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Electrostatics Chapter - Physics JEE Main

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Last updated date: 23rd Apr 2024
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Concepts of Electrostatics for JEE Main Physics

Electrostatics is a fundamental topic that spans concepts introduced in both Class 8 and Class 12. In Class 8, students are initiated into the basics of electrostatics, grasping the concept of electric charge and the fundamental law governing charges. Progressing to Class 12, Electrostatics delves deeper, exploring Coulomb's Law, electric field, and potential. Electrostatics in Class 12 students further delve into Gauss's law, capacitance, and the equilibrium of conductors and insulators in electrostatics. A comprehensive understanding of electrostatics is imperative for tackling advanced JEE Main problems related to electromagnetism and lays the groundwork for more intricate physics concepts.


The Electrostatics chapter commences by elucidating electrostatics, electric charge, and Coulomb’s law. It progresses to addressing problems related to Coulomb's law, electric potential, and electric field due to point charges and various charge distributions. Examples illustrating electrostatic effects in daily life are also examined. The chapter explores the torque and potential energy of an electric dipole, applying Gauss's law to calculate electric flux. JEE often features questions on capacitor types, alongside problems involving series and parallel combinations. Additionally, the impact of a dielectric medium in a capacitor is discussed. This section defines electrostatic force, outlines essential JEE Main and JEE Advanced formulas for electrostatics, and offers solved examples.


JEE Main Physics Chapters 2024


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.


Name of the Concept

Key Points of the Concept

1. Electric Charges: Conservation of Charge

  • Electric charge, also known as charge or electrostatic charge, is defined as the basic property of subatomic particles that causes them to experience a force when placed in an electromagnetic field. In general, electric charges are of two types – positive carried by the charge carriers named protons and negative by charge carriers termed as electrons. If the net charge of an object is equal to zero, i.e. neither positive nor negative, then it is said to be neutral. Electric charge is symbolized as Q and measured using a coulomb.  

  • S.I Unit of Charge is Coulomb. 

  • Positively Charged Particles:

In this type of particle, numbers of positive ions (protons) are larger than the numbers of negative ions (electrons). To neutralize positively charged particles, electrons from the surroundings come to this particle until the number of protons and electrons become equal.

  • Negatively Charged Particles:

Similarly, electrons are larger in number than that of protons. To neutralize negatively charged particles, electrons move to the ground or any other particle around until the number of protons and electrons become equal. 

2. Coulomb’s Law

  • Coulomb’s law gives us the magnitude of the force acting between two charges placed at a distance. Coulomb's law defines the magnitude of electrostatic force acting between point charges.

  • Coulomb’s law states that force acting between two charges is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between them.

$F\propto\dfrac{q_1q_2}{r^2}$

$F=k\dfrac{q_1q_2}{r^2}$

Where q1 and q2 are the two charges placed at a distance r between them.

$k=\dfrac{1}{4\pi\epsilon}$

  • K is the Coulomb’s constant and ε is the permittivity of the medium and the value of permittivity of vacuum is 8.85 × 10-12 Nm2/kg2

3. Coulomb’s First Law

  • Bodies with like charges repel each other, and bodies with unlike (opposite) charges, attract each other.

4. Coulomb’s Second Law

  • The force, whether of attraction or repulsion, between two charged bodies is directly proportional to the product of their charges and inversely to the square of the distance amid them. 

  • According to the second law, 

F ∝ Q1 Q2 and F ∝ 1/d2

Hence, F ∝ ((Q1 Q2) / d2)

F = k ((Q1/Q2)/d2)

Where,

Q1 and Q2 = Charges of the charged bodies

d = distance amid the centre of the two charged bodies

k = constant based on the medium in which the bodies are positioned

F = Force of attraction or repulsion between the charged bodies

Note: Both the S.I and M.K.S system, k = 1/4 εoεr and the value of εo =8.854 x 10⁻¹² C²/Nm². The value of εr that changes with change in medium is 1 in vacuum and air.

5. Principle of Coulomb's Law

  • To understand the principle of Coulomb's Law, suppose you have two bodies, out of which one is positively charged, and the other is negatively charged. In this case, the two bodies will attract each other as they have opposite charges. Now, if you increase the charge of one body, leaving the other one as it is, then the force of attraction will increase. Hence, we can conclude that the force amid the charged bodies is directly proportional to the charge of the bodies. Now, keeping the charge Q1 and Q2 of the two bodies constant, if you bring them closer, the force amid them will increase. However, if you place them far from each other, then the force will decrease. So, we can say that the force between two charged bodies is inversely proportional to the distance between them. 

  • Remember that the force developed between two charged bodies isn't the same in all the mediums, and vary with the mediums.

6. Electric field and electric lines of force

  • Electric field intensity (E) at a point is the electrostatic force acting on a unit test charge placed in the electric field.

$E=\dfrac{F}{q}$

$E=\dfrac{1}{4\pi\epsilon}\dfrac{q_1q_2}{r^2}$

  • Electric lines of force are the lines that represent the electric field at that point.

  • Electric field lines never intersect each other and the tangent to the electric field lines at a point gives the direction of the electric field at that point.

7. Electric field due to continuous charge distribution

  • Electric field due to infinitely long charged wire having linear charge density 𝜆 at a distance r is 

$E=\dfrac{1}{4\pi\epsilon}\dfrac{2\lambda}{r}$

  •  Electric field due to a uniformly charged  disc having radius R and surface charge density 𝜎 along its axis is


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$E=\dfrac{\sigma}{2\pi\epsilon_0}\left(1-\dfrac{z}{\sqrt{z^2+r^2}}\right)$

  • Electric field due to a charged infinite thin plane sheet having surface density 𝜎


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$E=\dfrac{\sigma}{2\pi\epsilon_0}$

8. Electric dipole and dipole moment

  • Electric dipole is a pair of equal and opposite charges separated by a small distance.

  • Magnetic dipole moment (P) is the product of the magnitude of one charge(q) and the distance between them(d). 


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$\vec P=q\times \vec d$


  • It is a vector quantity and its direction is from negative charge to positive charge.

  • Torque experienced by a dipole in a uniform electric field is 

$\vec\tau=\vec P\times \vec E$

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  • Potential energy of dipole in a uniform electric field is 

$U=-\vec P\cdot \vec E$

$U=-P E\cos\theta$

9. Torque on a Dipole in a Uniform Electric Field

  • When an electric dipole is placed in a uniform electric field, it experiences a torque. The torque is calculated using the formula:

  • τ = p * E * sin(θ)

Where 

τ is the torque

p is the dipole moment

E is the electric field strength, 

θ is the angle between the dipole moment and the electric field. 

10. Electric Flux

  • Electric flux is a fundamental concept in electrostatics that measures the flow of electric field lines through a given surface. It is denoted by the symbol Φ (phi) and is defined as the dot product of the electric field (E) and the surface area vector (A) over a closed surface:

Φ = ∫∫E · dA

  • In simpler terms, electric flux quantifies the number of electric field lines passing through a particular area. It can be positive, negative, or zero depending on the orientation of the electric field and the surface.

11. Gauss Law

  • According to Gauss Law, the net flux linked with a closed surface is equal to 1/ε0 times the net charge enclosed inside the surface.

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$\oint\vec E\cdot d\vec s=\dfrac{q_{enclosed}}{\epsilon_0}$



12. Electric potential and Equipotential surface

  • Electric potential at a point is the amount of work done (W) in bringing a unit positive charge from infinity to that point along any arbitrary path. It is a scalar quantity.

$V=\dfrac{W}{q}$

  • Electric potential due to a point charge (q) at a distance r is

$V=\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{r}$

  • Electric potential energy of two charge systems at a distance r between them is

$U=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r}$ 

  • In an equipotential surface, every point has the same potential.

Example: Surface of a charged surface

13. Electric potential due to various charge distribution

  • Electric potential due to a charged ring of radius R along its axis at a distance x from its centre

$V=\dfrac{1}{4\pi\epsilon_0}\left(\dfrac{Q}{\sqrt{x^2+R^2}}\right)$

  • Electric potential due to a charged sphere having radius R and charge Q

At any point Inside the sphere: 

$V=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R}$

On the surface of the sphere:

$V=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R}$

At a point outside the sphere at a distance r from the centre of the sphere.

$V=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r}$

14. Conductors and Insulators

  • In the world of electrostatics, materials can be broadly classified into two categories: conductors and insulators. Conductors are materials that allow electric charges to move freely within them. This is due to the presence of loosely bound electrons that can easily migrate in response to an applied electric field. Metals like copper and aluminum are excellent examples of conductors.

  • On the other hand, insulators are materials that do not allow the free movement of electric charges. In insulators, electrons are tightly bound to their atomic nuclei, making it difficult for them to move. Materials like rubber, plastic, and glass are common insulators.

15. Dielectrics and Electric Polarization

  • Dielectrics are a type of insulator that play a crucial role in the field of capacitors. When a dielectric material is placed between the plates of a capacitor, it becomes electrically polarized. This means that the electric charges within the dielectric material shift slightly in response to the electric field, creating temporary dipoles. This polarization helps to increase the capacitance of the capacitor and can store more electric charge.

16. Capacitance of a capacitor

  • The ability to store charge is called capacitance. Its unit is farad. The capacitance of a capacitor is given by

$C=\dfrac{Q}{V}$

  • Capacitance of parallel plate capacitor having area A and distance between the plate d is

$C=\dfrac{kA\epsilon_0}{d}$

  • Increase in dielectric constant (k) of a dielectric medium increases the capacitance.

17. Grouping of Capacitance

  • Capacitors in series:

The charge stored in capacitors in series combination is the same but the potential difference across each capacitor is different.


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$\dfrac{1}{c_{eq}}=\dfrac{1}{c_1}+\dfrac{1}{c_2}+\dfrac{1}{c_3}$

  • Capacitors in parallel

The potential across each capacitor in parallel is the same and the charge will be different.

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$c_{eq}=c_{1}+c_{2}+c_{3}$

18. Capacitance of a Parallel Plate Capacitor with and without Dielectric Medium

  • The capacitance of a parallel plate capacitor can be calculated using the formula 

C = ε₀ * (A / d)

Where,

C is the capacitance

ε₀ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates. 

  • When a dielectric material is introduced between the plates, the capacitance increases by a factor equal to the material's relative permittivity (k). This can be expressed as C = k * ε₀ * (A / d).


Characteristics of Electrically Charged Objects

The important characteristics of electrically charged objects are-


  1. Like charges repel which means + repels +, - repels -

  2. Unlike charges attract which means + attracts - and vice versa.

  3. 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

  1. Coulomb's law can be applied only for the point charges which are at rest.

  2. Coulomb’s law is applicable only in the cases where the inverse square law is obeyed.

  3. 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.

  4. 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

Sl. No

Name of the Concept

Formula

1.

Relation between electric field and electric potential

  • $\vec E=\dfrac{\partial V}{\partial x} \hat i+\dfrac{\partial V}{\partial y} \hat j+\dfrac{\partial V}{\partial z} \hat k$

  • $\vec E=E_x \hat i+E_y \hat j+E_z \hat k $

  • $V=\int_{x_1}^{x_2} E_x +\int_{y_1}^{y_2} E_y +\int_{z_1}^{z_2} E_z  $

2.

Electric field due to a charged conducting sphere having charge (Q) and radius R

  • At a point inside the sphere,

$E_{inside}=0$

  • At a point on the surface of the sphere,

$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R^2}$

  • At a point outside the sphere at a distance r from the centre of the sphere,

$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r^2}$

3.

Electric field due to a charged non-conducting sphere having volume  charge density (⍴) and radius R

  • At a point inside the sphere and at a distance r from the centre

$E=\dfrac{\rho r}{3\epsilon_0}$

  • At a point on the surface of the sphere

$E=\dfrac{\rho R}{3\epsilon_0}$

  • At a point outside the sphere and at a distance r from the centre of the sphere

$E=\dfrac{\rho R^3}{3\epsilon_0r^2}$

4.

Electric field due to a dipole having dipole moment P

  • Electric field due to a dipole at an axial point at a distance r from its centre

$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{2P}{r^3}$

  • Electric field due to a dipole at an equatorial point at a distance r from its centre

$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{P}{r^3}$

5.

Electric potential due to a dipole

  • Electric potential due to a dipole having dipole moment (P) at a distance r from the centre is

$V=\dfrac{1}{4\pi\epsilon_0}\dfrac{P\cos(\theta)}{r^2}$ 


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6.

Energy stored in a capacitor

$U=\dfrac{1}{2}CV^2$

$U=\dfrac{1}{2}QV$

$U=\dfrac{Q^2}{2C}$

Where Q is the charge stored, V is the potential difference and C is the capacitance.

7.

Energy density of a capacitor

$E.D=\dfrac{1}{2}\varepsilon_0E^2$

Where E is the electric field.

8.

Work done in rotating a dipole in an electric field from θ1 to θ2

$W=PE(\cos\theta_1-\theta_2)$


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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


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$\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$

$\dfrac{1}{C_{12}}=\dfrac{1}{C}+\dfrac{1}{C}$

$C_{12}=\dfrac{C}{2}$...(1)

For the capacitors in parallel connection, the equivalent capacitance is given by


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$C_{eq}=C_{3}+C_{4}$

$C_{34}=C+C$

$C_{34}=2C$...(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)


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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.


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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 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.

FAQs on Electrostatics Chapter - Physics JEE Main

1. What is the weightage of the electrostatics in JEE?

Electrostatics is a vast chapter compared to others. Around 2-3 questions covering about 8-12 marks are asked in JEE from various parts of the chapter. 

2. Which section has to be given more importance in the chapter on electrostatics?

Since electrostatic is a vast topic, there are many important concepts here in this chapter. However, more focus should be given to the electric fields of various charge distribution, Gauss law and capacitance to score better marks in JEE

3. How to score good marks in JEE from the chapter electrostatics?

If we go through the last 20 years’ question papers, we can understand the important concepts of the chapter from where the questions are asked frequently. Understanding these concepts in-depth and then solving practice problems is the best way to score good marks in JEE for the chapter electrostatics.

4. What is electrostatics in Class 8?

Electrostatics Class 8 introduces students to the fundamental principles of static electricity. In this grade, students explore the behavior of charged particles and learn about the concept of electrostatics class 8. They study how objects become charged, the types of charges (positive and negative), and the basic properties of charged objects. The curriculum in electrostatics class 8 typically covers topics like charging by friction, conduction, and induction. Students also delve into the concept of electric fields and gain an understanding of the attractive and repulsive forces between charged objects.

5. What is Electrostatic potential energy in JEE Main Physics?

In JEE Main Physics, Electrostatic Potential Energy denotes the stored energy within a configuration of charged particles in an electric field. The formula for electrostatic potential is $\(U = \frac{k \cdot |q_1 \cdot q_2|}{r}\)$, where $\(k\)$ is Coulomb's constant, it signifies the work done in arranging charges. Understanding this energy is pivotal in analyzing charged systems, electric potential, and diverse electromagnetic phenomena.

6. What is the electric potential formula for JEE Main?

The electric potential formula often used in JEE Main Physics is given by:

$V = \frac{k \cdot q}{r}$


  • V represents the electric potential.

  • k is Coulomb's constant.

  • q denotes the charge producing the electric potential.

  • r stands for the distance from the charge producing the potential.

7. How to calculate electric potential energy in JEE Main?

To calculate electric potential energy in JEE Main, you can use the formula:


$U = \frac{k \cdot |q_1 \cdot q_2|}{r}$


Where:

$U \text{ is the electric potential energy.}$


$k \text{ is Coulomb's constant } (8.9875 \times 10^9 \, \text{Nm}^2/\text{C}^2 \text{ in a vacuum).}$


$q_1 \text{ and } q_2 \text{ are the magnitudes of the charges involved.}

r \text{ is the distance between the charges.}$


Substitute the given values into the formula to determine the electric potential energy between two point charges. Understanding this calculation is essential for solving problems related to charges, electric fields, and interactions between charged particles in JEE Main Physics.