = capacitance (F)
Physically, this is the work an external force does to separate the charges and store them on the plates.
Where the ½ comes from: as you charge a capacitor, ΔV\Delta V grows in step with QQ (since ΔV=Q/C\Delta V = Q/C). The first bit of charge gets moved across almost no voltage, and the last bit gets moved across the full ΔV\Delta V. On average, the charge was moved across half the final voltage, so the work is 12QΔV\frac{1}{2}Q\Delta V.
On a graph of ΔV\Delta V vs. QQ, that work is the area under the line, a triangle. For example, a 2.0 μF capacitor charged to 12 V holds Q=(2.0×10−6)(12)=24Q = (2.0 \times 10^{-6})(12) = 24 μC:
2.6 - Capacitor
UC=12QΔV=12(24×10−6)(12)=1.4×10−4 JU_C = \frac{1}{2}Q\Delta V = \frac{1}{2}(24 \times 10^{-6})(12) = 1.4 \times 10^{-4} \text{ J}
The slope of the line is ΔV/Q=1/C\Delta V/Q = 1/C. A bigger capacitor gives a flatter line, so it stores less energy for the same charge but more energy for the same voltage.

Dielectrics

A dielectric is an insulating material placed between the capacitor plates. Adding one does three things: 1. It increases the capacitance by a factor of κ\kappa, so C=κC0C = \kappa C_0. 2. If the charge on the plates stays the same (an isolated capacitor), it reduces the field by a factor of κ\kappa, so E=E0/κE = E_0/\kappa. 3. It sets up an opposing field inside the dielectric material.
Here’s how it works: - The dielectric polarizes, meaning charges separate inside its molecules. - This creates an internal field opposite to the external field. - The net field drops. - More charge can be stored for the same voltage, so the capacitance goes up.
2.6 - Capacitor
In the picture, the plates’ field E⃗0\vec{E}_0 points from +Q+Q to −Q-Q. In each molecule of the dielectric, the electrons shift slightly toward the positive plate. That leaves a layer of negative charge on the dielectric’s left face and positive charge on its right face. Those induced charges make a field E⃗induced\vec{E}_{\text{induced}} that points the opposite way, so the net field inside the dielectric is weaker than E⃗0\vec{E}_0 alone.
Some common dielectric constants: - Vacuum: κ=1.0\kappa = 1.0 (by definition) - Air: κ≈1.0006\kappa \approx 1.0006, which is basically 1 - Paper: κ≈3.5\kappa \approx 3.5 - Glass: κ≈5\kappa \approx 5 to 1010 - Water: κ≈80\kappa \approx 80
With a dielectric added at constant voltage (still hooked to a battery): - CC rises to κC0\kappa C_0. - QQ rises to κQ0\kappa Q_0, so more charge is stored. - EE stays at E0E_0, since E=ΔV/dE = \Delta V/d and neither ΔV\Delta V nor dd changes. The battery pushes extra charge onto the plates, which makes up for the dielectric’s opposing field. - UU rises to κU0\kappa U_0, so more energy is stored.
With a dielectric added at constant charge (isolated, no battery): - CC rises to κC0\kappa C_0. - QQ stays the same, since it’s isolated. - ΔV\Delta V falls to V0/κV_0/\kappa. - EE falls to E0/κE_0/\kappa. - UU falls to U0/κU_0/\kappa.

Practice Problem 1: Basic Capacitance

Problem: A parallel-plate capacitor has plates of area 0.050 m² separated by 2.0 mm of air. Find: (a) the capacitance (b) the charge stored when it’s connected to a 12 V battery
Solution:
  1. Using C=κϵ0AdC = \frac{\kappa\epsilon_0 A}{d} with κ=1\kappa = 1 for air:
      2. C=(1)(8.85×10−12)(0.050)0.002C = \frac{(1)(8.85 \times 10^{-12})(0.050)}{0.002}C=2.21×10−10 F=221 pFC = 2.21 \times 10^{-10} \text{ F} = 221 \text{ pF}
  1. Using C=Q/ΔVC = Q/\Delta V:
      2. Q=CΔV=(2.21×10−10)(12)=2.65×10−9 C=2.65 nCQ = C\Delta V = (2.21 \times 10^{-10})(12) = 2.65 \times 10^{-9} \text{ C} = 2.65 \text{ nC}
Answer: (a) 221 pF, (b) 2.65 nC.

Practice Problem 2: Electric Field and Force

Problem: A capacitor has plates 5.0 mm apart with a potential difference of 100 V. An electron is placed between the plates. (a) What is the field magnitude? (b) What force does the electron feel? (c) What is its acceleration?
Solution:
  1. Field:
      2. E=ΔVd=1000.005=2.0×104 N/CE = \frac{\Delta V}{d} = \frac{100}{0.005} = 2.0 \times 10^4 \text{ N/C}
  1. Force on the electron: Direction: toward the positive plate, which is opposite to the field.
      2. F=qE=(1.60×10−19)(2.0×104)=3.2×10−15 NF = qE = (1.60 \times 10^{-19})(2.0 \times 10^4) = 3.2 \times 10^{-15} \text{ N}
  1. Acceleration:
      2. a=Fm=3.2×10−159.11×10−31=3.5×1015 m/s2a = \frac{F}{m} = \frac{3.2 \times 10^{-15}}{9.11 \times 10^{-31}} = 3.5 \times 10^{15} \text{ m/s}^2
Answer: (a) 2.0×1042.0 \times 10^4 N/C, (b) 3.2×10−153.2 \times 10^{-15} N toward the positive plate, (c) 3.5×10153.5 \times 10^{15} m/s².

Practice Problem 3: Energy Storage

Problem: A 10 μF capacitor is charged to 50 V. Find: (a) the charge stored (b) the energy stored
Solution:
  1. Charge:
      2. Q=CΔV=(10×10−6)(50)=5.0×10−4 C=0.50 mCQ = C\Delta V = (10 \times 10^{-6})(50) = 5.0 \times 10^{-4} \text{ C} = 0.50 \text{ mC}
  1. Energy:
      2. UC=12C(ΔV)2=12(10×10−6)(50)2U_C = \frac{1}{2}C(\Delta V)^2 = \frac{1}{2}(10 \times 10^{-6})(50)^2UC=12(10×10−6)(2500)=0.0125 J=12.5 mJU_C = \frac{1}{2}(10 \times 10^{-6})(2500) = 0.0125 \text{ J} = 12.5 \text{ mJ}
Answer: (a) 0.50 mC, (b) 12.5 mJ.

Practice Problem 4: Dielectric Effect

Problem: A capacitor with air between its plates has C0=50C_0 = 50 pF and is charged to ΔV=100\Delta V = 100 V. A dielectric with κ=4.0\kappa = 4.0 is inserted while the capacitor stays connected to the battery. (a) What is the new capacitance? (b) What is the new charge? (c) What is the new energy stored?
Solution:
  1. New capacitance:
      2. C=κC0=4.0×50=200 pFC = \kappa C_0 = 4.0 \times 50 = 200 \text{ pF}
  1. The voltage stays at 100 V since it’s still on the battery: Compared to the original, Q0=(50×10−12)(100)=5Q_0 = (50 \times 10^{-12})(100) = 5 nC, so the charge went up by a factor of 4. ✓
      2. Q=CΔV=(200×10−12)(100)=2.0×10−8 C=20 nCQ = C\Delta V = (200 \times 10^{-12})(100) = 2.0 \times 10^{-8} \text{ C} = 20 \text{ nC}
  1. New energy: Original: U0=12(50×10−12)(100)2=0.25U_0 = \frac{1}{2}(50 \times 10^{-12})(100)^2 = 0.25 μJ, so the energy went up by a factor of 4. ✓
      2. U=12C(ΔV)2=12(200×10−12)(100)2=1.0×10−6 J=1.0 μJU = \frac{1}{2}C(\Delta V)^2 = \frac{1}{2}(200 \times 10^{-12})(100)^2 = 1.0 \times 10^{-6} \text{ J} = 1.0 \text{ μJ}
Answer: (a) 200 pF, (b) 20 nC, (c) 1.0 μJ.

Practice Problem 5: An Electron Between the Plates

Problem: An electron enters the space between two parallel plates moving horizontally at v0=2.0×107v_0 = 2.0 \times 10^7 m/s, right along the midline. The plates are 4.0 cm long and 1.0 cm apart, with a potential difference of 50 V (top plate positive). Ignore edge effects. Find: (a) the electron’s acceleration (b) how far it gets pushed sideways (yy) by the time it leaves the plates (c) whether it hits a plate
2.6 - Capacitor
Solution:
  1. Acceleration:
      2. E=ΔVd=500.010=5.0×103 N/CE = \frac{\Delta V}{d} = \frac{50}{0.010} = 5.0 \times 10^3 \text{ N/C}a=∣q∣Em=(1.60×10−19)(5.0×103)9.11×10−31=8.8×1014 m/s2a = \frac{|q|E}{m} = \frac{(1.60 \times 10^{-19})(5.0 \times 10^3)}{9.11 \times 10^{-31}} = 8.8 \times 10^{14} \text{ m/s}^2
      The field points down (from + to −), so the force on the negative electron points up, toward the positive plate.
  1. Sideways push: treat it like a horizontal launch. There’s no horizontal force, so the time between the plates is just length over speed:
      2. t=Lv0=0.0402.0×107=2.0×10−9 st = \frac{L}{v_0} = \frac{0.040}{2.0 \times 10^7} = 2.0 \times 10^{-9} \text{ s}y=12at2=12(8.8×1014)(2.0×10−9)2=1.8×10−3 m=1.8 mmy = \frac{1}{2}at^2 = \frac{1}{2}(8.8 \times 10^{14})(2.0 \times 10^{-9})^2 = 1.8 \times 10^{-3} \text{ m} = 1.8 \text{ mm}
  1. It starts 5.0 mm from the top plate (half the gap) and only moves 1.8 mm toward it, so it gets out without hitting.
Answer: (a) 8.8×10148.8 \times 10^{14} m/s² toward the positive (top) plate, (b) 1.8 mm, (c) no, it clears the plate.
Why ignore gravity? The electric force on the electron is ∣q∣E=8.0×10−16|q|E = 8.0 \times 10^{-16} N, while its weight is mg=8.9×10−30mg = 8.9 \times 10^{-30} N, about 101410^{14} times smaller.

Common Mistakes

  • Confusing Q and ΔV. Capacitance links charge to voltage through C=Q/ΔVC = Q/\Delta V, so don’t mix them up.
  • Forgetting the absolute value. The QQ in C=Q/ΔVC = Q/\Delta V is the magnitude on each plate. Both plates have the same magnitude, just opposite signs.
  • Using the wrong area. Use the area of one plate, not the combined area of both.
  • Unit slips. Capacitance often comes in μF, nF, or pF. Convert to farads: 1 μF = 10−610^{-6} F.
  • Energy formula confusion. There are three equivalent forms, so pick the one that matches what you know. Know QQ and ΔV\Delta V? Use U=12QΔVU = \frac{1}{2}Q\Delta V. Know CC and ΔV\Delta V? Use U=12C(ΔV)2U = \frac{1}{2}C(\Delta V)^2. Know QQ and CC? Use U=Q22CU = \frac{Q^2}{2C}.
  • Mixing up the dielectric scenarios. With a battery connected (constant ΔV\Delta V), CC, QQ, and UU all rise by a factor of κ\kappa. Isolated (constant QQ), CC rises while ΔV\Delta V and UU fall.
  • Assuming a dielectric always weakens the field. It only does that when the charge is fixed (isolated). With the battery connected, E=ΔV/dE = \Delta V/d can’t change, so the battery adds charge until the field is back to E0E_0.
  • Getting the field direction wrong. The field points from the + plate to the − plate, so high to low potential.
  • Forgetting that edge effects are ignored. Assume a uniform field unless the problem says otherwise.
  • Getting the acceleration direction wrong for a negative charge. Electrons accelerate toward the positive plate, opposite to the field.
  • Using the wrong distance in the field formula. Use the separation dd, not 2d2d or anything else.

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