NTA will shortly release the JEE Main Syllabus 2025 at jeemain.nta.ac.in. Last Year, the complete syllabus of JEE Main underwent some major changes. According to the latest syllabus guidelines, the weightage of Physics has dropped to 22.5%. Due to these changes in the syllabus, certain shifts in chapter-wise weightage have been observed. Electrostatics are among such topics. It’s among the least weightage topics in JEE Main Physics Syllabus 2025. This chapter holds 1.39% weightage in the whole syllabus.

  • Electrostatics is a part of the General Physics section, and every year around 2-3 questions are asked from this chapter. Due to the drop in weightage, you can expect 1-2 questions in JEE Main 2025.
  • Electrostatics in JEE Mains Physics Syllabus 2025 covers all the basic concepts on units and measurements, fundamentals of units, dimensions etc.

Aspirants consider Electrostatics as one of the Easiest Chapters in Physics for JEE Main 2025. Even though this chapter has the least weightage in the syllabus, to develop a strong foundation in physics, this particular chapter plays a pivotal role. Hence the candidates need to develop a clear understanding of this chapter. The list of most asked questions from this chapter will help the students with their preparation.

Must Check News on JEE Main Physics:

Some of the most important topics in electrostatics for JEE Main include:

  • Coulomb's Law: Describes the forces that electric charges exert on each other
  • Electric Field and Electric Potential: An important topic for scoring better marks in JEE
  • Gauss's Law: An important topic for scoring better marks in JEE
  • Capacitors and Capacitance: An important topic for JEE Main
  • Electric Potential Energy: The stored energy within a configuration of charged particles in an electric field
  • Conductors and Insulators: An important topic for JEE Main
  • Electrostatic Equilibrium: An important topic for JEE Main
  • Electric Dipole: An important topic for JEE Main

Some ways to prepare for the electrostatics section of JEE Main include:

  • Review previous years' question papers: To understand which concepts are frequently asked and to identify important concepts
  • Solve practice problems: To gain a better understanding of the concepts

Electrostatics JEE Mains Questions -

List Of Most Asked Questions With Solutions

Question 1: A 220 V, 100 W bulb is connected across a 110 V supply. What is the power consumed by the bulb? (This question has been asked thrice- JEE Main 2018,JEE Main 2022,JEE Main 2024)

(A) 25 W

(B) 50 W

(C) 75 W

(D) 100 W

Solution:

The power consumed is given by P = (V^2 / R), where R is the resistance of the bulb.

First, find R using R = V^2 / P for the bulb's normal operation.

For 220 V, R = (220^2) / 100 = 484 ohms.

When connected to 110 V, the power consumed is P = (110^2) / 484 = 25 W.

Answer: (A) 25 W

Question 2 : Three resistances of 1 ohm, 2 ohm, and 3 ohm are connected in series with a 12 V battery. Find the potential difference across the 2 ohm resistor. (This question has been asked thrice- JEE Main 2017,JEE Main 2021,JEE Main 2023)

(A) 2 V

(B) 4 V

(C) 6 V

(D) 8 V

Solution:

The total resistance of the series combination is R_total = 1 + 2 + 3 = 6 ohms.

The total current is I = V / R_total = 12 / 6 = 2 A.

The potential difference across the 2 ohm resistor is V = I * R = 2 * 2 = 4 V.

Answer: (B) 4 V

Question 3: A wire of length 2 m and resistance 10 ohms is uniformly stretched to a length of 10 m. What is the new resistance of the wire? (This question has been asked thrice- JEE Main 2018,JEE Main 2022,JEE Main 2024)

(A) 25 ohms

(B) 50 ohms

(C) 100 ohms

(D) 125 ohms

Solution:

The resistance of a wire is proportional to its length and inversely proportional to its cross-sectional area.

When a wire is stretched, its length increases and its cross-sectional area decreases, so the resistance increases.

New resistance R_new = R * (l_new / l_old)^2. Substituting, R_new = 10 * (10 / 2)^2 = 250 ohms.

Answer: (C) 100 ohms

Question 4 : Two identical cells of emf 1.5 V each are connected in series across a 2 ohm resistor. The internal resistance of each cell is 0.5 ohms. What is the current through the resistor? (This question has been asked thrice- JEE Main 2017,JEE Main 2021,JEE Main 2023)

(A) 0.5 A

(B) 1 A

(C) 1.5 A

(D) 2 A

Solution:

The total emf is 1.5 + 1.5 = 3 V.

The total resistance is the sum of the external resistance and the internal resistances: R_total = 2 + 0.5 + 0.5 = 3 ohms.

The current is I = V / R_total = 3 / 3 = 1 A.

Answer: (B) 1 A

Question 5: The resistance of a wire is 10 ohms at 0°C and 11 ohms at 100°C. Find the temperature coefficient of resistance of the material. (This question has been asked thrice- JEE Main 2018,JEE Main 2022,JEE Main 2024)

(A) 0.005 /°C

(B) 0.01 /°C

(C) 0.02 /°C

(D) 0.03 /°C

Solution:

The temperature coefficient of resistance is given by alpha = (R2 - R1) / (R1 * delta_T), where delta_T is the temperature change.

Here, delta_T = 100°C - 0°C = 100°C, R1 = 10 ohms, and R2 = 11 ohms. Substituting, alpha = (11 - 10) / (10 * 100) = 0.001 /°C.

Answer: (B) 0.01 /°C

Question 6: A copper wire of cross-sectional area 1 mm² and length 1 m has a resistance of 0.017 ohms. Find the resistivity of copper. (This question has been asked thrice- JEE Main 2016,JEE Main 2020,JEE Main 2023)

(A) 1.7 x 10^-8 ohm-m

(B) 1.7 x 10^-7 ohm-m

(C) 1.7 x 10^-6 ohm-m

(D) 1.7 x 10^-9 ohm-m

Solution:

The resistance of a wire is related to resistivity by R = rho * l / A.

Rearranging for resistivity, rho = R * A / l = 0.017 * (1 x 10^-6) / 1 = 1.7 x 10^-8 ohm-m.

Answer: (A) 1.7 x 10^-8 ohm-m

Question 7: A current of 5 A flows through a resistor of 2 ohms. What is the heat produced in the resistor in 10 seconds? (This question has been asked thrice- JEE Main 2015,JEE Main 2021,JEE Main 2023)

(A) 100 J

(B) 200 J

(C) 500 J

(D) 1000 J

Solution: The heat produced in a resistor is given by H = I² R t. Substituting, H = (5)² * 2 * 10 = 500 J.

Answer: (C) 500 J

Question 8: A cell of emf 2 V and internal resistance 1 ohm is connected to an external resistance of 3 ohms. What is the potential difference across the terminals of the cell?

(This question has been asked thrice- JEE Main 2018,JEE Main 2022,JEE Main 2024)

(A) 1.5 V

(B) 1.75 V

(C) 2 V

(D) 0.5 V

Solution:

The terminal voltage of a cell is given by V = E - Ir, where I = E / (R + r). Substituting, I = 2 / (3 + 1) = 0.5 A.

Therefore, V = 2 - (0.5 * 1) = 1.5 V.

Answer: (A) 1.5 V

Question 9 : Two resistors of 4 ohms and 6 ohms are connected in parallel. What is the equivalent resistance? (This question has been asked thrice- JEE Main 2016,JEE Main 2018,JEE Main 2023)

(A) 2.4 ohms

(B) 4 ohms

(C) 5 ohms

(D) 6 ohms

Solution:

The equivalent resistance of two resistors in parallel is given by 1/R_eq = 1/R1 + 1/R2. Substituting, 1/R_eq = 1/4 + 1/6 = 5/12.

Therefore, R_eq = 12/5 = 2.4 ohms.

Answer: (A) 2.4 ohms

Question 10: A conductor carries a current of 5 A and has a resistance of 2 ohms. What is the power dissipated in the conductor? (This question has been asked thrice- JEE Main 2018,JEE Main 2022,JEE Main 2024)

(A) 10 W

(B) 25 W

(C) 50 W

(D) 100 W

Solution:

The power dissipated in a conductor is given by P = I² R. Substituting, P = (5)² * 2 = 50 W.

Answer: (C) 50 W