Capacitors: theory | Physics Circle

CONCEPT

Capacitance of a Conductor

In one line
Charge given to a conductor raises its potential in step. The ratio C = Q ÷ V depends only on the size and shape of the conductor.

CAPACITANCE

8.1.1 What capacitance means

Q ∝ V ⇒ Q = CV

C = QV

C is the ability of a conductor to hold charge.
It does not depend on the charge given or the potential raised.
It depends on shape, size and the medium around the conductor.

UNITS

8.1.2 Units and dimensions

1 farad = 1 coulomb1 volt

Unit	Value
1 mF	10⁻³ F
1 μF	10⁻⁶ F
1 nF	10⁻⁹ F
1 pF (μμF)	10⁻¹² F
1 F	9 × 10¹¹ stat farad

[C] = [M⁻¹ L⁻² T⁴ A²]

Q–V GRAPH

8.1.3 Reading a Q–V graph

V against Q is a straight line through the origin

Graph of V against Q: slope = 1 ÷ C. Steeper line means smaller C.
Graph of Q against V: slope = C.
The area under the V–Q line is the energy stored.

ISOLATED SPHERE

8.1.4 Capacitance of an isolated sphere

Charge Q spreads over the surface of radius R

V = 14πε₀ QR

C = 4πε₀R = R9 × 10⁹

C ∝ R: a bigger sphere holds more charge at the same potential.
Solid or hollow, the capacitance is the same: the charge sits on the surface.

EARTH

8.1.5 Capacitance of the earth

Earth as a sphere of radius 6400 km: C = 711 μF.
In practice the earth is taken as infinite capacitance at zero potential.

From volume V: R = (3V4π)^1⁄3 · from area A: R = √(A4π)

ENERGY

8.1.6 Energy of a charged conductor

U = ½ QV = ½ CV² = Q²2C

sphere: U = Q²8πε₀R

SPHERES AT ONE POTENTIAL

8.1.7 Two spheres at the same potential

Q = CV ∝ R, so the bigger sphere carries more charge.
Surface density σ = Q4πR² ∝ 1R: the smaller sphere has the larger σ.
Field at the surface E = σ ÷ ε₀, so E ∝ 1 ÷ R too.

σ₁σ₂ = R₂R₁ · Q₁Q₂ = R₁R₂

JEE · NEET

Corner

Combining n identical drops

n drops, each of radius r, charge q, capacitance c, potential v and energy u, merge into one big drop.

Quantity	Big drop
Charge	Q = nq
Radius	R = n^1⁄3 r
Capacitance	C = n^1⁄3 c
Potential	V = n^2⁄3 v
Energy	U = n^5⁄3 u

The big drop has more energy than all small drops together.

Spherical capacitor

Inner sphere +Q, outer sphere earthed

C = 4πε₀ abb − a

With a dielectric K between the spheres: C = 4πε₀K abb − a
If the outer sphere is charged and the inner one earthed: C = 4πε₀ b²b − a
An isolated sphere is a spherical capacitor with b = ∞.
Large radii and a small gap: it behaves like a parallel plate capacitor.

Cylindrical capacitor

Two coaxial cylinders of length l

C = 2πε₀lln(b ÷ a)

Practise PYQs on this topicCapacitance of a Conductor · 14 questions with solutions

CONCEPT

Parallel Plate Capacitor

In one line
Two plates carry +Q and −Q. The field between them is uniform, E = σ ÷ ε₀, and the capacitance is C = ε₀A ÷ d.

CAPACITOR

8.2.1 What a capacitor is

Two conductors close to each other carrying equal and opposite charge.
The charge Q of a capacitor means the charge on its positive plate.
Net charge on a capacitor is always zero.
A battery charges it until the plate voltage equals the battery voltage.
A capacitor actually stores energy in the field between the plates.

FIELD AND VOLTAGE

8.2.2 Field and voltage between the plates

Uniform field from the + plate to the − plate

E = σε₀ = QAε₀

V = Ed = σdε₀

E does not depend on the separation d.
Outside the plates the field is zero.
The plates are kept close to avoid fringing (bending of field at the edges).

CAPACITANCE

8.2.3 Capacitance of parallel plates

C = ε₀Ad

C ∝ A and C ∝ 1 ÷ d.
It does not depend on the charge, the voltage, the metal or the plate thickness.
For unequal plates use the overlapping area.
In CGS: C = A ÷ 4πd.

CHANGING A AND d

8.2.4 Changing area and separation

Change	New C
d halved	2C
d doubled, A halved	C4
radius × n, d same	n²C
radius × 2, d × 2	2C

Circular plates: A = πr², so C ∝ r².
Plates moved apart at speed v: dC ÷ dt ∝ 1 ÷ d².

UNEQUAL CHARGES

8.2.5 Plates given unequal charges

Charges settle on the four faces

V = Q₁ − Q₂2C

Inner faces carry ±(Q₁ − Q₂) ÷ 2; only these make the field between the plates.
Outer faces carry (Q₁ + Q₂) ÷ 2 each.
Plates with +2 μC and +4 μC on 1 μF: V = 2 ÷ 2 = 1 V.

CHARGE BETWEEN THE PLATES

8.2.6 A charge between the plates

F = qE = qV ÷ d

The force on q is the same everywhere between the plates.
One plate removed: the field becomes σ ÷ 2ε₀, so the force halves.
Flux through a closed surface around the + plate = Q ÷ ε₀ = CV ÷ ε₀.
An electron entering at right angles to E follows a parabola, like a projectile.
A charged pendulum between vertical plates hangs at tan θ = qE ÷ mg.

Practise PYQs on this topicParallel Plate Capacitor · 18 questions with solutions

CONCEPT

Energy Stored & Force between Plates

In one line
Work done in charging is stored as field energy, U = ½CV². The battery gives QV, so half its work is lost as heat.

ENERGY STORED

8.3.1 Energy stored in a capacitor

U = ½ CV² = ½ QV = Q²2C

The area under the V–Q graph.
Same C: U ∝ V² and U ∝ Q².
Charge increased by x %: energy increases by about 2x % for small x.

PERCENT QUESTIONS

8.3.2 Charge raised, energy raised

U′U = (Q + ΔQQ)²

Energy rises by 44%: ratio 1.44, so Q + 2 = 1.2Q and Q = 10 C.
Energy rises by 21%: ratio 1.21, so (Q + 2) ÷ Q = 1.1 and Q = 20 C.
Voltage from 5 V to 10 V needs W; 10 V to 15 V needs W × (15² − 10²) ÷ (10² − 5²) = 1.67 W.
Rate of storing energy: dU ÷ dt = CV (dV ÷ dt).

ENERGY IN TERMS OF E

8.3.3 Energy in terms of the field

U = ½ ε₀E² × Ad

Ad is the volume between the plates.

ENERGY DENSITY

8.3.4 Energy density

u = Uvolume = ½ ε₀E²

u = σ²2ε₀ = Q²2ε₀A²

Unit: J m⁻³.
With a dielectric: u = ½ ε₀KE².

FORCE BETWEEN PLATES

8.3.5 Force between the plates

F = QE2 = Q²2ε₀A = CV²2d

Each plate feels the field of the other plate only, E ÷ 2. So F = QE ÷ 2, not QE.
For an isolated capacitor (Q fixed) the force does not depend on d.
Force per unit area = u = σ² ÷ 2ε₀ (electrostatic pressure).

BATTERY WORK AND HEAT

8.3.6 Work done by the battery

W_battery = QV = CV²

energy storedwork by battery = 12

The other half, ½CV², is lost as heat in the wires, whatever the resistance.
Discharging through a resistor: all of ½CV² becomes heat.
Heat warms a block: ½CV² = mcΔT.

PULLING THE PLATES APART

8.3.7 Work to pull the plates apart

Quantity	Battery removed (Q fixed)	Battery connected (V fixed)
C	falls	falls
V	rises	same
Q	same	falls
E	same	falls
U	rises	falls

Q fixed, d → nd: W = U′ − U = Q²2C(n − 1)

Battery removed: you do work against the attraction, so U rises.
Example: separation × 4 needs W = 3 × Q²2C = 3ε₀AV²2d

Practise PYQs on this topicEnergy Stored & Force between Plates · 41 questions with solutions

CONCEPT

Dielectrics: Full Filling, Battery On or Off

In one line
A dielectric weakens the field between the plates, so the same charge needs less voltage. Filled fully, C becomes KC.

DIELECTRICS

8.4.1 What a dielectric is

An insulator that transmits electric effect without conducting.
Polar molecules have a permanent dipole moment: water, alcohol, CO, NH₃, HCl.
Non-polar molecules get a dipole only in a field: N₂, O₂, benzene, methane.

Polar molecules line up in a field

POLARISATION

8.4.2 Polarisation of a slab

Bound charges on the faces make a field Eᵢ opposite to E₀

E = E₀ − Eᵢ = E₀K

K = field with airfield with medium = ε_r = εε₀

K is the relative permittivity (also called SIC).
K = 1 for vacuum and K = ∞ for a metal.
Function of a dielectric: it reduces the effective potential on the plates, so C rises.

induced charge Q_i = Q(1 − 1K)

FULL FILLING

8.4.3 Gap filled completely

C′ = Kε₀Ad = KC

Permittivity of the medium: ε = Kε₀ = (C′ ÷ C) ε₀.
Oil of K removed: C becomes C ÷ K.
Separation × 2 and K filled: C′ = KC ÷ 2.
Two charged plates dipped in kerosene (isolated): E falls to E ÷ K.

BATTERY ON OR OFF

8.4.4 Battery removed or connected

Q is fixed in one case, V in the other

Quantity	Battery removed	Battery connected
C	KC	KC
Q	Q	KQ
V	VK	V
E	EK	E
U	UK	KU

Battery removed and slab put in: U falls; the field does work on the slab and pulls it in.
Work done on the slab = U − U ÷ K = ½CV²(1 − 1K).
Battery connected: the battery sends (K − 1)CV more charge.
Slab put in and taken out with the battery connected: everything returns to the start.

DIELECTRIC STRENGTH

8.4.5 Dielectric strength

Too strong a field makes the dielectric conduct: dielectric breakdown.
The largest field it can stand is its dielectric strength. SI unit V m⁻¹; practical unit kV mm⁻¹.
Air ionises above about 3 × 10⁶ N C⁻¹.

Q_max = Kε₀A × E_max · V_max = E_maxd

IN A COMBINATION

8.4.6 Dielectric in one capacitor of a pair

Series pair C and C on V, one filled with K: Ceq changes from C ÷ 2 to KC ÷ (K + 1).
The other capacitor now takes the larger share of V: V₂ = KV ÷ (K + 1).
Parallel pair, one filled: Ceq = C + KC; change = (K − 1)C.
Parallel 2C and C, battery removed, C filled with K: common V = 3V ÷ (K + 2).

Practise PYQs on this topicDielectrics · 114 questions with solutions

CONCEPT

Partial Slabs & Compound Dielectrics

In one line
A slab that fills only part of the gap acts as capacitors joined together. Layers along the gap are in series; strips side by side are in parallel.

PARTIAL SLAB

8.5.1 Slab of thickness t

Slab of thickness t and constant K in a gap d

C′ = ε₀Ad − t + (t ÷ K)

The slab adds as if the air gap became shorter by t(1 − 1 ÷ K).
The slab can be anywhere in the gap; C is the same.
t = d gives KC.

METAL SHEET

8.5.2 Metal sheet in the gap

A metal sheet removes its thickness from the gap

C′ = ε₀Ad − t

Metal: K = ∞, so t ÷ K = 0.
t = d ÷ 2: C becomes 2C.
A foil of negligible thickness: C does not change.
Sheet filling the whole gap (or plates joined by a wire): C becomes infinite.

KEEPING V THE SAME

8.5.3 Extra separation to keep V or C

d′ = t(1 − 1K) ⇒ K = tt − d′

A 2 mm slab needs 1.6 mm extra gap: K = 2 ÷ 0.4 = 5.

LAYERS IN SERIES

8.5.4 Several slabs along the gap

Each layer is a capacitor; the layers are in series

C′ = ε₀A(t₁ ÷ K₁) + (t₂ ÷ K₂) + (t₃ ÷ K₃)

Two equal layers d ÷ 2 each: C′ = 2K₁K₂K₁ + K₂ C.
Some air left over: add (d − t₁ − t₂ …) to the bottom.
The field is weakest in the slab with the largest K.

STRIPS IN PARALLEL

8.5.5 Slabs side by side

Each strip spans the full gap; the strips are in parallel

C′ = ε₀d(K₁A₁ + K₂A₂)

Half the area air, half K: C′ = K + 12 C.
Equal halves K₁ and K₂: C′ = K₁ + K₂2 C.

MIXED ARRANGEMENTS

8.5.6 Mixed arrangements

Find which blocks are in series (one behind the other) and which are side by side.
First combine the series pairs, then add the parallel strips.
Half the area with K₁ over the full gap, the other half with K₂ and K₃ in layers: C = (ε₀A ÷ 2d) × [K₁ + 2K₂K₃ ÷ (K₂ + K₃)].

JEE · NEET

Corner

Dielectric constant that changes across the gap

Cut the gap into thin layers dx in series.
Each layer has dC = ε₀K(x)A ÷ dx; add the reciprocals.

1C = ∫ dxε₀K(x)A

K varying across the plate width instead: the strips are in parallel, so add the capacitances.

Force pulling a slab in

A slab partly inside a charged capacitor is pulled further in by the fringing field.

F = ε₀b(K − 1)V²2d

b is the width of the plates. With the battery connected this force is constant.

Practise PYQs on this topicDielectrics · 114 questions with solutions

CONCEPT

Series & Parallel Combination

In one line
In series every capacitor carries the same charge and the voltages add. In parallel every capacitor has the same voltage and the charges add.

SERIES

8.6.1 Series combination

Same charge Q on every capacitor

1C_s = 1C₁ + 1C₂ + 1C₃

two capacitors: C_s = C₁C₂C₁ + C₂

C_s is less than the smallest capacitor.
V = V₁ + V₂ + V₃, and V ∝ 1 ÷ C: the smallest C takes the biggest voltage.
Energy also divides as U ∝ 1 ÷ C.
n identical capacitors: C_s = C ÷ n, each takes V ÷ n.
Breakdown voltage V each, n in series: the combination can stand nV.

V₁ = C₂C₁ + C₂ V · V₂ = C₁C₁ + C₂ V

PARALLEL

8.6.2 Parallel combination

Same V on every capacitor

C_p = C₁ + C₂ + C₃

C_p is more than the largest capacitor.
Q = Q₁ + Q₂ + Q₃, and Q ∝ C; energy also U ∝ C.
n identical capacitors: C_p = nC.
Plates of area A at gaps d, 2d, 3d in parallel: C_p = (ε₀A ÷ d)(1 + ½ + ⅓).

Q₁ = C₁C₁ + C₂ Q · Q₂ = C₂C₁ + C₂ Q

FINDING TWO CAPACITORS

8.6.3 From the series and parallel values

C₁ + C₂ = C_p · C₁C₂ = C_sC_p

C_s = 3 μF and C_p = 16 μF: C₁C₂ = 48 and C₁ + C₂ = 16, so 12 μF and 4 μF.
Two equal capacitors: C_s : C_p = 1 : 4.

MAKING A TARGET VALUE

8.6.4 Building a required value

Seven 10 μF capacitors for 10 ÷ 11 μF: put 4 in parallel (40 μF) and 3 in series with them.
n identical in parallel, charged to V, then rejoined in series: the total voltage becomes nV, the energy stays the same.
Three 4 μF to make 6 μF: two in series, one in parallel with them.

ENERGY IN COMBINATIONS

8.6.5 Energy in series and parallel

U = ½ C_eqV²

Same supply: U_s : U_p = C_s : C_p = 1 : n² for n identical capacitors.
Equal energy: V_sV_p = √(C_pC_s) = n for n identical capacitors.
Work done by the battery in charging a combination = C_eqV².

Practise PYQs on this topicCombination of Capacitors · 95 questions with solutions

CONCEPT

Capacitor Networks

In one line
Redraw the network so that the two terminals sit at the ends. Then reduce it step by step, or use symmetry and the bridge rule.

STEP BY STEP

8.7.1 Reducing a network

Mark every junction with a letter. Points joined by a plain wire get the same letter.
Combine pairs that are clearly in series or in parallel, then redraw.
A capacitor whose both ends are at the same point carries no charge: remove it.
Capacitors are opposite to resistors: series uses reciprocals, parallel adds.

FINDING Q AND V

8.7.2 Charge and voltage on one capacitor

Find C_eq and the total charge Q = C_eqV.
Go back through the network: series parts share the same Q, parallel parts share the same V.
Potential of a point: start from a known point and subtract Q ÷ C across each capacitor.

WHEATSTONE BRIDGE

8.7.3 Bridge of capacitors

C₅ carries no charge when the bridge is balanced

C₁C₂ = C₃C₄

Balanced: remove C₅ and solve two series arms in parallel.
Identical capacitors in a symmetric diamond are always balanced.
Not balanced: use the Kirchhoff method below.

KIRCHHOFF FOR CAPACITORS

8.7.4 Kirchhoff method

Junction rule: the total charge on the plates joined to an isolated junction stays zero (or its initial value).
Loop rule: ΣV = 0 around any loop.
Across a capacitor from + plate to − plate the potential falls by q ÷ C.
Across a battery from − to + the potential rises by E.

MULTI-PLATE

8.7.5 Many plates in a stack

n plates in a row with the ends joined to the source: (n − 1) capacitors in series, C_eq = ε₀A(n − 1)d

Alternate plates joined: odd plates to A, even plates to B

alternate plates joined: C_eq = (n − 1)ε₀Ad

Each inner plate serves two neighbouring capacitors.
Four plates with plates 1 and 3 joined: work out each gap as a capacitor and see which pairs share the same two points.

INFINITE AND REPEATED

8.7.6 Repeated sections

An infinite ladder: adding one more section does not change C_eq; write the equation and solve the quadratic.
Unknown C from a given C_eq: write C_eq in terms of C, then equate.

Practise PYQs on this topicCombination of Capacitors · 95 questions with solutions

CONCEPT

Charge Sharing & Loss of Energy

In one line
Join two charged conductors and charge flows until they reach a common potential. Charge is conserved, but some energy is always lost as heat.

COMMON POTENTIAL

8.8.1 Common potential

Plates of the same sign joined together

V = total chargetotal capacitance = C₁V₁ + C₂V₂C₁ + C₂

Charge flows from the higher potential to the lower one.
Flow stops when both are at the same potential.
Charged C₁ at V₀ joined to uncharged C₂: V = C₁V₀ ÷ (C₁ + C₂).
An identical uncharged capacitor halves the voltage.

NEW CHARGES

8.8.2 Charge on each after sharing

Q₁′ = C₁V · Q₂′ = C₂V

Q₁′Q₂′ = C₁C₂

Find the unknown C: C₂ = C₁(V₀ − V) ÷ V.
10 μF at 50 V joined to C, final 20 V: C = 10 × 30 ÷ 20 = 15 μF.

TWO SPHERES

8.8.3 Two charged spheres joined

Charges after joining are in the ratio of the radii

Q₁′ = Q r₁r₁ + r₂ · Q₂′ = Q r₂r₁ + r₂

Q is the total charge Q₁ + Q₂.
Two identical spheres touched: each gets half the total.

ENERGY LOSS

8.8.4 Loss of energy

ΔU = C₁C₂2(C₁ + C₂)(V₁ − V₂)²

The loss appears as heat (and a spark) in the joining wire.
Identical capacitor, one uncharged: half the energy is lost; energy left = U ÷ 2.
Loss is zero only if V₁ = V₂.
% lost when C₁ charged is joined to uncharged C₂ = C₂ ÷ (C₁ + C₂) × 100.

REVERSE POLARITY

8.8.5 Positive plate joined to negative plate

V = C₁V₁ − C₂V₂C₁ + C₂

ΔU = C₁C₂2(C₁ + C₂)(V₁ + V₂)²

Negative ends joined first, then positive ends: the same sign is joined, use the + formula.
Final potential can be zero when C₁V₁ = C₂V₂.
C at V and 2C at 2V joined + to −: V′ = (CV − 4CV) ÷ 3C = −V, size V.

Practise PYQs on this topicCharge Sharing & Loss of Energy · 37 questions with solutions

CONCEPT

Capacitors in DC Circuits & RC

In one line
In a DC circuit a capacitor charges and then stops the current in its own branch. In steady state treat it as a break and find the voltage across it.

STEADY STATE

8.9.1 Capacitor in a DC circuit

Once charged, no current flows through the capacitor branch.
Remove the capacitor branch, solve the resistor circuit for currents.
Voltage across C = potential difference between its two ends; then Q = CV.
A resistor in series with C carries no current, so no voltage drops across it.
A capacitor directly across a battery (internal resistance r) gets the full emf in steady state.

Q = C × V_{across C}

AT THE START

8.9.2 At the instant the switch closes

An uncharged capacitor acts as a plain wire at t = 0.
So at the start the current is largest; in steady state it acts as an open switch.
Resistors in parallel with C get no current at t = 0.

DISPLACEMENT CURRENT

8.9.3 Current through the gap

I_d = ε₀ dφ_Edt = C dVdt

No charge crosses the gap, but the changing field acts as a current.
Displacement current in the gap = conduction current in the wires.
Through a part of the plate area A′: I′ = I × A′ ÷ A.
2 A through 1 μF needs dV ÷ dt = 2 × 10⁶ V s⁻¹.

TIME CONSTANT

8.9.4 Time constant

τ = RC

Unit: second. In time RC the charge reaches 63% of its final value.
While discharging it falls to 37% in time RC.
Full charging and discharging are in the JEE · NEET Corner.

LC CIRCUIT

8.9.5 Charged capacitor across a coil

½ CV² = ½ LI_max² ⇒ I_max = V √(CL)

A capacitor across a switch prevents sparking: ½LI² = ½CV², so C = LI² ÷ V².

JEE · NEET

Corner

Charging through R

Switch S closed at t = 0

q = Q₀(1 − e^{−t ÷ RC}) · V = V₀(1 − e^{−t ÷ RC})

i = V₀R e^{−t ÷ RC}

At t = 0 the capacitor acts as a plain wire: the current is largest.
Finally the capacitor acts as a break: the current is zero.

Discharging and time constant

In one time constant q reaches 63% of Q₀ or falls to 37%

q = Q₀ e^{−t ÷ RC}

τ = RC

Unit of RC is second.
Charge falls to half in t = 0.693 RC.
Energy ∝ q², so energy falls to half in 0.693 RC2
In charging, the battery gives QV₀; half is stored, half is lost as heat in R.

Leaky capacitor

A dielectric of resistivity ρ between the plates leaks charge.

RC = ρε₀K

Practise PYQs on this topicCapacitors in DC Circuits & RC · 46 questions with solutions

EXAM-DAY CARD

Just see this before the exam

C = QV

capacitance

C = ε₀Ad

parallel plate

U = ½CV²

energy

C′ = KC

dielectric

1C_s = Σ1C

series

V = ΣCVΣC

sharing

FORMULA SHEET

Whole chapter on one page

Conductor

C = Q ÷ V, 1 μF = 10⁻⁶ F

sphere: C = 4πε₀R

U = ½CV² = Q² ÷ 2C

same V: σ ∝ 1 ÷ R

Parallel plate

E = σ ÷ ε₀, V = Ed

C = ε₀A ÷ d

unequal: V = (Q₁ − Q₂) ÷ 2C

u = ½ε₀E²

Force & work

F = QE ÷ 2 = Q² ÷ 2ε₀A

W_battery = CV², stored ½CV²

heat lost = ½CV²

dU ÷ dt = CV dV ÷ dt

Dielectric

full: C′ = KC

slab: ε₀A ÷ (d − t + t ÷ K)

metal: ε₀A ÷ (d − t)

removed: V, E, U ÷ K · kept: Q, U × K

Combination

series: C₁C₂ ÷ (C₁ + C₂), same Q

parallel: C₁ + C₂, same V

V₁ = C₂V ÷ (C₁ + C₂)

alternate plates: (n − 1)ε₀A ÷ d

Sharing & RC

V = (C₁V₁ + C₂V₂) ÷ (C₁ + C₂)

ΔU = C₁C₂(V₁ − V₂)² ÷ 2(C₁ + C₂)

steady state: no current in C branch

τ = RC, I_d = C dV ÷ dt

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

Force between plates taken as QE

Series and parallel rules copied from resistors

Battery removed vs connected mixed up

Energy loss forgotten when capacitors are joined

CORRECT ✓

F = QE ÷ 2 (field of the other plate)

Capacitors: parallel adds, series uses reciprocals

Removed: Q fixed. Connected: V fixed

Some energy is always lost on joining