How much of JEE Main comes from Electrostatics?

JEEGenie estimates about 7% of JEE Main marks come from Electrostatics, based on recent papers. The share varies from year to year.

Which formulas matter most in Electrostatics?

The Electrostatics sheet on this page has 23 formula entries, starting with Coulomb's Law, Electric Field, Electric Field Due to a Point Charge, Electric Dipole, Electric Field Due to a Dipole. Each one comes with when to use it and the trap to avoid.

What topics does Electrostatics cover for JEE?

Electrostatics covers Coulomb's Law, Electric Field, Gauss's Law, Electric Potential, Capacitance, Dielectrics.

How should I revise Electrostatics before the exam?

Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Electrostatics questions on JEEGenie.

How many previous-year questions are there on Electrostatics?

JEEGenie has 680 previous-year questions on Electrostatics, and this page shows a few with worked explanations.

Is practice on Electrostatics free on JEEGenie?

Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.

Electrostatics for JEE Main & Advanced — formulas, what is asked, and traps

JEEGenie estimates about 7% of JEE Main marks come from Electrostatics. This page covers 23 formula entries, each with when to use it, what JEE asks and the traps. A common slip: Thinking the field direction is the same as the force on a negative charge; it's opposite.

Key points to know

Coulomb's Law. The electrostatic force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Electric Field. The electric field at a point is the force per unit positive test charge placed at that point.
Electric Dipole. An electric dipole consists of two equal and opposite charges separated by a small distance. It has a dipole moment vector from negative to positive charge.
Electric Flux. Electric flux through a surface is the dot product of electric field and area vector, representing the number of field lines passing through.
Gauss's Law. The total electric flux through a closed surface is equal to the net charge enclosed divided by $ϵ_{0}$ .
Electric Potential. Electric potential at a point is the work done per unit positive charge to bring it from infinity to that point against the electric field.
Equipotential Surface. An equipotential surface is a surface where the electric potential is constant. No work is done to move a charge on it.
Conductor in Electrostatic Equilibrium. In a conductor at electrostatic equilibrium, the electric field inside is zero, and all excess charge resides on the surface. The potential is constant throughout.
Capacitance. Capacitance is the ability of a conductor to store charge per unit potential difference.
Dielectric and Polarization. A dielectric is an insulating material that becomes polarized in an electric field, reducing the net field inside and increasing capacitance.

Formula sheet

Name	Formula	When to use
Coulomb's Law	$F = \frac{1}{4 π ϵ_{0}} \frac{q_{1} q_{2}}{r^{2}}$	Use to calculate the magnitude of electrostatic force between two point charges in vacuum or a medium.
Electric Field	$\vec{E} = \frac{\vec{F}}{q_{0}}$	Use to find the electric field due to a point charge or a system of charges.
Electric Field Due to a Point Charge	$\vec{E} = \frac{1}{4 π ϵ_{0}} \frac{q}{r^{2}} \hat{r}$	Use to compute electric field magnitude and direction from a single point charge.
Electric Dipole	$\vec{p} = q \vec{d}$	Use to characterize a dipole in electric fields, especially for torque and potential energy.
Electric Field Due to a Dipole	$E_{axial} = \frac{1}{4 π ϵ_{0}} \frac{2 p}{r^{3}}$ , $E_{equatorial} = \frac{1}{4 π ϵ_{0}} \frac{p}{r^{3}}$	Use to find electric field at large distances from a dipole along its axis or perpendicular to it.
Electric Flux	$Φ_{E} = \vec{E} \cdot \vec{A} = E A \cos θ$	Use to calculate flux through a surface in uniform fields, or as a step in Gauss's law.
Gauss's Law	$\oint \vec{E} \cdot d \vec{A} = \frac{Q_{enc}}{ϵ_{0}}$	Use to find electric field for symmetric charge distributions (spherical, cylindrical, planar).
Applying Gauss's Law for Spherical Symmetry	$\oint \vec{E} \cdot d \vec{A} = E \cdot 4 π r^{2} = \frac{Q_{enc}}{ϵ_{0}}$	Use for uniformly charged spheres, spherical shells, or point charges.
Applying Gauss's Law for Cylindrical Symmetry	$\oint \vec{E} \cdot d \vec{A} = E \cdot 2 π r L = \frac{Q_{enc}}{ϵ_{0}}$	Use for infinite line charges, infinite cylindrical shells, or infinite cylinders with uniform charge density.
Applying Gauss's Law for Planar Symmetry	$\oint \vec{E} \cdot d \vec{A} = 2 E A = \frac{σ A}{ϵ_{0}}$ so $E = \frac{σ}{2 ϵ_{0}}$	Use for infinite plane sheets of charge, or for parallel plate capacitors.
Electric Potential	$V = \frac{W}{q_{0}} = - \int_{\infty}^{r} \vec{E} \cdot d \vec{r}$	Use to find potential due to point charges or continuous distributions, and to find work done.
Potential Due to a Point Charge	$V = \frac{1}{4 π ϵ_{0}} \frac{q}{r}$	Use to find potential at a point due to a single point charge.
Potential Due to a Dipole	$V = \frac{1}{4 π ϵ_{0}} \frac{p \cos θ}{r^{2}}$	Use to find potential at large distances from a dipole.
Equipotential Surface	$Δ V = 0$ on the surface	Use to visualize electric fields and to find work done in moving charges.
Relation Between Electric Field and Potential	$\vec{E} = - \nabla V = - (\frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k})$	Use to find electric field from a given potential function, or to find potential from field by integration.
Conductor in Electrostatic Equilibrium	${\vec{E}}_{inside} = 0$ , $V_{inside} = constant$	Use to analyze conductors in electric fields, such as in capacitors or shielding.
Method of Images for Conductors	$q_{image} = - q$ at mirror location	Use to find electric field and potential for a point charge near a grounded conducting plane.
Capacitance	$C = \frac{Q}{V}$	Use to characterize capacitors in circuits.
Capacitance of a Parallel Plate Capacitor	$C = \frac{ϵ_{0} A}{d}$ (vacuum), $C = \frac{ϵ_{0} ϵ_{r} A}{d}$ (with dielectric)	Use to find capacitance of a parallel plate capacitor.
Energy Stored in a Capacitor	$U = \frac{1}{2} C V^{2} = \frac{1}{2} Q V = \frac{Q^{2}}{2 C}$	Use to find energy stored in a capacitor or to find work done in charging.
Combination of Capacitors: Series and Parallel	$\frac{1}{C_{eq}} = \sum \frac{1}{C_{i}}$ (series), $C_{eq} = \sum C_{i}$ (parallel)	Use to find equivalent capacitance in circuits with multiple capacitors.
Dielectric and Polarization	$\vec{P} = ϵ_{0} χ_{e} \vec{E}$ , $ϵ_{r} = 1 + χ_{e}$	Use to understand the effect of dielectrics on capacitance and electric fields.
Gauss's Law in Dielectrics	$\oint \vec{D} \cdot d \vec{A} = Q_{free, enc}$ , where $\vec{D} = ϵ_{0} ϵ_{r} \vec{E}$	Use to find electric field in dielectrics when free charge distribution is known.

Open the Electrostatics formula page

What JEE asks

Coulomb's Law. JEE often asks to find net force on a charge due to multiple charges using vector addition, or to find equilibrium position of a charge in a system.
Electric Field. JEE often asks to find net electric field at a point due to multiple charges, or to find the point where field is zero.
Electric Field Due to a Point Charge. JEE often combines this with superposition principle for multiple charges, or uses it in problems with dipoles.
Electric Dipole. JEE often asks for torque on a dipole in uniform field, or electric field due to a dipole on axial and equatorial lines.
Electric Field Due to a Dipole. JEE often tests the ratio of fields on axial and equatorial lines, or asks to find net field at a point due to multiple dipoles.
Electric Flux. JEE often asks flux through a surface in a uniform field, or through a closed surface using Gauss's law.
Gauss's Law. JEE often asks to find E due to a uniformly charged sphere, infinite line, or infinite plane using Gauss's law.
Applying Gauss's Law for Spherical Symmetry. JEE often asks to find E inside and outside a uniformly charged sphere, or for a spherical shell with charge on surface.
Applying Gauss's Law for Cylindrical Symmetry. JEE often asks to find E due to an infinite line charge or a cylindrical shell, or to compare with spherical case.
Applying Gauss's Law for Planar Symmetry. JEE often asks to find E due to a single sheet or between two oppositely charged plates, or to find force between plates.
Electric Potential. JEE often asks to find potential due to multiple charges, or to find potential difference between two points.
Potential Due to a Point Charge. JEE often asks to find net potential due to multiple charges, or to find potential at a point on the axis of a dipole.
Potential Due to a Dipole. JEE often asks to find potential at a point on the axis or equatorial line, or to find work done to move a charge in dipole field.
Equipotential Surface. JEE often asks to identify the shape of equipotential surfaces for different charge distributions, or to find potential difference from field.
Relation Between Electric Field and Potential. JEE often asks to find E from V or vice versa, or to find the direction of field from equipotential surfaces.
Conductor in Electrostatic Equilibrium. JEE often asks about charge distribution on conductors, or about field inside a cavity in a conductor.
Method of Images for Conductors. JEE often asks to find the force on a charge near a conducting plane, or the potential at a point.
Capacitance. JEE often asks to find capacitance of parallel plate, spherical, or cylindrical capacitors.
Capacitance of a Parallel Plate Capacitor. JEE often asks to find capacitance with a dielectric slab inserted, or to find the force between plates.
Energy Stored in a Capacitor. JEE often asks to find energy change when a dielectric is inserted or when capacitors are connected.
Combination of Capacitors: Series and Parallel. JEE often asks to find equivalent capacitance of complex networks, or to find charge and voltage on each capacitor.
Dielectric and Polarization. JEE often asks to find the new capacitance when a dielectric is inserted, or to find the induced charge on dielectric surface.
Gauss's Law in Dielectrics. JEE often asks to find E in a dielectric slab between capacitor plates, or to find bound charge density.

Common traps

Forgetting that the force is along the line joining the charges and can be attractive or repulsive. Coulomb's Law
Thinking the field direction is the same as the force on a negative charge; it's opposite. Electric Field
Forgetting the direction: positive charge gives outward field, negative gives inward. Electric Field Due to a Point Charge
Thinking dipole moment direction is from positive to negative; it's from negative to positive. Electric Dipole
Forgetting that these are approximations for $r ≫ d$ . Electric Field Due to a Dipole
Forgetting that area vector is perpendicular to the surface. Electric Flux
Forgetting that the surface must be closed (Gaussian surface). Gauss's Law
Forgetting to consider the enclosed charge correctly for points inside the sphere (only charge inside radius r). Applying Gauss's Law for Spherical Symmetry
Forgetting that the ends of the cylinder contribute zero flux because field is parallel to them. Applying Gauss's Law for Cylindrical Symmetry
Forgetting that the field is independent of distance from the plane. Applying Gauss's Law for Planar Symmetry
Thinking potential is a vector; it's a scalar. Electric Potential
Forgetting that potential is a scalar, so signs of charges matter. Potential Due to a Point Charge
Forgetting that this is valid only for $r ≫ d$ . Potential Due to a Dipole
Thinking electric field is zero on an equipotential surface; it's perpendicular to it. Equipotential Surface

Topics in Electrostatics

Coulomb's Law
Electric Field
Gauss's Law
Electric Potential
Capacitance
Dielectrics

Previous-year questions on Electrostatics

JEEGenie has 680 previous-year questions on Electrostatics. A few worked examples follow.

Year	Questions
2026	39
2025	55
2024	51
2023	53
2022	55
2021	57
2020	42
2019	75
2018	19
2017	12
2016	16
2015	12
2014	13
2013	16
2012	16
2011	8
2010	10
2009	10
2008	10
2007	10
2006	5
2005	8
2004	9
2003	9
2002	7
2001	4
2000	3
1999	3
1998	6
1997	5
1996	4
1995	2
1994	1
1993	2
1992	3
1991	2
1990	2
1989	2
1988	4
1987	3
1986	2
1985	2
1984	2
1983	3
1982	1
1981	3
1980	1
1979	2
1978	1

Question 1 medium

Two charged conducting spheres

S_{1}

and

S_{2}

of radii 8 cm and 18 cm are connected to each other by a wire. After equilibrium is established, the ratio of electric fields on

S_{1}

and

S_{2}

spheres are

E_{S 1}

and

E_{S 2}

respectively. The value of

\frac{E_{S 1}}{E_{S 2}}

is ________.

$\frac{3}{2}$
$\frac{2}{3}$
$\frac{4}{9}$
$\frac{9}{4}$

Answer: D

Explanation. At equilibrium, the potentials of the two spheres are equal:

V_{1} = V_{2} \Rightarrow \frac{k Q_{1}}{R_{1}} = \frac{k Q_{2}}{R_{2}}

So,

\frac{Q_{1}}{Q_{2}} = \frac{R_{1}}{R_{2}}

The surface electric field is

E = \frac{k Q}{R^{2}}

Thus,

\frac{E_{S 1}}{E_{S 2}} = \frac{Q_{1}}{Q_{2}} \cdot \frac{R_{2}^{2}}{R_{1}^{2}} = \frac{R_{1}}{R_{2}} \cdot \frac{R_{2}^{2}}{R_{1}^{2}} = \frac{R_{2}}{R_{1}}

Given

R_{1} = 8 cm

R_{2} = 18 cm

\frac{E_{S 1}}{E_{S 2}} = \frac{18}{8} = \frac{9}{4}

\frac{9}{4}

So the correct option is **D**.

From NTA/IIT papers: JEE Main 2026 · 2 Apr Shift 1

Question 2 easy

A point charge

+ q

is placed at the origin. A second point charge

+ 9 q

is placed at (

d, 0, 0

) in Cartesian coordinate system. The point in between them where the electric field vanishes is:

$(3 d / 4, 0, 0)$
$(d / 4, 0, 0)$
$(4 d / 3, 0, 0)$
$(d / 3, 0, 0)$

Answer: B

Explanation. Let the point where the electric field vanishes be at a distance

x

from the origin along the positive

x

-axis, between the two charges. Thus, its coordinate is

(x, 0, 0)

, where

0 < x < d

. The electric field due to the charge

+ q

at the origin points away from the origin, i.e., in the

+ x

direction at this point. Its magnitude is

E_{1} = \frac{k q}{x^{2}} .

The electric field due to the charge

+ 9 q

(d, 0, 0)

points away from that charge, i.e., toward the origin, so in the

- x

direction at this point. Its magnitude is

E_{2} = \frac{k (9 q)}{(d - x)^{2}} .

For the net electric field to vanish, the magnitudes must be equal:

\frac{k q}{x^{2}} = \frac{9 k q}{(d - x)^{2}} .

Cancel

k q

\frac{1}{x^{2}} = \frac{9}{(d - x)^{2}} .

Taking the positive square root (since distances are positive),

\frac{1}{x} = \frac{3}{d - x} .

Cross-multiply:

d - x = 3 x .

d = 4 x,

x = \frac{d}{4} .

Thus the point is

(\frac{d}{4}, 0, 0) .

Answer: (B)

From NTA/IIT papers: JEE Main 2025 · 2 Apr Shift 1

Question 3 hard

Two beads, each with charge

q

and mass

m

, are on a horizontal, frictionless, non-conducting, circular hoop of radius

R

. One of the beads is glued to the hoop at some point, while the other one performs small oscillations about its equilibrium position along the hoop. The square of the angular frequency of the small oscillations is given by [

ε_{0}

is the permittivity of free space.]

$q^{2} / (4 π ε_{0} R^{3} m)$
$q^{2} / (32 π ε_{0} R^{3} m)$
$q^{2} / (8 π ε_{0} R^{3} m)$
$q^{2} / (16 π ε_{0} R^{3} m)$

Answer: B

Explanation. Free bead equilibrium is diametrically opposite fixed bead (max separation). Hoop center at origin, fixed bead at A=(R,0), free bead at angle pi+

θ

: B=(-Rcos(

θ

),-Rsin(

θ

)). For small

θ

: B-A~=(-2R,-R*

θ

), d~=2R. Coulomb force F=

q^{2}

/(4*pi*eps0*

d^{2}

)~=

q^{2}

/(16*pi*eps0*

R^{2}

), direction ~=(-1,-

θ

/2). Tangent at B ~=(

θ

,-1). Tangential component F_t=F[(-1,-

θ

/2).(

θ

,-1)]=F(-

θ

θ

/2)=-F*

θ

/2, a restoring force. m*R*

θ

''=-F*

θ

/2=-(

q^{2}

/(32*pi*eps0*

R^{2}

))*

θ

θ

''=-(

q^{2}

/(32*pi*eps0*m*

R^{3}

))*

θ

. Comparing to

θ

''=-

w^{2}

θ

w^{2}

q^{2}

/(32*pi*eps0*m*

R^{3}

). Answer: (B)

From NTA/IIT papers: JEE Advanced 2024

Question 4 medium

A dipole comprises of two charged particles of identical magnitude

q

and opposite in nature. The mass 'm' of the positive charged particle is half of the mass of the negative charged particle. The two charges are separated by a distance '

l

'. If the dipole is placed in a uniform electric field '

\overset{ˉ}{E}

'; in such a way that dipole axis makes a very small angle with the electric field, '

\overset{ˉ}{E}

'. The angular frequency of the oscillations of the dipole when released is given by:

$\sqrt{\frac{3 q E}{2 m l}}$
$\sqrt{\frac{4 q E}{m l}}$
$\sqrt{\frac{8 q E}{3 m l}}$
$\sqrt{\frac{8 q E}{m l}}$

Answer: A

Explanation. For small angular displacement

θ

from equilibrium, the restoring torque on the dipole is

τ = - q E l \sin θ \approx - q E l θ .

The moment of inertia about the centre of mass must be used. Let the positive charge (mass

m

) be at distance

x_{1}

from the CM and the negative charge (mass

2 m

) at distance

x_{2}

from the CM. Since the CM is fixed:

m x_{1} = 2 m x_{2}, x_{1} + x_{2} = l ⟹ x_{1} = \frac{2 l}{3}, x_{2} = \frac{l}{3} .

Thus,

I = m x_{1}^{2} + 2 m x_{2}^{2} = m (\frac{4 l^{2}}{9}) + 2 m (\frac{l^{2}}{9}) = \frac{6 m l^{2}}{9} = \frac{2 m l^{2}}{3} .

The equation of rotational motion is

I \ddot{θ} = - q E l θ ⟹ \ddot{θ} + \frac{q E l}{I} θ = 0.

ω^{2} = \frac{q E l}{I} = \frac{q E l}{2 m l^{2} / 3} = \frac{3 q E}{2 m l} .

Therefore,

ω = \sqrt{\frac{3 q E}{2 m l}} .

A

From NTA/IIT papers: JEE Main 2023 · 6 Apr Shift 2

Question 5 easy

The volume charge density of a sphere of radius

6 m

2 μ C {c m}^{- 3}

. The number of lines of force per unit surface area coming out from the surface of the sphere is _______________

\times 10^{10} {N C}^{- 1}

. [Given : Permittivity of vacuum

ϵ_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} - m^{- 2}

)

Answer: 45

Explanation. **Principle:** Number of lines of force per unit surface area is the electric field

E

at the surface (since lines per unit area

= E

in SI, with

ϵ_{0}

normalization). **Charge of sphere:**

Q = ρ V = ρ \cdot \frac{4}{3} π R^{3}

Convert:

ρ = 2 μ C c m^{- 3} = 2 \times 10^{- 6} \times 10^{6} C m^{- 3} = 2 C m^{- 3}

Q = 2 \times \frac{4}{3} π (6)^{3} = 2 \times \frac{4}{3} π (216) = 576 π C

**Field at surface** (Gauss's law, shell point):

E = \frac{Q}{4 π ϵ_{0} R^{2}} = \frac{576 π}{4 π ϵ_{0} (36)} = \frac{4}{ϵ_{0}}

E = \frac{4}{8.85 \times 10^{- 12}} = 4.52 \times 10^{11} N C^{- 1}

**Express in required form:**

E = 45.2 \times 10^{10} \approx 45 \times 10^{10} N C^{- 1}

So lines of force per unit area

= 45 \times 10^{10} N C^{- 1}

. Answer: 45

From NTA/IIT papers: JEE Main 2022 · 25 Jul Shift 1

Frequently asked questions

How much of JEE Main comes from Electrostatics?: JEEGenie estimates about 7% of JEE Main marks come from Electrostatics, based on recent papers. The share varies from year to year.
Which formulas matter most in Electrostatics?: The Electrostatics sheet on this page has 23 formula entries, starting with Coulomb's Law, Electric Field, Electric Field Due to a Point Charge, Electric Dipole, Electric Field Due to a Dipole. Each one comes with when to use it and the trap to avoid.
What topics does Electrostatics cover for JEE?: Electrostatics covers Coulomb's Law, Electric Field, Gauss's Law, Electric Potential, Capacitance, Dielectrics.
How should I revise Electrostatics before the exam?: Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Electrostatics questions on JEEGenie.
How many previous-year questions are there on Electrostatics?: JEEGenie has 680 previous-year questions on Electrostatics, and this page shows a few with worked explanations.
Is practice on Electrostatics free on JEEGenie?: Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.