JEE Main Electrostatics Important Questions

JEE Main 2024: Electrostatics is an important chapter in JEE Main 2024 Syllabus Physics in terms of weightage. The JEE or Joint Entrance Examination (Main) is conducted for admission to undergraduate engineering and architecture courses in various colleges in India. Electrostatics is a fundamental concept in Physics that is commonly covered in IIT JEE. It is a vital chapter for understanding other related topics in Physics, such as capacitors and electric fields. Candidates preparing for JEE Main 2024 can find a list of JEE Main Electrostatics Important Questions along with the solutions on this page, which will further help them practice and ace their exam preparation.

The National Testing Agency (NTA) has released the JEE Main 2024 exam dates. Students should know that the JEE Main 2024 Session 1 exam will be conducted on January 24, 25, 27, 28, 29, 30, 31 & February 1 , 2024, while the JEE Main 2024 Session 2 exam is scheduled to take place between April 1 & 15, 2024 .

Latest: JEE Main 2024 Daily Practice Questions for 22 September 2023

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Read this article to learn more about JEE Main Electrostatics Important Questions. We have listed down the major topics and chapter-wise weightage to help the students understand the difficulty level better. Alongside, JEE Main Previous Year’s Question Papers are also available here.

JEE Main 2024 Electrostatics – Important Topics

While practising the JEE Main Electrostatics questions, students must be aware of the important topics. These topics have been shared based on the previous year’s question paper analysis and trends. The important topics for JEE Main Electrostatics are also covered in the curriculum of Classes 11 & 12, so students are not completely new to the chapters. Check out the important topics to study for IIT JEE Electrostatics here –

• Coulomb's law

• Conservation of Charge

• Forces between multiple charges

• Electric Field lines

• Electric Field

• Electric Dipole

• Electric charges due to dipole

• Electric Flux

• Torque

• Gauss’s Law

• Electrostatic field

• Electrical Potential energy of the system

• Conductors and insulators

• Capacitors and a combination of capacitors

• Dielectrics and electric polarization

• Capacitance

• Energy

Also Read: JEE Main 2024: Strategy to Score 90+ in Physics

JEE Main 2024 Electrostatics – Year-wise Weightage, No. of Questions

In the last five years, questions have been consistently asked from IIT JEE Electrostatics topics, such as Coulomb’s Law, Electric Charge, Conductors and insulators, and so on. Keeping up with the trend, candidates can expect at least 2 questions from the list of JEE Main Electrostatics important topics this year. Since each question carries 4 marks as per the JEE Main marking scheme, candidates can score 4-8 marks easily in this section. Take a look at the year-wise weightage of questions and marks here –

Year	Average No. of Questions Asked	Marks Allotted to Each Question	Total Marks Allotted
2023	2	4	8
2022	1	4	4
2021	1	4	4
2020	2	4	8
2019	1	4	4

JEE Main 2024 Electrostatics Important Questions

JEE Main Electrostatics important questions include Integer Questions and Multiple-Choice Questions (MCQs), allowing candidates to assess their knowledge at different levels. The comprehensive explanations and solutions PDF help the students understand the correct techniques and strategies when solving the questions. These questions on Electrostatics incorporate the JEE Main previous year’s questions, providing candidates with enough opportunities to practice harder and excel in the upcoming exam.

Go through the JEE Main Electrostatics important questions and solutions here –

Q1: Three charges +Q, q, +Q are placed respectively, at distance 0, d/2 and d from the origin, on the x-axis. If the net force experienced by +Q placed at x = 0 is zero, then value of q is
(a) +Q/4
(b) –Q/2
(c) +Q/2
(d) –Q/4

Solution
QQ/d2 + Qq/(d/2)2 =0
Q + 4q = 0
or q = -Q/4

Answer: (d) –Q/4

Q2: An electric field of 1000 V/m is applied to an electric dipole at an angle of 45°. The value of the electric dipole moment is 10–29 Cm. What is the potential energy of the electric dipole?
(a) –10 × 10–29 J
(b) –7 × 10–27 J
(c) –20 × 10–18 J
(d) –9 × 10–20 J

Solution
E = 1000 V/m , p = 10-29 cm, θ = 450
Potential energy stored in the dipole,
U = -p.Ecos θ = – 10-29 x 1000 x cos450
�=−12×10−26
U = – 0.707 x 10-26 J= -7 x 10 -27 J

Answer: (b) –7 × 10–27 J

Also Read: JEE Main 2024: List of Chapters & Topics Covered Under Physics Section

Q3: A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is
(a) 508 pJ
(b) 692 pJ
(c) 560 pJ
(d) 600 pJ

Solution
Initial Energy of the capacitor, Ui = (1/2) CV2
= (1/2) x 12pF x 10 x 10
= 600 pJ
After the slab, the energy of the slab, Uf = (1/2) Q2/C’
Q = CV = (12 pF)(10 V) = 120 p C
C’ = kC = 6.5 x 120 x 10-12 F
Therefore, Uf = [(1/2) (120 x 10-12)2]/[6.5 x 120 x 10-12]
Uf = 92 pJ
W + Uf= Ui
⇒ W = Ui – Uf
= 600 pJ – 92 pJ
= 508 pJ

Answer: (a) 508 pJ

Q4: Voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of 106 V m–1. The plate area is 10–4 m2. What is the dielectric constant if the capacitance is 15 pF? (given ε0 = 8.86 × 10–12 C2 N–1 m–2)
(a) 3.8
(b) 8.5
(c) 6.2
(d) 4.5

Solution
C = Kε0A/d and V = Ed
Or K = CV/ε0AEmax
K = (15 x 10-12 x 500)/(8.86 x 10-12 x 10-4 x 106) = 8.5

Answer: (b) 8.5

Q5: The bob of a simple pendulum has a mass of 2 g and a charge of 5.0 C. It is at rest in a uniform horizontal electric field of intensity 2000 V m–1. At equilibrium, the angle that the pendulum makes with the vertical is (take g = 10 m s–2)
(a) tan–1 (0.2)
(b) tan–1 (0.5)
(c) tan–1 (2.0)
(d) tan–1 (5.0)

Solution
The forces acting on the bob are its weight and the force due to field.
At equilibrium,
Tcosθ = mg ——-(1)
Tsinθ = qE ——–(2)
Dividing (2) by (1)
tanθ = qE/mg
θ = tan-1((5 x 10-6 x 2 x 103) / (2 x 10-3 x 10)) = tan-1(0.5)

Answer: (b) tan-1(0.5)

Also Read: List of Best Books & Online Study Materials for JEE Main 2024 Physics

JEE Main Electrostatics Previous Year Question Papers – Download PDF

Candidates can review and download the JEE Main Electrostatics questions and solutions from the previous years in PDF format to practice regularly. Click on the ‘Download PDF’ link shared below to access the questions and answers for free.

JEE Main Electrostatics Questions with Solutions – Download PDF

(Credit: Byju’s)

JEE Main Electrostatics Previous Year Question Papers – Download PDF

JEE Main 2024 aspirants are also advised to check the JEE Main Physics question papers from the previous year to gain a comprehensive idea of the probable questions. We have provided the day-wise and shift-wise PDF links for reference below –

JEE Main 2023 Exam Date	Shift 1 Download Link	Shift 2 Download Link
Session 1 – January 24, 2023	JEE Main Physics Question Paper Shift 1	JEE Main Physics Question Paper Shift 2
Session 1 – January 25, 2023	JEE Main Physics Question Paper Shift 1	JEE Main Physics Question Paper Shift 2
Session 1 – January 29, 2023	JEE Main Physics Question Paper Shift 1	JEE Main Physics Question Paper Shift 2
Session 1 – January 30, 2023	JEE Main Physics Question Paper Shift 1	JEE Main Physics Question Paper Shift 2
Session 2 – April 6, 2023	JEE Main Physics Question Paper Shift 1	JEE Main Physics Question Paper Shift 2
Session 2 – April 8, 2023	JEE Main Physics Question Paper Shift 1	JEE Main Physics Question Paper Shift 2

(Credit: Aakash Byju’s; Collegedunia)

Importance of Solving JEE Main Electrostatics Important Questions

Solving important questions for JEE Main Electrostatics can prove beneficial in more ways than one. Some of the major advantages of practising the previous year’s or JEE Main Sample Papers 2024 are listed below –

• Helps you gauge the topic-wise weightage, difficulty level ad nature of the questions

• Helps you identify your shortcomings and learn from your mistakes

• Makes you familiar with the actual question pattern so you know what to expect during the test

• Helps you improve accuracy and speed as you learn the art of time management while practising the papers

• Helps boost confidence

We hope this article has been helpful in providing you with insights into the JEE Main Electrostatics Important Questions. Candidates are also suggested to take JEE Main 2024 Mock Test and refer to the best books for JEE Main Electrostatics to nail their exam preparation. All the best for JEE Main 2024!

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