

Comprehensive Guide to Electrostatics for JEE Main 2025 Preparation
Electrostatics is a crucial chapter in Physics for JEE Main 2025, forming the foundation for understanding electric forces, fields, and potential. This topic explores the behaviour of charges at rest, covering fundamental principles like Coulomb's law, electric field intensity, electric potential, and the concept of capacitance. Electrostatics not only tests your conceptual understanding but also strengthens problem-solving skills critical for competitive exams.
In this guide, we will break down the essential concepts, formulas, and tips to help you master Electrostatics according to the latest JEE Main 2025 Syllabus. This article provides you step-by-step explanations of each concepts and practice problems to boost your preparation.
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Important Topics of Electrostatics Chapter
Coulomb’s Law.
Electric field and electric lines of force.
Electric field due to continuous charge distribution.
Electric dipole and dipole moment.
Gauss Law.
Electric potential and Equipotential surface.
Electric potential due to the various charge distributions.
Capacitance of a capacitor.
Grouping of Capacitance.
Electrostatics Important Concepts for JEE Main
What is Electrostatic?
Electrostatics is the study of electric charges at rest. It involves the interaction between charged particles and the forces and fields they create. Coulomb's law is a fundamental principle in electrostatics that describes the force between two-point charges.
Characteristics of Electrically Charged Objects
The important characteristics of electrically charged objects are-
Like charges repel which means + repels +, - repels -
Unlike charges attract which means + attracts - and vice versa.
There is no net charge on a neutral object which means the charge is conserved. If the fur and plastic rod were neutral at first. After the rod becomes charged by the fur, the negative charge of the fur will be transferred to the plastic rod. The net negative charge on both fur and rod are equal.
Limitations of Coulomb's Law
Coulomb's law can be applied only for the point charges which are at rest.
Coulomb’s law is applicable only in the cases where the inverse square law is obeyed.
Where the charges are in an arbitrary shape, it is difficult to imply Coulomb's law as it is not possible to determine the location between two objects.
Coulomb’s law cannot be directly used to calculate charge on big planets.
Applications of Gauss's Law:
Field Due to Infinitely Long Uniformly Charged Straight Wire: Using Gauss's law, it can be shown that the electric field (E) due to an infinitely long uniformly charged straight wire is inversely proportional to the distance from the wire. Mathematically, E = λ / (2πε₀r), where λ is the linear charge density and r is the distance from the wire.
Uniformly Charged Infinite Plane Sheet: For a uniformly charged infinite plane sheet, the electric field is uniform and perpendicular to the sheet. The electric field (E) can be calculated as E = σ / (2ε₀), where σ is the surface charge density.
Uniformly Charged Thin Spherical Shell: Within a uniformly charged thin spherical shell, the electric field is zero. This result highlights the power of Gauss's law in establishing the electric field behavior in a symmetric system.
List of Important Formulas for Electrostatics Chapter
JEE Main Electrostatics Solved Examples
1. The electric potential in a region is given by V=(6x-8xy2-8y+6yz-4x2) volt. Then the electric field acting on a point charge of 2C placed at the origin will be …
Ans: Given,
The electric potential in a region , V=(6x-8xy2-8y+6yz-4x2) volt
The electric field in the region can be obtained from the electric potential using the formula
$\vec E=\dfrac{\partial V}{\partial x} \hat i+\dfrac{\partial V}{\partial y} \hat j+\dfrac{\partial V}{\partial z} \hat k$
$\vec E=(6-8y^2-8x) \hat i+(-16xy-8+6z) \hat j+6y \hat k$...(1)
To find the electric field at origin, put x=0, y=0 and z=0 in equation (1)
$\vec E=6 \hat i+-8 \hat j+0 \hat k$
$|\vec E|=\sqrt{6^2+(-8)^2}$
$|\vec E|=10~N/C$
The force acting on the charge q due to electric E is given by
$F=qE$
$F=2C\times 10~N/C=20~N$
Key point: Electric field at a point can be obtained by the differentiation of electric potential with respect to distance.
2. A parallel plate capacitor with oil between plates (dielectric constant of oil k=2) has a capacitance C. If the oil is removed, then the capacitance of the capacitor becomes…
Ans: Let A be the area of the capacitor and d be the distance between the plates of the capacitor.
The formula for the capacitance of the parallel plate capacitor with oil as a dielectric is given by,
$C=\dfrac{kA\epsilon_0}{d}$...(1)
The capacitance of the parallel plate capacitor when oil is removed is
$C'=\dfrac{A\epsilon_0}{d}$...(2)
Divide equation (2) by (1) to obtain the new capacitance C’ when oil is removed in terms of initial capacitance C.
$\dfrac{C'}{C}=\dfrac{\left(\dfrac{A\epsilon_0}{d}\right)}{\left(\dfrac{kA\epsilon_0}{d}\right)}$
$\dfrac{C'}{C}=\dfrac{1}{k}$
$C'=\dfrac{C}{k}$
When the oil is removed from the parallel plate capacitor, its capacitance is reduced by a factor of k.
Key point: The capacitance of a parallel plate capacitor depends on the dielectric medium and the dimensions of the capacitor. It does not depend on the charge and potential difference across the capacitor.
Previous Year Questions from JEE Paper
1. Two equal capacitors are first connected in series and then in parallel. The ratio of equivalent capacitances in the two cases will be (JEE 2021)
Sol:
Let the capacitances of each capacitor be C.
In the series connection of two capacitors, the equivalent capacitances is given by

$\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$
$\dfrac{1}{C_{12}}=\dfrac{1}{C}+\dfrac{1}{C}$
$C_{12}=\dfrac{C}{2}$...(1)
For the capacitors in parallel connection, the equivalent capacitance is given by

$C_{eq}=C_{3}+C_{4}$
$C_{34}=C+C$
$C_{34}=2C$...(2)
The ratio of the equivalent capacitances in series and the parallel connection is obtained by dividing equation (1) and equation (2).
$\dfrac{C_{12}}{C_{34}}=\dfrac{\left(\dfrac{C}{2}\right)}{2C}$
$\dfrac{C_{12}}{C_{34}}=\dfrac{1}{4}$
The ratio of the equivalent capacitances of series and parallel connection is 1:4
Key point: In series connection, equivalent capacitance is equal to the reciprocal of the sum of the reciprocal of each capacitance. For parallel connection, equivalent capacitance is equal to the sum of each individual capacitance.
2. A point charge of +12 μC is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in the figure. The magnitude of electric flux through the square will be _____ ✕ 103 Nm2/C. (JEE 2021 Feb)

Sol. To find the electric flux, we use Gauss law and assume a Gaussian surface in the form of a cube of side length 12 cm in such a way that charge is located at the centre of the cube.

Applying Gauss law, total flux linked with the cube is given by
$\phi_{total}=\dfrac{q}{\epsilon_0}$
The electric flux(ɸ) through the square we want to find is 1/6 times the total flux.
$\phi=\dfrac{q}{6\epsilon_0}$
$\phi=\dfrac{12\times10^{-6}}{6\times8.85\times10^{-12}}$
$\phi=226\times10^{3}~Nm^2/C$
Therefore, the flux through the square will be 226 × 103 Nm2/C.
Key point: Gauss Law is applicable to any Gaussian surface irrespective of size and shape.
Practice Questions
1. There is 10 mC of charge at the centre of a circle of radius 10 cm. The work done in moving a charge of 1mC around the circle once is
Answer: 0 J
2. A parallel plate capacitor having a plate separation of 2 mm is charged by connecting it to a 300 V supply. The energy density is
Ans: 0.1 J/m3
JEE Main Physics Electrostatics Study Materials
Here, you'll find a comprehensive collection of study resources for Electrostatics designed to help you excel in your JEE Main preparation. These materials cover various topics, providing you with a range of valuable content to support your studies. Simply click on the links below to access the study materials of Electrostatics and enhance your preparation for this challenging exam.
JEE Main Physics Chapters 2025
JEE Main Physics Study and Practice Materials
Explore an array of resources in the JEE Main Physics Study and Practice Materials section. Our practice materials offer a wide variety of questions, comprehensive solutions, and a realistic test experience to elevate your preparation for the JEE Main exam. These tools are indispensable for self-assessment, boosting confidence, and refining problem-solving abilities, guaranteeing your readiness for the test. Explore the links below to enrich your Physics preparation.
Conclusion
In this article, we're diving into the fascinating world of Electrostatics in physics, a key topic in JEE Main. We'll explore the fundamental concepts, problem-solving strategies, and essential equations. You'll uncover the secrets of electric charges, fields, and forces. Everything you need is right here, so you don't have to search high and low. We provide simple, downloadable PDFs filled with clear explanations and practice questions to supercharge your exam prep. Get ready to ace your exams with a solid understanding of Electrostatics.
Important Study Materials Links for JEE Exams
Electrostatics for JEE Main 2025: Key Concepts and Applications

FAQs on Electrostatics for JEE Main 2025: Key Concepts and Applications
1. What is the expected weightage of Electrostatics in the JEE Main 2026 Physics paper?
Electrostatics is a high-weightage unit in the JEE Main Physics syllabus. Typically, you can expect 2-3 questions from this chapter, which covers topics from electric charges and fields to electric potential and capacitance. This translates to roughly 8-12 marks, making it a crucial chapter for scoring well.
2. Which topics within Electrostatics are most important for JEE Main?
While the entire chapter is important, certain topics are frequently tested in JEE Main. For your 2026 preparation, focus on:
- Gauss's Law and its applications: Finding the electric field for symmetric charge distributions (sphere, cylinder, plane sheet).
- Electric Dipoles: Calculating torque, potential energy, and electric fields in axial and equatorial positions.
- Capacitors: Problems on series/parallel combinations, energy stored, and effects of dielectrics.
- Electric Potential: Calculating potential and potential energy for systems of charges and continuous distributions.
3. What are the most critical Electrostatics formulas for rapid problem-solving in JEE Main?
For efficient problem-solving, you must have these formulas at your fingertips:
- Coulomb's Law: F = k|q₁q₂|/r²
- Electric Field (Point Charge): E = kq/r²
- Gauss's Law: ∮E⋅dA = q_enclosed/ε₀
- Potential Energy of a Dipole: U = -p⋅E
- Capacitance with Dielectric: C = Kε₀A/d
- Energy Stored in Capacitor: U = Q²/2C = ½CV²
Mastering the application of these formulas in various scenarios is key.
4. What types of questions from Electrostatics are commonly asked in JEE Main?
JEE Main questions on Electrostatics often test your conceptual understanding and application skills. Common types include:
- Conceptual MCQs: Based on properties of electric field lines, equipotential surfaces, and behaviour of conductors.
- Numerical Problems: Involving calculation of net force or field due to multiple charges (using the principle of superposition).
- Application of Gauss's Law: Finding the electric field for symmetrical charge distributions.
- Capacitor Circuits: Finding equivalent capacitance, charge, or potential difference in complex circuits, often involving dielectric slabs.
5. How are the concepts of Electrostatics connected to other chapters in the JEE Physics syllabus?
Electrostatics forms the foundation for several other key chapters in Physics. A strong understanding is crucial for:
- Current Electricity: The concept of electric potential difference (voltage) drives the flow of current.
- Magnetism: Moving charges (currents) create magnetic fields, directly linking the two domains.
- Electromagnetic Induction (EMI): Changing magnetic flux induces an electromotive force (EMF), a concept rooted in electric fields and potential.
- Modern Physics: Concepts like the motion of charged particles in electric fields are fundamental to understanding experiments in atomic and nuclear physics.
6. What is a common conceptual trap when calculating the electric field of a charged sphere in JEE Main problems?
A frequent trap is failing to distinguish between a conducting sphere and a non-conducting sphere. For a uniformly charged solid non-conducting sphere, the electric field inside (at distance r from the center) is proportional to r (E ∝ r). However, for a conducting sphere, the electric field inside is always zero, as all excess charge resides on the surface. Misidentifying the type of sphere leads to completely incorrect results.
7. Why is Gauss's Law often preferred over Coulomb's Law for certain JEE problems?
Gauss's Law is preferred for problems with a high degree of symmetry (like spheres, infinite planes, or long wires). While Coulomb's Law requires integrating the vector contributions of infinitesimal charge elements, which can be mathematically complex, Gauss's Law simplifies the calculation by relating the net electric flux through a closed surface to the net charge enclosed. It allows you to find the electric field magnitude without complex vector integration, saving significant time in an exam setting.
8. How should I approach JEE Main problems involving a capacitor with multiple dielectric slabs?
When a capacitor has multiple dielectric slabs, analyse their arrangement:
- If the slabs are placed one after another, filling the space between plates (area is constant, distance is divided), treat the setup as a series combination of capacitors.
- If the slabs are placed side-by-side (distance is constant, area is divided), treat it as a parallel combination of capacitors.
Calculate the capacitance of each section (C = Kε₀A/d) and then find the equivalent capacitance using the rules for series or parallel combinations.
9. What is the physical significance of the negative sign in the formula for the potential energy of a dipole, U = -p·E?
The negative sign signifies the relationship between potential energy and stability. The potential energy is minimum (most negative) when the dipole moment p is aligned with the electric field E (θ = 0°). This is the position of stable equilibrium. The potential energy is maximum when p is anti-parallel to E (θ = 180°), which is the position of unstable equilibrium. The negative sign ensures that the system's potential energy decreases as it moves towards its most stable state.
10. Are there any shortcuts or tricks for solving questions on equipotential surfaces in JEE Main?
Yes, remembering key properties can act as a shortcut:
- The electric field is always perpendicular to the equipotential surface.
- The work done in moving a charge between any two points on the same equipotential surface is zero.
- Equipotential surfaces are closer together where the electric field is stronger and farther apart where it is weaker.
- For a point charge, equipotential surfaces are concentric spheres.
Using these properties, you can often deduce the correct answer without extensive calculations.











