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19. Capacitance

Syllabus
9702–2028–2029
Section
19
Level
A2

19.1 Capacitors and capacitance

Syllabus
9702–2028–2029
Topic
19.1
Level
A2

Capacitance is charge per potential for a conductor or capacitor

capacitanceC=Q/V,measuredinfarads(F),where1F=1CV−1capacitance C=Q/V, measured in farads (F), where 1 F=1 C V⁻¹

Arrangement Q means V means
isolated conducting sphere charge on the sphere potential of its surface relative to infinity
parallel-plate capacitor magnitude of charge on either one plate (+Q or -Q) potential difference between the plates

Applying a p.d. to separated conducting plates moves charge from one plate to the other, creating equal and opposite plate charges and storing energy in the electric field.

Capacitance depends on conductor geometry and the material between conductors. A larger isolated sphere has greater C; parallel-plate C changes with plate area, separation and dielectric.

Capacitance is not a fixed number of coulombs that a device can hold. For unchanged geometry and dielectric, increasing V increases Q proportionally but does not itself increase C.

Capacitance is C=Q/V, so stored charge rises linearly with voltage for a fixed capacitor

For a fixed capacitor, Q=CV. The slope of a Q–V graph is C.

Use farads, coulombs and volts; identify whether voltage is across the capacitor and account for dielectric or geometry changes before treating C as constant.

A 220 μF capacitor charged to 5.0 V stores 1.1×10⁻³ C.

The capacitor does not store a fixed amount of charge independent of voltage; C is the proportionality constant.

Derive capacitor combination formulas from charge and voltage constraints

For parallel capacitors, every capacitor has the same p.d. V, while charge supplied by the source divides: Q_T=Q_1+Q_2+... .

CTV=C1V+C2V+...ThereforeCT=C1+C2+...C_TV=C_1V+C_2V+... Therefore C_T=C_1+C_2+...

For initially uncharged capacitors in series, each capacitor acquires the same charge magnitude Q, while p.d.s add: V_T=V_1+V_2+... .

Q/CT=Q/C1+Q/C2+...Therefore1/CT=1/C1+1/C2+...Q/C_T=Q/C_1+Q/C_2+... Therefore 1/C_T=1/C_1+1/C_2+...

Fortwoinseries:CT=C1C2/(C1+C2)For two in series: C_T=C_1C_2/(C_1+C_2)

Connection Shared/conserved fact Bound check
parallel same V; charges add C_T is greater than the largest C
series same charge magnitude; p.d.s add C_T is less than the smallest C

The formulas are consequences, not resistor rules to swap by memory. A derivation must state the common charge or voltage and the corresponding additive total before substituting Q=CV.

Reduce capacitor networks, then recover each charge and p.d.

Connection Combined capacitance Charge relation p.d. relation
parallel C_T=ΣC Q_T=ΣQ_i V_T=V_1=V_2=...
series 1/C_T=Σ(1/C) Q_T=Q_1=Q_2=... V_T=ΣV_i
Step Action
1 identify one innermost series or parallel group
2 replace it by its equivalent capacitance
3 repeat to find C_T
4 use Q_T=C_TV_T, then expand backward using shared Q or V

A 20 μF and 10 μF pair in parallel gives 30 μF. In series with 45 μF, C_T=(1/30+1/45)⁻¹=18 μF.

Across a 12 V supply, that network stores series charge Q_T=(18 μF)(12 V)=216 μC. The 30 μF branch group has V=216 μC/30 μF=7.2 V, so its 20 μF and 10 μF capacitors carry 144 μC and 72 μC respectively.

Check topology at each step: equal charge applies only to elements in the same series chain, and equal p.d. only to branches across the same two nodes. Also check that series C_T is below the smallest member and parallel C_T above the largest.

19.2 Energy stored in a capacitor

Syllabus
9702–2028–2029
Topic
19.2
Level
A2

Stored capacitor energy is the area under a potential–charge graph

ForasmalladdedchargedQatpotentialV:dW=VdQThereforestoredenergyWistheareaunderagraphofV(verticalaxis)againstQ(horizontalaxis).For a small added charge dQ at potential V: dW=V dQ Therefore stored energy W is the area under a graph of V (vertical axis) against Q (horizontal axis).

For a fixed linear capacitor, V=Q/C, so the V–Q graph is a straight line from (0,0) to (Q,V). The area is a triangle.

W=area=(1/2)×base×height=(1/2)QVW=area=(1/2)×base×height=(1/2)QV

If the graph reaches V=8.0 V at Q=1.2×10⁻⁴ C, W=(1/2)(1.2×10⁻⁴)(8.0)=4.8×10⁻⁴ J.

For charging from (Q1,V1) to (Q2,V2) on a straight line, added energy is the trapezium area ΔW=(1/2)(V1+V2)(Q2-Q1). For a curved graph, estimate or integrate the area under the curve.

Check axes before taking area. On V against Q, area has units V C=J. The final product QV is a rectangle and is twice the triangular area when charging a linear capacitor from zero.

Capacitor energy is W=½QV=½CV²=Q²/(2C)

For a linear capacitor, stored energy W=½QV=½CV²=Q²/(2C). The half factor comes from the average voltage during charging.

Choose the form matching known Q, V or C and keep units in farads, volts and coulombs.

A 100 μF capacitor at 20 V stores 0.020 J.

Using QV without the half factor doubles the energy for a capacitor charged from zero.

19.3 Discharging a capacitor

Syllabus
9702–2028–2029
Topic
19.3
Level
A2

Voltage, charge and current magnitudes decay exponentially during capacitor discharge

Qfalls→V=Q/Cfalls→currentmagnitudeI=V/Rfalls→∣dQ/dt∣fallsSoequaltimeintervalsremoveequalfractions,notequalamounts.Q falls → V=Q/C falls → current magnitude I=V/R falls → |dQ/dt| falls So equal time intervals remove equal fractions, not equal amounts.

Quantity against time Initial value Shape and final behaviour
charge Q Q0 steepest fall initially; exponential curve asymptotic to 0
p.d. V V0=Q0/C same fractional exponential decay as Q; asymptotic to 0
current magnitude I I0=V0/R maximum initially; same fractional decay; asymptotic to 0

If current is defined positive in the charging direction, discharge current is negative and rises toward zero from -I0. If the graph shows current magnitude, it starts at +I0 and falls toward zero. State the convention.

Each decay graph has a negative gradient whose magnitude decreases with time. The initial tangent is steepest because V and I are greatest at t=0; the curve never reaches zero at a finite ideal time.

ln(Q/Q0)=ln(V/V0)=ln(I/I0)=−t/(RC):alog−ratioagainsttgraphisastraightlinethroughtheoriginwithgradient−1/(RC).ln(Q/Q0)=ln(V/V0)=ln(I/I0)=-t/(RC): a log-ratio against t graph is a straight line through the origin with gradient -1/(RC).

Do not sketch a straight-line discharge or a non-zero final plateau for an ideal RC circuit. Q and V are proportional, while I is the rate at which Q changes—not a constant.

The RC time constant τ=RC sets the timescale of capacitor charging or discharging

For a resistor R and capacitor C, τ=RC. After one time constant in a discharge, the relevant quantity has fallen to e⁻¹≈37% of its initial value.

Larger R or C makes the response slower. Identify the effective resistance seen by the capacitor, not every resistor in the diagram.

R=2.0 kΩ and C=100 μF gives τ=0.20 s.

τ is not the time to reach exactly zero; exponential decay approaches zero asymptotically.

Use exponential equations for capacitor discharge current, charge or voltage

Quantity Discharge equation Initial value
charge Q=Q0e^(-t/RC) Q0=CV0
p.d. V=V0e^(-t/RC) V0=Q0/C
current magnitude I=I0e^(-t/RC) I0=V0/R

At t=0, e^0=1 so x=x0. At t=RC, x/x0=e^-1=0.368. As t becomes large, x approaches zero. These limits quickly expose a wrong sign or charging formula.

x/x0=e(−t/RC)ln(x/x0)=−t/(RC)t=−RCln(x/x0)x/x0=e^(-t/RC) ln(x/x0)=-t/(RC) t=-RC ln(x/x0)

For τ=RC=3.6 s, the time for current to fall to 15% is t=-(3.6)ln(0.15)=6.83 s.

lnx=lnx0−t/(RC):agraphoflnxagainsttisstraight,withinterceptlnx0andgradient−1/(RC).HalvingRdoublesthegradientmagnitude.ln x = ln x0 - t/(RC): a graph of ln x against t is straight, with intercept ln x0 and gradient -1/(RC). Halving R doubles the gradient magnitude.

Use consistent time units so t/(RC) is dimensionless. These equations describe discharge; do not substitute the charging form x0(1-e^(-t/RC)). For signed current, attach the direction sign separately from the decaying magnitude.