Capacitance C = Q/V, Energy Stored ½CV², Time Constant RC & Exponential Decay Graphs
A capacitor consists of two conducting plates separated by an insulating dielectric material, storing equal and opposite charges (+Q and -Q) when connected across a potential difference V according to Capacitance C = Q / V (in Farads, F = C V⁻¹). Inserting a polar dielectric of relative permittivity ε_r between the plates aligns molecular dipoles to create an opposing internal electric field, increasing capacitance by C = Aε₀ε_r / d. Because p.d. V increases linearly from 0 to V₀ as charge Q accumulates on the plates, the electrical energy stored is given by the triangular area under the V-Q graph: E = ½QV = ½CV² = ½Q²/C - meaning 50% of the energy supplied by a constant-voltage battery (QV) is inevitably dissipated as heat in the circuit resistance. Examiners heavily test the RC time constant τ = RC (37% / 63% rules), discharging vs charging exponential graphs, and linear ln(V) vs t plots.
- Capacitance (C = Q / V) & Parallel-Plate Permittivity (C = Aε₀ε_r / d)
- Capacitance C is the charge stored per unit potential difference across the plates: C = Q / V (units: Farads, 1 F = 1 C V⁻¹, typically µF = 10⁻⁶ F, nF = 10⁻⁹ F, pF = 10⁻¹² F). For a parallel-plate capacitor of plate overlap area A and plate separation d filled with a dielectric of relative permittivity (dielectric constant) ε_r: C = A ε₀ ε_r / d, where ε₀ = 8.85 × 10⁻¹² F m⁻¹ is the permittivity of free space.
- How a Polar Dielectric Increases Capacitance
- A dielectric consists of insulating polar molecules. When an electric field E is applied between the charged capacitor plates, the polar dielectric molecules rotate and align with their positive ends (δ⁺) toward the negative plate and their negative ends (δ⁻) toward the positive plate. This creates an internal induced electric field opposing the applied field: 1. If connected to a constant-voltage battery (V constant): the opposing dipole field attracts more charge Q from the battery onto the plates, so C = Q / V increases by factor ε_r. 2. If isolated/disconnected from the battery (Q constant): the opposing field reduces the net electric field E and reduces the p.d. V = Ed across the plates, so C = Q / V again increases by factor ε_r (and stored energy E = ½Q²/C decreases because work is done pulling the dielectric in).
- Energy Stored in a Capacitor (E = ½QV = ½CV² = ½Q²/C) & 50% Heat Loss
- As charge Q builds up on a capacitor, each additional electron must be pushed onto the negative plate against the electrostatic repulsion of the charge already present, so V is directly proportional to Q (straight line through origin). The work done (energy stored) is the triangular area under the V-Q (or Q-V) graph: E = ½QV = ½CV² = ½Q² / C. Key exam fact: When a battery of EMF V₀ transfers charge Q to charge a capacitor from 0 to V₀, the total energy supplied by the battery is W_battery = Q V₀, whereas the capacitor only stores E = ½Q V₀; the remaining 50% (½Q V₀) is wasted as thermal energy in the resistance of the wires/resistor.
Electric Field Simulator & Interactive NPN Transistor Builder
Adjust charges and distances to visualize Coulomb’s Law field lines, or switch to the Transistor Lab to wire and test an NPN circuit.
1. Coulomb’s Law Point-Charge Calculation
Formula: F = k · |Q₁ · Q₂| / r² where k = 1/(4πε₀) = 8.99 × 10⁹ N·m²·C⁻²
3.60 N
Attractive (towards each other)7.19e+6 N/C
At r/2 = 0.10 m2. Uniform Parallel-Plate Field Calculator
Formula: E = V / d (units: V m⁻¹ or N C⁻¹)
Key Concepts & Exam Definitions (6)
- Capacitance (C = Q / V) & Parallel-Plate Permittivity (C = Aε₀ε_r / d)
- Capacitance C is the charge stored per unit potential difference across the plates: C = Q / V (units: Farads, 1 F = 1 C V⁻¹, typically µF = 10⁻⁶ F, nF = 10⁻⁹ F, pF = 10⁻¹² F). For a parallel-plate capacitor of plate overlap area A and plate separation d filled with a dielectric of relative permittivity (dielectric constant) ε_r: C = A ε₀ ε_r / d, where ε₀ = 8.85 × 10⁻¹² F m⁻¹ is the permittivity of free space.
- How a Polar Dielectric Increases Capacitance
- A dielectric consists of insulating polar molecules. When an electric field E is applied between the charged capacitor plates, the polar dielectric molecules rotate and align with their positive ends (δ⁺) toward the negative plate and their negative ends (δ⁻) toward the positive plate. This creates an internal induced electric field opposing the applied field: 1. If connected to a constant-voltage battery (V constant): the opposing dipole field attracts more charge Q from the battery onto the plates, so C = Q / V increases by factor ε_r. 2. If isolated/disconnected from the battery (Q constant): the opposing field reduces the net electric field E and reduces the p.d. V = Ed across the plates, so C = Q / V again increases by factor ε_r (and stored energy E = ½Q²/C decreases because work is done pulling the dielectric in).
- Energy Stored in a Capacitor (E = ½QV = ½CV² = ½Q²/C) & 50% Heat Loss
- As charge Q builds up on a capacitor, each additional electron must be pushed onto the negative plate against the electrostatic repulsion of the charge already present, so V is directly proportional to Q (straight line through origin). The work done (energy stored) is the triangular area under the V-Q (or Q-V) graph: E = ½QV = ½CV² = ½Q² / C. Key exam fact: When a battery of EMF V₀ transfers charge Q to charge a capacitor from 0 to V₀, the total energy supplied by the battery is W_battery = Q V₀, whereas the capacitor only stores E = ½Q V₀; the remaining 50% (½Q V₀) is wasted as thermal energy in the resistance of the wires/resistor.
- Capacitor Discharging Equations (Q, V, and I All Decay Exponentially)
- When a capacitor charged to Q₀ and V₀ discharges through a fixed resistor R, charge Q, potential difference V, and current I all decrease exponentially from their initial maximum values toward zero: Q = Q₀ e^(-t / RC), V = V₀ e^(-t / RC), and I = I₀ e^(-t / RC) (where initial current I₀ = V₀ / R). Taking natural logarithms linearises the equation to y = mx + c: ln(V) = (-1 / RC) t + ln(V₀), so a graph of ln(V) (or ln(Q) or ln(I)) against time t is a straight line with a negative gradient equal to -1 / RC.
- The Time Constant (τ = RC) & Half-Life (T₁/₂ = RC ln(2) = 0.693 RC)
- The time constant τ = RC (units: Ω × F = seconds, s) is the time taken (at t = RC) for the charge Q, p.d. V, or current I of a discharging capacitor to fall to e⁻¹ = 1/e ≈ 0.368 (37%) of its initial value (or to fall BY 63%). During charging, t = RC is the time taken for Q and V to rise to (1 - e⁻¹) ≈ 0.632 (63%) of their final maximum value. The half-life (time to fall to 50%, ½Q₀) is T₁/₂ = RC ln(2) ≈ 0.693 RC (after ~5RC, a capacitor is >99% charged or discharged).
- Capacitor Charging Curves (Q and V Rise vs Current I Decays)
- When an uncharged capacitor is connected in series with a resistor R and battery of EMF V₀: 1. Current I starts at a maximum I₀ = V₀ / R (since V_C = 0 initially) and decays exponentially to zero as the capacitor charges and opposes the battery: I = I₀ e^(-t / RC) (identical shape to the discharging current curve, though flowing in the opposite direction). 2. Charge Q and capacitor p.d. V_C start at zero and rise exponentially toward Q₀ = CV₀ and V₀: Q = Q₀(1 - e^(-t / RC)) and V_C = V₀(1 - e^(-t / RC)) (while resistor p.d. V_R = V₀ - V_C = V₀ e^(-t / RC) decays to zero).
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
Explain how inserting a polar dielectric material between the plates of a capacitor increases its capacitance.
Why does a capacitor store energy E = ½QV rather than E = QV, and what happens to the other ½QV supplied by the battery during charging?
Define the time constant (τ = RC) of a discharging capacitor circuit, and state its value as a percentage for both discharging and charging.
Contrast the shapes of the Voltage-Time (V-t) and Current-Time (I-t) graphs during capacitor charging versus discharging.
On a straight-line graph of ln(V) (y-axis) plotted against time t (x-axis) for a discharging capacitor, what do the y-intercept and gradient represent?
How do you discharge a capacitor at a constant current I instead of an exponential decay?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
A parallel-plate capacitor consists of two square metal plates of side length 0.150 m separated by a sheet of mica of thickness 0.080 mm and relative permittivity ε_r = 6.0 (ε₀ = 8.85 × 10⁻¹² F m⁻¹). The capacitor is charged using a 120 V DC supply and then disconnected from the supply. Calculate: (a) the capacitance of the capacitor, (b) the charge and energy stored on the capacitor, and (c) the new energy stored if the mica sheet is pulled out from between the plates while the capacitor remains isolated. (6 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Calculate plate area A = (0.150 m)² = 0.0225 m², and substitute into C = A ε₀ ε_r ÷ d: C = (0.0225 × 8.85 × 10⁻¹² × 6.0) ÷ (0.080 × 10⁻³ m) [1 mark]. 2. C = (1.195 × 10⁻¹²) ÷ (8.0 × 10⁻⁵) = 1.49 × 10⁻⁸ F (14.9 nF) [1 mark]. 3. (b) Charge stored Q = CV = (1.493 × 10⁻⁸ F) × 120 V = 1.79 × 10⁻⁶ C (1.79 µC) [1 mark]. 4. Energy stored E₁ = ½QV = 0.5 × (1.792 × 10⁻⁶ C) × 120 V = 1.08 × 10⁻⁴ J [1 mark]. 5. (c) Because the capacitor is disconnected (isolated) from the supply, the charge Q remains constant at 1.79 × 10⁻⁶ C while removing the dielectric (ε_r = 6.0 → 1.0) reduces the capacitance by a factor of 6 to C₂ = C ÷ 6 = 2.49 × 10⁻⁹ F [1 mark]. 6. Using E = ½Q² ÷ C (with Q constant), dividing C by 6 multiplies the stored energy by 6: E₂ = 6 × E₁ = 6 × 1.075 × 10⁻⁴ = 6.45 × 10⁻⁴ J (the extra 5.38 × 10⁻⁴ J came from the mechanical work done pulling the dielectric out against electrostatic attraction) [1 mark].
- Q2
A 470 µF capacitor is charged to a potential difference of 12.0 V and then discharged through a fixed resistor R. After 15.0 s, the potential difference across the capacitor has fallen to 3.20 V. Calculate: (a) the time constant of the circuit, (b) the resistance R of the resistor, (c) the time taken for the capacitor voltage to fall from 12.0 V to 6.0 V (half-life), and (d) the percentage of the initial stored energy remaining in the capacitor at t = 15.0 s. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Substitute V = 3.20 V, V₀ = 12.0 V, and t = 15.0 s into V = V₀ e^(-t / τ): 3.20 = 12.0 e^(-15.0 / τ) → e^(-15.0 / τ) = 3.20 ÷ 12.0 = 0.2667 [1 mark]. 2. Take natural logs: -15.0 ÷ τ = ln(0.2667) = -1.3218 → τ = 15.0 ÷ 1.3218 = 11.3 s (11.35 s) [1 mark]. 3. (b) From τ = RC, calculate resistance R = τ ÷ C = 11.35 s ÷ (470 × 10⁻⁶ F) = 2.41 × 10⁴ Ω (24.1 kΩ) [1 mark]. 4. (c) Half-life T₁/₂ = τ ln(2) = 11.35 × 0.6931 = 7.87 s [1 mark]. 5. (d) Since stored energy E = ½CV² is proportional to V², the fraction of energy remaining at t = 15.0 s is (V ÷ V₀)² = (3.20 ÷ 12.0)² = (0.2667)² = 0.0711 = 7.11% [1 mark].
- Q3
An uncharged 220 µF capacitor is connected in series with a 50.0 kΩ resistor, a switch, and a 9.00 V battery of negligible internal resistance. When the switch is closed at t = 0, calculate the initial current I₀, the current and potential difference across the capacitor at t = 8.00 s, and the charge on the capacitor at t = 8.00 s. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. At t = 0, the uncharged capacitor has V_C = 0 V, so the entire 9.00 V is across the 50.0 kΩ resistor: initial current I₀ = V₀ ÷ R = 9.00 V ÷ (50.0 × 10³ Ω) = 1.80 × 10⁻⁴ A (180 µA) [1 mark]. 2. Calculate the time constant τ = RC = (50.0 × 10³ Ω) × (220 × 10⁻⁶ F) = 11.0 s [1 mark]. 3. At t = 8.00 s, calculate current using I = I₀ e^(-t / RC): I = (1.80 × 10⁻⁴) × e^(-8.00 / 11.0) = (1.80 × 10⁻⁴) × 0.4832 = 8.70 × 10⁻⁵ A (87.0 µA) [1 mark]. 4. Calculate p.d. across the charging capacitor at t = 8.00 s using V_C = V₀(1 - e^(-t / RC)) (or V_C = V₀ - IR): V_C = 9.00 × (1 - 0.4832) = 4.65 V [1 mark]. 5. Calculate charge on the capacitor at t = 8.00 s: Q = C V_C = (220 × 10⁻⁶ F) × 4.651 V = 1.02 × 10⁻³ C (1.02 mC) [1 mark].
- Q4
Explain why the energy stored in a discharging capacitor decays with a time constant equal to ½RC rather than RC. (3 marks)
Click “Check Mark Scheme” to compare your answer, 1. During discharging, the potential difference across the capacitor decays according to V = V₀ e^(-t / RC) [1 mark]. 2. Substitute V into the energy equation E = ½CV²: E = ½C (V₀ e^(-t / RC))² = (½CV₀²) e^(-2t / RC) = E₀ e^(-t / (½RC)) [1 mark]. 3. Because the exponent contains -2t / RC = -t / (½RC), the stored energy falls to 1/e (37%) of its initial value E₀ in half the time constant, t = ½RC (and at t = RC, the energy has fallen to (1/e)² = 1/e² ≈ 13.5% of E₀) [1 mark].
Common Exam Questions (4)
What happens if the dielectric in a capacitor is connected to a constant-voltage battery when the dielectric is pulled out, versus when the capacitor is isolated (disconnected)?
Removing the dielectric always reduces capacitance C = Aε₀ε_r / d by factor ε_r: 1. If still connected to the battery, V stays constant, so charge Q = CV decreases (charge flows back into the battery) and energy E = ½CV² decreases. 2. If disconnected (isolated), charge Q stays constant, so V = Q / C increases and energy E = ½Q² / C increases.
In a charging capacitor circuit (battery V₀ in series with resistor R and capacitor C), how are V_R and V_C related at any instant t?
By Kirchhoff's Second Law around the loop, the sum of the p.d. across the resistor and the p.d. across the capacitor always equals the battery EMF: V_R + V_C = V₀. Therefore, if you know the current I at time t, you can immediately find V_R = IR and V_C = V₀ - IR without even needing the exponential charging formula.
Why are capacitors used in camera flashes, defibrillators, and AC-to-DC smoothing circuits?
Unlike chemical batteries (which have internal resistance and slow electrochemical reaction rates), capacitors have almost zero internal resistance and can release all their stored energy (½CV²) in milliseconds (small RC) to deliver huge peak power (P = E / t) for a camera flash or heart defibrillator. In rectifier circuits, a capacitor with a large time constant (RC ≫ period of AC ripple) charges at voltage peaks and discharges slowly through the load between peaks to smooth pulsating DC into steady DC.
How do I choose between E = ½QV, E = ½CV², and E = ½Q²/C in exam questions?
Pick the formula that contains the two quantities given in the question (or the quantity that stays constant): if V is kept constant (connected to battery), use E = ½CV²; if Q is kept constant (isolated capacitor), use E = ½Q²/C; if reading from a V-Q graph, use E = ½QV.
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