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Manzil 2026: ELECTRIC CHARGES & FIELDS in 1 Shot | JEE Main & Advanced

1. Summary

Neeraj Kumar Chaudhary Sir covers the Electric Charges and Fields chapter in a single shot, focusing on concepts and problem-solving for JEE Main & Advanced. The video covers Coulombs Law, electric field, electric dipole, electric flux, and Gauss's Law with relevant problems from JEE Main & Advanced.

2. Key Takeaways

* **Focus**: The session is designed for the JEE Main and Advanced exams, covering key concepts and problem-solving strategies.

* **Content Coverage**: The video covers Coulomb's Law, electric field, electric dipole, electric flux, and Gauss's Law.

* **Approach**: The lecture emphasizes conceptual understanding and problem-solving, with a focus on high-probability topics for the exams.

* **Resources**: Manzil lecture notes and DPP are available.

* **Benefit of Using Pi**: App benefits are highlighted, like a distraction-free environment and access to class notes, DPPs, and PYQ sessions.

* **Emphasis**: Emphasis on building conceptual clarity, applying formulas effectively, and solving problems related to electric charges, fields and Gauss law.

* **Time Duration**: The total duration may be seven to eight hours.

3. Detailed Notes

3.1. Introduction (00:00:00-00:00:06)

* Brief intro and warm up with students.

* Chapter focus is electric charges and fields, a foundational topic for JEE Main and Advanced exams.

* Importance of the chapter with focus on scoring +8 number.

3.2. Topics to be Covered (00:00:04-00:00:09)

* Overview of the topics to be covered in the session.

* Dividing the topics into sections to perfect them.

* Aiming to help students succeed in JEE exams.

* Five Main topics for the section.

* Coulomb's Law.

* Electric Field and Field Line.

* Motion of a Charged Particle in an Electric Field.

* Electric Dipole

* Electric Flux & Gauss Law

3.3. Coulomb's Law (00:00:10-00:01:11)

* Coulomb's Law: Describes the force between two point charges.

* *Statement of Coulomb's Law*: If there are two charges q1 and q2, then force will be exerted by both charges each other as per Newton's Law (Action - Reaction).

* *Statement*: Force on one due to charge 2 is equal and opposite to force on charge 2 due to 1.

* F12 = -F21 (Direction reversed).

* *The force equation is*: F = k * q1 * q2 / r²

* *Magnitude equation is*: F = kq1q2/r².

* *The value of coefficient k is*: k= 9 x 10^9.

* *Vector Form* : F21 = k * q1 * q2 / r² * r (Unit Vector) and F12= -k * q1 * q2 / r² * r (Unit Vector).

* Newton's Gravitational Law: Comparison with Coulomb's Law and the Always Attractive nature of the Gravitational Force.

* *Force equation for Gravitational Force*: gm1m2 / r²

3.4. Question 1 (00:00:20-00:00:24)

* **Question** : *[JEE Main]* Two point charges have same distance in air. The Individual charges are doubled, What will happen to the force between them?

* **Answer:** The force will remain same.

3.5. Question 2 & 3 (00:00:24-00:00:31)

* **Question** : *[JEE Main]* Two identical conducting spheres with charge Q repel each other with a force of 60 N. A third identical uncharged conducting sphere R is successively brought into contact with both spheres. Find the new force.

* **Solution:**

* The final answer is 2.

* **Question** : Two point charges of 20 and -5 and are placed in some distance. At what position we place the third charge so that net force will be zero?

* **Solution:** The Answer is Q in external (left side) of Q1 (20).

3.6. Concept: (00:00:31-00:00:38)

* *Concept* If the force acts on q, then to make the force = 0 , the force should cancel out each other.

3.7. Effect of Medium on Electric Force (01:03:50-01:04:40)

* Coulomb's Law stays true.

* *Effect of Medium* : Mediums introduce a force that counteracts electric force. (Force reduces in a medium.

* *Electric Field is also known as*: e = F / q Not

* *Dielectric Constant and the Medium Effect*: The Electric Field in any medium reduces to E / K , where K is the dielectric constant.

3.8. Question 6 (01:14:30-01:15:40)

* **Question** : Two point charges are in air. When in a different medium (dielectric constant), The electrostatic force reduce to 6N from 16N. Find the dielectric constant?

* **Solution**:

* Answer: e = kq1q2/r² and fm = fe/k so k= fe/ fm = 16/6 = 8/3.

3.9. Important Concept: (01:19:50-01:20:00)

* Electric Forces depend on r and the magnitude of q1 and q2.

3.10. Question 7 (01:31:00-01:31:31)

* **Question**: If two identical ball with charge q and hanging distance l, then which of the following statements correct?

* **Solution**: The Answer will be q1 is small then q2 .

3.11. SHM in Electrostatics (01:42:20-01:42:35)

* If you give a small displacement, the net force will lead to equilibrium.

* *Condition of Equilibrium is*: when the force is zero.

3.12. Question 8 (01:57:00-02:01:45)

* **Question** : *[JEE Main]* When to balls with mass m. In the electric field and hanging with a silk thread , find Q .

* **Solution**:

* Tan Theta = fe / mg.

* Fe = kq1q2 / r2.

* X = sin theta * 2l.

* Answer q / 4*pi*e 0* .

3.13. Question 9 (02:01:45-02:09:00)

* **Question** : *[JEE Main]* Find the value of q. Where electrical field is zero.

3.14. Question 10 & 11 (02:09:00-02:11:30)

* **Question** : Two point charges of Q and 3Q are kept at the distance r. Find the point where field intensity = 0?

* **Answer:** d / 1+ sqrt of 3

3.15. Question 12 (Concept) (02:16:30-02:16:43)

* *Concept of a Cylindrical Region*: Draw the line. Then q comes on the same line. It will be the Equipotential surface.

3.16. Electric Field (02:30:00-02:31:00)

* Electric Field

* *Electric field direction* - Positive charge, electric field outward and in case of negative charge electric field inward.

* *Formula of electric field* - e = q / 4π 0 * R² .

3.17. Electric Field due to point charge (02:45:25-02:45:54)

* *Electric Field Due to Point Charge* - E= kq/r^2 and * Direction of electric field* = along the direction of Force.

3.18. Electric Field due to Rod (02:52:04-02:53:10)

* Electric Field due to Rod.

* *Important Concept*: The formula for electric field is given by E = k* Lambda / R* (Sin(Alpha) + sin (Beta).

3.19. Ring (03:04:05-03:04:17)

* Electric Field Due to Ring.

* *Formula* : E = k * q * x / (r² + x²)^3/2

3.20. Circular Arc (03:19:40-03:20:00)

* Electric Field due to Circular Arc.

* *Formula* : E = 2k * lambda / r * sin(phi / 2)

3.21. Circular Arc (03:19:40-03:20:00)

* *Electric Field due to Circular arc is*: E= 2kλ / R (Sin alpha/2)

* *Important Point*: The total electric field due to arc is horizontal.

3.22. Disc (03:27:30-03:27:54)

* If the area of the disc is A. The electric field will be 2π.

3.23. Questions (03:45:56-04:10:00)

* **Question**: *[JEE Main]* Two charges are separated by d. In a horizontal plane with two springs.

* *If q and 3q*

* *Solution*: x = a/2

3.24. Electric Field Lines (04:35:14-04:35:37)

* Electric Field Lines.

* *Field lines are always start from the positive and end on the negative charge*.

3.25. Question (04:55:54-05:07:28)

* *Question* : Two charges are placed in different positions and the lines suggested that determine the answer and q1> q2.

3.26. Motion of a Charged Particle (05:07:28-05:07:40)

* *If the charge is positive it will move in the same direction of electric field*.

3.27. Question (05:11:55-05:11:59)

* If a negative charge is placed near the metal. Find the correct option.

* *Answer: A is correct.*

3.28. Electric Dipole (05:27:02-05:27:06)

* *The formula to calculate dipole* - q* d

3.29. Dipole in Uniform Electric Field (05:35:45-05:35:47)

* *Dipole has a tendency to align in the direction of the field*

3.30. Electric Field due to a Dipole (05:48:40-05:49:12)

* The Electric field is 2 * P / R²

* *The axis of a dipole* - Negative to Positive.

* *The direction of Electric Field* - In same direction of Force.

3.31. Questions (05:59:32-06:04:19)

* **Question**: If a charge is kept at the origin then which of the following is equal?

* **Answer**: This (D is correct).

3.32. Electric Flux (06:25:10-06:25:25)

* *Electric Field* = e.a.

* *Electric field means*: E.A

3.33. Number of Field lines is Flux (06:36:14-06:36:38)

* The number of field lines equals the flux.

* *Therefore*: number of field lines = flux.

3.34. Question (06:49:14-06:50:21)

* **Question**: The flux through the total closed surface. If any.

* *Answer*: (Q/E 0) .

3.35. Gauss Law (07:28:20-07:28:24)

* *The value of Gauss Law is*: = Q / 0.