Electrostatics Formula Sheet — JEE Main Physics

Every key Electrostatics formula, definition and theorem for JEE Main Physics in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 26 sub-topics

Electric charges
Conservation of charge
Coulomb's law - force between two point charges
Forces between multiple charges
Superposition principle
Continuous charge distribution
Electric field
Electric field due to a point charge
Electric field lines
Electric dipole
Electric field due to a dipole
Torque on a dipole in uniform electric field
Electric flux
Statement of Gauss's theorem and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell
Electric potential
Potential difference
Electric potential due to a point charge, a dipole and system of charges
Equipotential surfaces
Electrical potential energy of a system of two point charges and of electric dipole in an electrostatic field
Conductors and insulators
Free charges and bound charges inside a conductor
Dielectrics and electric polarisation
Capacitors and capacitance
Combination of capacitors in series and in parallel
Capacitance of a parallel plate capacitor with and without dielectric medium between the plates
Energy stored in a capacitor

Electric Charge & Coulomb's Law

Properties of charge

▸ $Quantised$ : $Q = \pm n e$ ( $e = 1.6 \times 1 0^{- 19}$ C).
▸ $Conserved$ and $additive$ (algebraic, scalar).
▸Relativistically $invariant$ (independent of frame/speed).
▸Conductors have free electrons; insulators do not.

Coulomb's law

F = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r ^{2}} = k \frac{q _{1} q _{2}}{r ^{2}}

k = 9 \times 1 0^{9} N m^{2} C^{- 2}

ε_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}

;

r =

separation

In a medium

F_{m e d} = \frac{F _{v a c}}{K} = \frac{1}{4 π ε _{0} K} \frac{q _{1} q _{2}}{r ^{2}}

K = ε_{r} =

dielectric constant of the medium (

K = 1

vacuum)

Two like charges separated by r repelling each other along the line joining them — Like charges repel, unlike attract — along the line joining.

Comparative: Coulomb vs gravitation

	Coulomb	Gravitation
Law	$k q_{1} q_{2} / r^{2}$	$G m_{1} m_{2} / r^{2}$
Sign	$\pm$ (attract/repel)	always attractive
Strength	$\sim 1 0^{39} \times$ stronger	weaker
Medium-dependent?	yes ( $/ K$ )	no

⚡ Shortcut · Superposition

Net force

F = \sum_{i} F_{i}

— each pair's Coulomb force is computed as if the others were absent, then added as

vectors

. For three charges in a line, watch the directions before summing magnitudes.

🚫 Examiner Trap · Charge & Coulomb's law

(1) r is the

separation

between charges (centre-to-centre for spheres), not a coordinate. (2) In a medium, force drops by K — don't forget it. (3) Add forces as

vectors

, never scalar magnitudes. (4) Charge is quantised (

Q = n e

) and conserved — a body can't carry a fractional e.

Electric Field & Field Lines

Electric field

E

: Force per unit positive test charge

E = \frac{F}{q _{0}}

(in the limit

q_{0} \to 0

); point charge

E = \frac{k q}{r ^{2}}

. Unit N C

^{- 1} =

m^{- 1}

. Obeys superposition

E = \sum E_{i}

Comparative: field of standard bodies

Body	$E$	Note
Point charge	$k q / r^{2}$	radial
Ring (axis)	$\frac{k Q x}{( x ^{2} + R ^{2} ) ^{3/2}}$	max at $x = R / 2$
Shell / sphere (out)	$k Q / r^{2}$	like point charge
Shell inside / solid inside	$0 / kQr / R^{3}$	—
Infinite wire	$\frac{λ}{2 π ε _{0} r}$	$\propto 1/ r$
Infinite sheet	$\frac{σ}{2 ε _{0}}$	uniform

Field lines of a positive charge (outward), negative charge (inward), a dipole, and two like charges with a null point — Field-line patterns.

Field-line properties

▸Start at $+$ , end at $-$ ; never form closed loops (electrostatic field is conservative).
▸Tangent gives $E$ direction; line density $\propto ∣ E ∣$ .
▸Two lines never cross; meet a conductor $⊥$ ; none inside a conductor.

Continuous charge (line charge

λ

)

E_{a rc} = \frac{2 kλ sin ( θ /2 )}{R}; E_{se mi c i rc l e} = \frac{2 kλ}{R}; E_{ro d, a x i s} = \frac{k Q}{d ( d + L )}

arc of angle

θ

and radius R, field at its centre (full ring:

0

); rod of length L, point on its axis at distance d from the near end. Rod seen at perpendicular distance r, ends at angles

α, β

E_{⊥} = \frac{kλ}{r} (sin α + sin β)

E_{∥} = \frac{kλ}{r} (cos α - cos β)

; infinite rod

\to \frac{2 kλ}{r}

🎯 Exam · Charged bob in a field

A bob of mass m, charge q on a thread in a horizontal field E hangs at

tan θ = \frac{qE}{m g}

with tension

T = (m g)^{2} + (qE)^{2}

. Two identical charged bobs from one point:

tan θ = \frac{F _{e}}{m g}

for each.

🚫 Examiner Trap · Electric field & lines

(1) Isolated sheet gives

\frac{σ}{2 ε _{0}}

, but a

conductor

surface gives

\frac{σ}{ε _{0}}

(twice) — don't confuse. (2)

E = 0

inside

a uniform shell, but

V \neq = 0

there. (3) Field lines never cross and never close on themselves. (4) E is max on a ring's axis at

x = R / 2

not

at the centre (

E_{ce n t re} = 0

Electric Dipole & Gauss's Law

Dipole field

p = q (2 a); E_{a x ia l} = \frac{2 k p}{r ^{3}}, E_{e q u a t or ia l} = \frac{k p}{r ^{3}}

p

points

-

+

; general

E = \frac{k p}{r ^{3}} 1 + 3 cos^{2} θ

for

r ≫ a

Torque & energy

τ = p \times E (τ = pE sin θ); U = - p \cdot E

θ =

angle between

p

and

E

; stable at

θ = 0

, unstable at

18 0^{\circ}

; no net force in a

uniform

field

Comparative: axial vs equatorial

	Axial	Equatorial
Field $E$	$2 k p / r^{3}$	$k p / r^{3}$
Direction	along $p$	opposite $p$
Potential $V$	$k p / r^{2}$	$0$

Gauss's law

Φ = \oint E \cdot d A = \frac{q _{e n c}}{ε _{0}}

net flux depends

only

on enclosed charge, not on its position or outside charges

A point charge enclosed by a spherical Gaussian surface with radial field lines passing through it — Total flux $= q / ε_{0}$ .

Gauss applications

E_{w i re} = \frac{λ}{2 π ε _{0} r}, E_{s h ee t} = \frac{σ}{2 ε _{0}}, E_{co n d u c t or} = \frac{σ}{ε _{0}}

λ =

linear,

σ =

surface charge density

Flux shortcuts

▸Uniform field through a flat area: $Φ = E A cos θ$ ; through any CLOSED surface in a uniform field: $0$ .
▸Charge q at the centre of a cube: $\frac{q}{6 ε _{0}}$ per face. At a corner: $\frac{q}{8 ε _{0}}$ through the whole cube, $\frac{q}{24 ε _{0}}$ through each of the 3 far faces, $0$ through the 3 faces that touch it.
▸At the centre of a face: $\frac{q}{2 ε _{0}}$ through the cube; at the midpoint of an edge: $\frac{q}{4 ε _{0}}$ .
▸Point charge and a disc that subtends a half-angle $α$ : $Φ = \frac{q}{2 ε _{0}} (1 - cos α)$ .

🚫 Examiner Trap · Dipole & Gauss

(1) Axial field (

2 k p / r^{3}

) is

twice

the equatorial; both fall as

1/ r^{3}

, faster than a point charge. (2) A uniform field exerts

torque but no net force

on a dipole. (3) Gauss's

Φ

depends only on

enclosed

charge, but

E

on the surface depends on

all

charges. (4) Gauss is only

useful

for high symmetry (sphere/cylinder/sheet).

Electric Potential

Electric potential

V

: Work per unit positive charge to bring it from

\infty

to a point:

V = - \int_{\infty}^{r} E \cdot d r

. Scalar, unit volt (J

C^{- 1}

); only

differences

are physical.

Potential due to

V_{p o in t} = \frac{k q}{r}; V_{sys t e m} = \sum \frac{k q _{i}}{r _{i}}; V_{d i p o l e} = \frac{k p cos θ}{r ^{2}}

add V as a

scalar with sign

; dipole axial

k p / r^{2}

, equatorial

0

Field

\leftrightarrow

potential

E = - \frac{d V}{d r}; V_{B} - V_{A} = - \int_{A}^{B} E \cdot d r

E

points toward

decreasing

E =

-

(slope of V-r)

Concentric equipotential circles around a point charge, perpendicular to the radial field lines — Equipotentials $⊥$ field lines.

Equipotential surfaces

▸Always $⊥$ to field lines; $no work$ to move a charge along one.
▸Closer surfaces $\Rightarrow$ stronger field; a conductor (surface + interior) is one equipotential.
▸Two equipotentials never intersect.

Potential of spheres

V_{s h e ll} = \frac{k Q}{R} (r \leq R); V_{so l i d} = \frac{k Q ( 3 R ^{2} - r ^{2} )}{2 R ^{3}} (r \leq R)

outside both:

k Q / r

. A uniformly charged SOLID (non-conducting) sphere has

V_{ce n t re} = \frac{3}{2} V_{s u r f a ce}

; a conducting sphere behaves like the shell

🎯 Exam · Conducting spheres joined by a wire

They reach the SAME potential:

\frac{q _{1}}{R _{1}} = \frac{q _{2}}{R _{2}}

, so

q \propto R

σ \propto \frac{1}{R}

and surface field

E \propto \frac{1}{R}

— charge crowds at sharp points (corona discharge, lightning conductors). Total charge is conserved; some energy is lost as heat.

🚫 Examiner Trap · Electric potential

(1) V is a

scalar

— add contributions with sign, not as vectors. (2)

E = 0

does

not

imply

V = 0

(e.g. inside a charged shell

E = 0

V \neq = 0

);

V = 0

does not imply

E = 0

(dipole equatorial). (3)

E = - d V / d r

: field points

down

the potential gradient. (4) Work by you

= q Δ V

; a

+

charge moves spontaneously from high to low V.

Potential Energy, Conductors & Dielectrics

Potential energy

U = \frac{k q _{1} q _{2}}{r}; U_{sys t e m} = pairs \sum \frac{k q _{i} q _{j}}{r _{ij}}

charge in external potential:

U = q V

; dipole in field:

U = - p \cdot E

, so the work to turn it from

θ_{1}

θ_{2}

W = pE (cos θ_{1} - cos θ_{2})

(

2 pE

from aligned to anti-aligned)

Comparative: conductor vs dielectric

	Conductor	Dielectric
Free charges	yes	no (bound)
Field inside	$0$	$E_{0} / K$ (reduced)
Net charge sits	on surface	polarised, in bulk
In a capacitor	$K \to \infty$ effect	raises $C$ by $K$

A conductor in an external field develops induced surface charges so that the field inside is zero — Induced charges cancel the field inside.

Dielectrics

E_{in} = \frac{E _{0}}{K}; K = \frac{C}{C _{0}}; P = ε_{0} (K - 1) E

E_{0} =

applied field; polarisation

P

reduces the internal field;

D = ε_{0} E + P

🚫 Examiner Trap · PE, conductors & dielectrics

(1) Inside a conductor

E = 0

and

V =

const (even in a cavity — electrostatic shielding); charge lives on the

surface

. (2) A dielectric only

reduces

the field (

E_{0} / K

), never to zero (unlike a conductor). (3)

U_{sys t e m}

sums over

pairs

— don't double-count. (4) Polar molecules (

H_{2}

O) have permanent dipoles; non-polar (

N_{2}

) get induced ones.

Capacitance & Energy Stored

Capacitance

C

: Charge stored per unit potential difference

C = \frac{Q}{V}

. Unit farad (C

V^{- 1}

). Depends

only

on geometry and the medium, not on Q or V.

Parallel-plate

C = \frac{ε _{0} A}{d} dielectric C = \frac{K ε _{0} A}{d}

A =

plate area,

d =

separation,

K =

dielectric constant

Partial slab (thickness

t

)

C = \frac{ε _{0} A}{d - t + \frac{t}{K}}

conducting slab (

K \to \infty

C = \frac{ε _{0} A}{d - t}

Parallel-plate capacitor with +Q and -Q plates, a dielectric slab K between them and a uniform field — Dielectric raises $C$ by a factor $K$ .

Comparative: capacitor geometries

Type	$C$
Parallel-plate	$\frac{K ε _{0} A}{d}$
Spherical	$\frac{4 π ε _{0} ab}{b - a}$
Cylindrical	$\frac{2 π ε _{0} l}{ln ( b / a )}$
Isolated sphere	$4 π ε_{0} R$

Energy stored

U = \frac{1}{2} Q V = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C}

energy density

u = \frac{1}{2} ε_{0} E^{2}

(in vacuum;

\frac{1}{2} K ε_{0} E^{2}

in a dielectric)

Force between the plates

F = \frac{Q ^{2}}{2 ε _{0} A} = \frac{1}{2} ε_{0} E^{2} A

each plate feels the field of the OTHER plate only (

σ /2 ε_{0}

), hence the

\frac{1}{2}

; attractive

Two dielectrics in one capacitor

▸Side by side (each fills area $A /2$ , full gap d): parallel, $C = \frac{ε _{0} A}{2 d} (K_{1} + K_{2})$ .
▸Layered (each thickness $d /2$ , full area): series, $C = \frac{2 ε _{0} A}{d} \cdot \frac{K _{1} K _{2}}{K _{1} + K _{2}}$ .

🚫 Examiner Trap · Capacitance & energy

(1) C depends on geometry/medium

only

— not on Q or V. (2) Pick the right energy form: with charge fixed use

Q^{2} /2 C

, with voltage fixed use

\frac{1}{2} C V^{2}

. (3) Inserting a dielectric with the

battery on

raises Q and energy; with the

battery off

lowers V and energy. (4) Force on a slab pulls it

in

(energy lowers).

Combination of Capacitors & RC Circuits

Comparative: series vs parallel

	Series	Parallel
Effective $C$	$\frac{1}{C} = \sum \frac{1}{C _{i}}$	$C = \sum C_{i}$
Same for all	charge $Q$	voltage $V$
Divides	$V \propto 1/ C$	$Q \propto C$
$C_{e ff}$ vs members	$<$ smallest	$>$ largest

Series / Parallel

\frac{1}{C _{s}} = \sum \frac{1}{C _{i}}; C_{p} = \sum C_{i}

series: same

Q

; parallel: same

V

Three capacitors in series and three in parallel with their effective-capacitance formulas — Series (same $Q$ ) vs parallel (same $V$ ).

RC charging / discharging — not in JEE Main 2026 syllabus

q = C V (1 - e^{- t / RC}); q = Q_{0} e^{- t / RC}

time constant

τ = RC

: reaches

63%

(charge) / falls to

37%

(discharge) in one

τ

Slab / battery rules

▸Battery $connected$ (V fixed): Q and energy change with C.
▸Battery $disconnected$ (Q fixed): V and energy change instead.

🚫 Examiner Trap · Combinations & RC

(1) Series C is

smaller

than the smallest member — opposite to resistors-in-series intuition. (2) Sharing charge between two capacitors

loses

energy (heat/radiation): common

V = \frac{C _{1} V _{1} + C _{2} V _{2}}{C _{1} + C _{2}}

and loss

Δ U = \frac{C _{1} C _{2} ( V _{1} - V _{2} ) ^{2}}{2 ( C _{1} + C _{2} )}

(plates of opposite sign joined: use

V_{1} + V_{2}

). (3) In a steady-state network a capacitor carries NO current — treat it as a break and find the voltage across it. (4) RC transients (

t = 0

short,

t = \infty

open,

τ = RC

) are not in the 2026 syllabus.

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What are the most important Electrostatics formulas for JEE Main?

This Electrostatics formula sheet covers all the high-yield Physics formulas, definitions and theorems you need for JEE Main, across Electric charges, Conservation of charge, Coulomb's law - force between two point charges, Forces between multiple charges, Superposition principle — each shown with the key result and, where useful, a worked example.

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