Electrostatics: theory | Physics Circle

CONCEPT

Electric Charge & Coulomb’s Law

In one line
Charge comes in whole-number packets of e; two point charges push or pull each other with a force that falls as the square of the distance.

CHARGE

8.1.1 Properties of charge

Like charges repel; unlike charges attract.
Quantised: Q = ±ne, n = 1, 2, 3 …; e = 1.6 × 10⁻¹⁹ C.
Conserved: it can be moved from one body to another, never created or destroyed.
Invariant: it does not depend on speed (mass does).
Charging changes mass a little: losing electrons makes a body positive and lighter.

n = Qe

SI unit: coulomb (C) = A s. Dimension: [AT].
1 C = 3 × 10⁹ statcoulomb (esu).
A charge of ±⅔e or 1.5e is impossible on any body.

CHARGING

8.1.2 Methods of charging

Friction: rubbing transfers electrons. Glass with silk: glass +, silk −. Ebonite or polythene with wool: ebonite −, wool +.
Conduction: touching a charged body; both get the same sign.
Induction: a nearby charge separates charges; earth the far side, then remove the earth and the charge. The body gets the opposite sign.

Charging by induction gives the opposite sign

DENSITY

8.1.3 Charge densities

Line, surface and volume distributions

Distribution	Density	SI unit
line	λ = QL	C m⁻¹
surface	σ = QA	C m⁻²
volume	ρ = QV	C m⁻³

Sphere of radius r: σ = Q4πr², so for the same charge σ ∝ 1r².
On a conductor σ is largest where the surface is most curved (sharp points), so charge leaks from points.

COULOMB

8.1.4 Coulomb’s law

Equal and opposite forces along the line joining the charges

F = 14πε₀ q₁q₂r² = k q₁q₂r²

k = 9 × 10⁹ N m² C⁻²; ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻².
Valid for point charges at rest; F₁₂ = −F₂₁ (Newton’s third law).
Doubling one charge and halving r makes F eight times.
Electric force between electron and proton is about 10³⁹ times the gravitational force.

TOUCHING

8.1.5 Touching identical spheres

Two identical conductors in contact share the total charge equally: each gets q₁ + q₂2.
Opposite charges first cancel, then the rest is shared.
Charge Q split into q and Q − q: the force is largest when q = Q2.

MEDIUM

8.1.6 Effect of a medium

F_m = FK = 14πε₀K q₁q₂r²

K = dielectric constant (relative permittivity), ε = ε₀K.
Same force in a medium as in air at distance r: the medium distance is r√K.
Same force in air as in the medium at distance d: the air distance is d√K.

Slab of thickness t between the charges

F = 14πε₀ q₁q₂(r − t + t√K)²

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CONCEPT

Superposition & Equilibrium of Charges

In one line
The force on a charge is the vector sum of the forces from every other charge; a charge is in equilibrium when that sum is zero.

SUPERPOSITION

8.2.1 Principle of superposition

F = F₁ + F₂ + F₃ + … (vector sum)

Two forces at angle θ

F = √F₁² + F₂² + 2F₁F₂ cos θ

Two equal forces F at	Resultant
0°	2F
60°	√3 F
90°	√2 F
120°	F
180°	0

Charges on a square or triangle: find each force, use symmetry, cancel the equal and opposite parts, then add.
Three charges q at the corners of an equilateral triangle: net force on each = √3 F, F = force between two of them.
Square with q at each corner: force on one corner = F(√2 + ½), F = force from the next corner.

THREE CHARGES

8.2.2 Three charges on a line

Third charge q between Q₁ and Q₂

For q to be in equilibrium, the two forces on it must be equal and opposite.
Like charges Q₁, Q₂: the point lies between them, nearer the smaller charge.
Unlike charges: the point lies outside, beyond the smaller charge.
The value or sign of q itself does not matter for its own equilibrium.

Q₁x₁² = Q₂(x − x₁)² ⇒ x₁ = x√Q₂Q₁ + 1

WHOLE SYSTEM

8.2.3 Whole system in equilibrium

Equal charges Q at the ends and q at the midpoint: every charge balances when q = −Q4.
Four charges Q at the corners of a square with q at the centre: q = −Q4(1 + 2√2).
Charges +e and +4e a distance d apart: a third charge is in equilibrium at x = d3 from +e.

HANGING BALLS

8.2.4 Charged balls hanging from threads

Forces on one ball: T, mg and the electric force F

T sin θ = F, T cos θ = mg ⇒ tan θ = Fmg

Small angle: tan θ ≈ r2l, so r³ = q²l2πε₀mg and r ∝ q^⅔.
Two identical balls of different charges still hang at equal angles (equal and opposite forces).

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Stable or unstable equilibrium

Displace the middle charge a little along the line joining the charges.
Like end charges and an opposite middle charge: unstable along the line.
Middle charge of the same sign as the ends: stable along the line, unstable sideways.
With potential energy U(x): d²Udx² > 0 means stable, < 0 means unstable.

Charged balls in a liquid

K = ρρ − σ

Same angle in air and in a liquid of density σ: the electric force falls by K while the weight falls by the buoyancy.
ρ = density of the balls.

Leaking charge

x³ ∝ q² ⇒ dxdt ∝ x^−½

For small angles x³ = q²l2πε₀mg.
If charge leaks at a constant rate, the balls approach with speed v ∝ x^−½.

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CONCEPT

Electric Field of Point Charges

In one line
Electric field is the force on a unit positive charge; it points away from + charges and towards − charges.

FIELD

8.3.1 Electric field intensity

E = Fq₀ (q₀ → 0)

Vector quantity; SI unit N C⁻¹ = V m⁻¹.
Dimension: [MLT⁻³A⁻¹].
Force on a charge in a field: F = qE. On a + charge F is along E; on a − charge F is opposite to E.

POINT CHARGE

8.3.2 Field of a point charge

E = 14πε₀ Qr²

Radially outward for +Q, inward for −Q

Field due to many charges: add the fields as vectors (superposition).
Equal charges at the corners of a regular polygon: field at the centre is zero.
Regular hexagon with one charge removed or changed: the field at the centre equals that of the missing (or extra) charge alone.

NEUTRAL POINT

8.3.3 Neutral point (E = 0)

Where the two fields cancel

q₁x₁² = q₂x₂² ⇒ x₁x₂ = √q₁q₂

Like charges: N between them, x₁ = x√q₂q₁ + 1 from q₁.
Unlike charges: N outside, on the side of the smaller charge, l = x√q₂q₁ − 1 from q₁ (|q₁| < |q₂|).

FIELD LINES

8.3.4 Electric lines of force

Lines start on + and end on −

Start on a positive charge and end on a negative charge (or go to infinity).
The tangent at any point gives the direction of E.
Two lines never cross (E has one direction at a point).
Lines do not form closed loops (unlike magnetic lines).
Closer lines mean a stronger field; parallel, equally spaced lines mean a uniform field.
Lines meet a conductor at right angles and do not enter it.
Number of lines from a charge q in a medium = qε₀K.

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CONCEPT

Field of Continuous Charges

In one line
Split the charge into tiny bits, write the field of each bit as a point charge, and add them with symmetry.

METHOD

8.4.1 Continuous distribution

dq = λ dl, dE = kdqr², E = ∫ dE

Use symmetry first: components that cancel in pairs need not be calculated.

WIRE

8.4.2 Straight wire (line charge λ)

Point P at distance r from a finite wire

E_⊥ = kλr(sin α + sin β)

E_∥ = kλr(cos β − cos α)

P on the perpendicular bisector (α = β): E = 2kλr sin α.
Infinitely long wire: E = 2kλr = λ2πε₀r.
Point on the axis of a rod of length L, at distance d from its near end: E = kQd(d + L).

ARC

8.4.3 Charged arc at its centre

Semicircular arc: field at the centre O

arc of angle φ: E = 2kλr sin(φ2)

Shape	E at centre
semicircle	2kλr = Q2π²ε₀r²
quarter circle	√2 kλr
120° arc	√3 kλr
full ring	0

Removing a small arc from a ring: the field at the centre has the size of the removed part’s field and points towards the gap (positive ring).

RING

8.4.4 Ring on its axis

Point P on the axis at distance x

E = kQx(x² + R²)^3/2

Centre (x = 0): E = 0. Far away (x ≫ R): E = {F("kQ","x²")}, like a point charge.
E is maximum at x = R√2: E_max = Q6√3 πε₀R².

E along the axis of a ring

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Uniformly charged disc (on its axis)

E = σ2ε₀(1 − x√x² + R²)

Very near the disc (x → 0): E = σ2ε₀, like an infinite sheet.
Far away: the disc behaves like a point charge.

Hemispherical shell

centre: E = σ4ε₀

Field of a uniformly charged hemispherical shell at its centre.

Point near one end of a long wire

E_x = E_y = kλr, E = √2 kλr

The net field makes 45° with the wire.

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CONCEPT

Electric Flux & Gauss’s Law

In one line
Flux counts the field lines through a surface; the total flux out of any closed surface is the charge inside divided by ε₀.

FLUX

8.5.1 Electric flux

Area vector A along the normal

φ = E · A = EA cos θ

Scalar. SI unit N m² C⁻¹ = V m.
θ = 0° (surface ⊥ E): φ = EA, maximum. θ = 90° (surface ∥ E): φ = 0.
Field E = aî + bĵ + ck̂ through an area A in the YZ plane (A = Aî): φ = aA.
For a closed surface: outward flux +, inward flux −.

GAUSS

8.5.2 Gauss’s law

φ = ∮ E · dA = q_encε₀

Only charges inside the surface give net flux

Depends only on the charge enclosed: not on the shape, size or radius of the surface, nor on where the charge is inside.
A balloon with charge on it blown up: flux stays the same.
Charges outside give zero net flux (lines enter and leave), but they do change E at points of the surface.
Dipole inside: q_enc = 0, so φ = 0.
Flux entering φ₁ and leaving φ₂: q = ε₀(φ₂ − φ₁).
In a medium: φ = qε₀K.

CUBE

8.5.3 Charge and a cube

Use imaginary cubes to surround the charge

Charge q at	Flux through cube	Per face
centre	qε₀	q6ε₀
centre of a face	q2ε₀	—
an edge (middle)	q4ε₀	—
a corner	q8ε₀	q24ε₀ (3 far faces)

Two opposite faces with q at the centre: q3ε₀.

OTHER SHAPES

8.5.4 Cylinder and hemisphere

Charge inside a closed cylinder

Closed cylinder with q inside, flux φ through the curved surface: each flat end gets 12(qε₀ − φ).
Hemisphere in a uniform field E (flat face ⊥ E): curved surface φ = πR²E (equal to the flat face).
Charge at the centre of the flat face of a hemisphere: curved surface gets q2ε₀.
Square of side a at height a2 above a charge q (the charge at the centre of an imaginary cube): φ = q6ε₀.

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CONCEPT

Gauss’s Law Applications & Conductors

In one line
With a symmetric Gaussian surface, Gauss’s law gives E at once for wires, sheets and spheres; inside a conductor E is zero.

LINE

8.6.1 Infinite line charge

E = λ2πε₀r = 2kλr

E ∝ 1r. The same holds outside a long charged cylinder.
Long cylinder of volume charge ρ, radius R: inside E = ρr2ε₀; outside E = ρR²2ε₀r.

SHEET

8.6.2 Infinite plane sheet

Field on both sides, normal to the sheet

E = σ2ε₀ (does not depend on distance)

Two sheets with +σ and −σ

Two sheets +σ and −σ: E = σε₀ between them, 0 outside.
Two sheets both +σ: E = 0 between, σε₀ outside.

SPHERE

8.6.3 Charged sphere and shell

Point	Shell or conductor	Uniform solid sphere
outside (r > R)	kQr²	kQr²
surface	kQR² = σε₀	kQR² = ρR3ε₀
inside (r < R)	0	kQrR³ = ρr3ε₀

E against r

Outside, every sphere acts like a point charge at its centre.
Solid sphere: E rises linearly inside and is maximum on the surface.

CONDUCTORS

8.6.4 Conductors in electrostatics

E = 0 everywhere inside the material of a conductor.
All excess charge sits on the outer surface; the interior has no net charge.
Just outside a conductor: E = σε₀, normal to the surface (in a medium: σKε₀).
The whole conductor is at one potential (an equipotential).
σ is larger where the surface curves more sharply.

Charge +q inside a cavity of a conducting shell with charge Q

Charge q in a cavity induces −q on the cavity wall and +q on the outer surface.
Electrostatic shielding: a closed metal box keeps outside fields out; a car or aircraft body protects people from lightning.
Tankers carry metal chains touching the ground to leak charge away.
Van de Graaff generator: charge given to the inside of a hollow conductor moves to its outer surface; corona discharge at points. It gives a very high voltage at a small current.
Air breaks down at about 3 × 10⁶ V m⁻¹; largest charge on a sphere: Q = 4πε₀R²E_max.

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Non-uniform charge density ρ(r)

Find the charge inside radius r: q = ∫ ρ(r) 4πr² dr.
Then E = q4πε₀r².
ρ = kr gives E ∝ r² inside; ρ = kr gives E constant inside.

Cavity in a charged sphere

Field inside a spherical cavity is uniform: E = ρ3ε₀ × (vector from sphere centre to cavity centre).

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CONCEPT

Motion of Charges in a Field

In one line
In a uniform field a charge has a constant acceleration qEm, so it moves exactly like a body under gravity.

FROM REST

8.7.1 Charge released from rest

a = qEm

v = qEtm = √2qEdm = √2qVm

KE = qV = qEd = q²E²t²2m

V = potential difference crossed, d = distance moved along E.
Same V, same mass: v ∝ √q. Same V, same charge: v ∝ 1√m.
Momentum p = qEt = √(2mqV).
Work done by the field = qV, whatever the path.

BALANCED DROP

8.7.2 Charged drop at rest

qE = mg ⇒ q = mgE = mgdV

Between plates a distance d apart with potential difference V: E = Vd.
Number of extra electrons: n = qe.
Drop of radius r and density ρ: m = 43πr³ρ.
A negative drop is held up when the upper plate is positive.

PENDULUM

8.7.3 Charged pendulum

Bob in a horizontal field

Horizontal field: the bob rests at tan θ = qEmg.
Effective gravity g′ = √g²1 + (qEm)²; field down on + charge: g′ = g + qEm; field up: g′ = g − qEm.

T = 2π√lg′

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Charge entering across a uniform field

Path between the plates is a parabola

y = qE2m (xu)²

Velocity along the plates stays u; the field gives a constant acceleration across them.
Deflection on leaving the plates: y = qEL²2mu².
tan β = qELmu² for the angle of the velocity on leaving.

Small oscillations between two fixed charges

Like charges Q at x = ±a and a charge q (same sign) at the midpoint: along the line it does SHM with T = 2π√4πε₀ma³2Qq.
Displaced at right angles to the line it is unstable (for like charges).

Charged block on a spring

A field E shifts the equilibrium by qEk.
The period T = 2π√mk does not change.

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CONCEPT

Electric Potential

In one line
Potential is the work done per unit charge to bring a test charge from infinity; it is a scalar, so potentials simply add.

POTENTIAL

8.8.1 Electric potential

V = Wq₀

Scalar; SI unit volt = J C⁻¹. Dimension [ML²T⁻³A⁻¹].
Potential difference: V_B − V_A = W_ABq₀.

POINT CHARGES

8.8.2 Point charges

V = 14πε₀ Qr

Several charges: V = k Σ Q_ir_i (algebraic sum, use signs).
Charge q at each of the n corners of a regular polygon: V = nkqr at the centre, while E there is zero.
+q, −q, +q, −q round a square: V = 0 and E = 0 at the centre.
Charges at the eight corners of a cube of side a: centre distance = √3 a2.

ZERO POTENTIAL

8.8.3 Points of zero potential

Two points on the line of two unlike charges

inside: q₁x = q₂d − x; outside: q₁x = q₂d + x

Only for unlike charges; both points lie nearer the smaller charge.
Like charges: V is never zero at a finite point.

RING AND SPHERE

8.8.4 Ring, sphere and shell

Ring on its axis: V = kQ√x² + R²; at the centre V = kQR (but E = 0).
Half ring or any arc of radius r with total charge Q: V = kQr at the centre; with λ: semicircle V = λ4ε₀.

Point	Shell or conductor	Uniform solid sphere
outside	kQr	kQr
surface	kQR	kQR
inside	kQR (same everywhere)	kQ(3R² − r²)2R³
centre	kQR	32 V_surface

V against r

CONCENTRIC

8.8.5 Concentric shells

Inner shell a, outer shell b

Each shell gives kQ ÷ (its own radius) inside it and {F("kQ","r")} outside it; add the contributions.
Shells with equal σ: Q ∝ R², so the potential at the common centre = {F("σ","ε₀")} × (sum of radii).
Inner shell earthed: its charge changes until its potential is zero.

CONNECTED

8.8.6 Spheres joined by a wire

Charge flows until both are at the same potential.
Q ∝ R, so Q₁Q₂ = R₁R₂ and σ₁σ₂ = R₂R₁; the smaller sphere has the larger σ and E.
Common potential V = Q₁ + Q₂4πε₀(R₁ + R₂).

DROPS

8.8.7 n small drops form one big drop

Quantity	Big drop
radius	R = n^⅓ r
charge	Q = nq
potential	V = n^⅔ v
surface density	σ_big = n^⅓ σ
energy	U = n^5/3 u

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CONCEPT

E–V Relation & Equipotentials

In one line
The field is the negative slope of the potential; it always points towards falling potential and cuts equipotentials at right angles.

RELATION

8.9.1 E from V

E = −dVdr

E_x = −∂V∂x, E_y = −∂V∂y, E_z = −∂V∂z

Potential gradient is the rate of change of V with distance; unit V m⁻¹.
E points in the direction in which V decreases fastest.
V constant in a region ⇒ E = 0 there.

Flat parts of V–x have zero field

V FROM E

8.9.2 V from E

V_B − V_A = −∫ E · dr

Uniform field: V = Ed, with d measured along E.
Points along E have falling potential; points at right angles to E have the same potential.
E = (aî + bĵ): from origin to (x, y) the potential falls by ax + by.
Conductor: V(R) − V(nR) = kQ(1R − 1nR); use it to find Q, then E at nR.

V(r) TO CHARGE

8.9.3 From V(r) to the charge density

Find E(r) = −dVdr, then use Gauss’s law for a sphere of radius r.
V = a r² + b inside a ball gives E = −2ar and ρ = −6aε₀.

EQUIPOTENTIALS

8.9.4 Equipotential surfaces

Spheres for a point charge, planes for a uniform field

Every point of the surface has the same potential.
E is at right angles (90°) to every equipotential surface.
No work is done in moving a charge along an equipotential.
Two equipotentials never cross.
Closer equipotentials mean a stronger field.
Point charge: concentric spheres. Uniform field: parallel planes. Every conductor surface is an equipotential.

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CONCEPT

Electrostatic Potential Energy

In one line
Potential energy of a set of charges is the work done to bring them together from infinity; it is a scalar sum over every pair.

TWO CHARGES

8.10.1 A pair of charges

U = 14πε₀ q₁q₂r

Use the signs of the charges: like charges give U > 0, unlike give U < 0.
Bringing two electrons closer: U increases. Electron towards a proton: U decreases.

SYSTEM

8.10.2 Several charges

Add U for every pair

U = k(q₁q₂r₁₂ + q₂q₃r₂₃ + q₁q₃r₁₃)

n charges have n(n − 1)2 pairs; three equal charges on a triangle of side a: U = 3kq²a.
For zero total PE, set the sum of the pair terms to zero and solve for the unknown charge.
Two charges q at the ends and Q at the midpoint: U = k({F("q²","2a")} + {F("2qQ","a")}), with 2a the separation.

WORK

8.10.3 Work done

W = U_final − U_initial

W = q(V_B − V_A)

Moving q₂ from r₁ to r₂ near q₁: W = kq₁q₂(1r₂ − 1r₁).
Moving a charge between points at the same distance from a charge (e.g. corners of a square around a centre charge): W = 0.
Work by the field = −ΔU; work by an external agent (slowly) = +ΔU.

ENERGY

8.10.4 Energy conservation

Released charges fly apart: loss of U = gain of KE. Two equal masses m released from r₁ to r₂: v = q√km(1r₁ − 1r₂).
Closest approach of a charge shot with speed v at a fixed charge: 12mv² = kQqr₀, so r₀ ∝ 1v².

ELECTRON VOLT

8.10.5 Electron volt

1 eV = 1.6 × 10⁻¹⁹ J

KE gained by charge q through V volt = qV; a proton through V volt gets V eV.

PRESSURE

8.10.6 Charged bubble and pressure

electrostatic pressure p = σ²2ε₀

Always outwards: a charged soap bubble or balloon expands.
Force pushing the halves of a charged shell apart: F = σ²2ε₀ πR², so F ∝ σ²R².
Energy density of a field: u = 12ε₀E².

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Self-energy of a charged sphere

shell: U = Q²8πε₀R

solid sphere: U = 3Q²20πε₀R

Energy density in a field: u = 12ε₀E².

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CONCEPT

Electric Dipole & Polar Molecules

In one line
Two equal and opposite charges a small distance apart form a dipole; its field falls as {F("1","r³")} and its potential as {F("1","r²")}.

DIPOLE MOMENT

8.11.1 Dipole moment

p = q × 2l

Vector from −q to +q.
SI unit C m; 1 debye = 3.33 × 10⁻³⁰ C m. Dimension [LTA].
For several charges: p = Σ q_ir_i (net charge zero).

Axial and equatorial points of a short dipole

FIELD

8.11.2 Field of a short dipole

axial: E_a = 14πε₀ 2pr³ (along p)

equatorial: E_e = 14πε₀ pr³ (opposite to p)

At the same distance E_a = 2E_e.
Equal fields: r_a : r_e = 2^⅓ : 1.
General point (angle θ from the axis): E = kpr³√1 + 3 cos²θ.

POTENTIAL

8.11.3 Potential of a dipole

V = 14πε₀ p cos θr²

Axial point: V = kpr². Equatorial point: V = 0 (but E ≠ 0).
Total charge is zero, so the flux through a surface around a dipole is zero.

POLAR MOLECULES

8.11.4 Polar and non-polar molecules

Centres of + and − charge apart in H₂O, together in CO₂

Polar: centres of positive and negative charge do not coincide, so a permanent dipole moment. H₂O, HCl, NH₃, CO, alcohol.
Non-polar: the centres coincide, p = 0 without a field. O₂, N₂, H₂, CO₂, CH₄, CCl₄, benzene.
In a field a non-polar molecule gets an induced dipole moment.
Polar molecules are randomly oriented with no field, so the net dipole moment of the material is zero.

DIELECTRIC

8.11.5 Dielectric in a field

Polarisation (dipole moment per unit volume) sets up a field opposite to the applied field.
Field inside the dielectric is reduced: E = E₀K, less than the outside field.
Polarisation P: unit C m⁻², dimension [L⁻²TA].

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CONCEPT

Dipole in an External Field

In one line
A uniform field gives a dipole a torque but no net force; it turns the dipole until p lies along E.

TORQUE

8.12.1 Torque in a uniform field

Equal and opposite forces qE form a couple

τ = p × E, τ = pE sin θ

Net force = 0 (qE − qE); only a turning effect.
Maximum torque pE at θ = 90°; zero at 0° and 180°.
Charge on the dipole from torque: q = {F("τ","2lE sin θ")}.

ENERGY

8.12.2 Potential energy

U = −p · E = −pE cos θ

θ	U	τ	Equilibrium
0°	−pE (minimum)	0	stable
90°	0	pE	none
180°	+pE (maximum)	0	unstable

WORK

8.12.3 Work in turning a dipole

W = pE(cos θ₁ − cos θ₂)

From θ = 0: W = pE(1 − cos θ).
Turned 90° from the field direction: W = pE. Turned 180°: W = 2pE.
From 0° to 60° needs W; from 0° to 180° needs 4W.

NON-UNIFORM

8.12.4 Non-uniform field

Net force ≠ 0 and torque ≠ 0: the dipole turns and also moves.
Along a stronger field (p along E): it is pulled towards the stronger field.
Two dipoles along one line attract (p in the same direction); the field between two opposite dipoles is zero where {F("p₁","x³")} = {F("p₂","(d − x)³")}.

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Oscillation of a dipole

T = 2π√IpE

For a small turn from θ = 0, the torque −pEθ gives angular SHM.
I = moment of inertia about the centre.

Two dipoles (dipole–dipole)

Position	Force	PE
along one line	6kp₁p₂r⁴ (attraction)	−2kp₁p₂r³
side by side, parallel	3kp₁p₂r⁴ (repulsion)	kp₁p₂r³
perpendicular	3kp₁p₂r⁴	0

Practise PYQs on this topicElectric Dipole · 79 questions with solutions

EXAM-DAY CARD

Just see this before the exam

F = kq₁q₂r²

Coulomb

E = kQr²

field

φ = qε₀

Gauss

V = kQr

potential

U = kq₁q₂r

energy

τ = pE sin θ

dipole

FORMULA SHEET

Whole chapter on one page

Charge & force

Q = ne

F = 14πε₀q₁q₂r²

medium: FK

x₁ = x√Q₂Q₁ + 1

Field

E = kQr²

wire: λ2πε₀r

sheet: σ2ε₀

ring: kQx(x² + R²)^3/2

Gauss & spheres

φ = qε₀

cube face: q6ε₀

shell inside: E = 0

solid inside: kQrR³

Potential

V = kQr

E = −dVdr

drops: V = n^⅔v

joined: σ₁σ₂ = R₂R₁

Energy & motion

U = kq₁q₂r

W = qΔV

a = qEm

KE = qV

Dipole

p = q·2l

E_a = 2kpr³, E_e = kpr³

τ = pE sin θ

U = −pE cos θ

TOP MISTAKES

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MISTAKE ✗

Field inside a charged shell = kQR²

Flux depends on the size of the surface

Potential adds as a vector

Medium distance = r × K

CORRECT ✓

E = 0 inside; V = kQR inside

φ = qε₀ for any closed surface

V is a scalar: add with signs

Equivalent air distance = r√K