||ELECTRIC FIELD DUE TO CHARGED “DISC” || #jee #jeemains #jeeadvanced #study #motivation

📚 Electric Field Due to a Uniformly Charged Disc | Class 12 Physics | Electrostatics In this lecture, we derive and understand the electric field intensity due to a uniformly charged disc using the concept of continuous charge distribution. This topic is extremely important for CBSE Class 12 Physics, JEE Main & Advanced, NEET, and other competitive examinations. 🔹 Concept of surface charge density (σ) 🔹 Elemental ring method 🔹 Step-by-step mathematical derivation 🔹 Electric field on the axis of a charged disc 🔹 Important assumptions and formulae 🔹 Special cases and applications 🔹 Exam-oriented tips and numerical approach This lecture will help you build a strong foundation in Electrostatics and understand how integration is applied in physics problems involving continuous charge distributions. 📌 Important Formula Covered: E=\frac{\sigma}{2\varepsilon_0}\left(1-\frac{x}{\sqrt{x^2+R^2}}\right) where: σ = Surface charge density R = Radius of the disc x = Distance from the center along the axis ε₀ = Permittivity of free space 🎯 Useful for: CBSE Class 12 Physics NEET Aspirants JEE Main & Advanced School & Board Exam Preparation Competitive Physics Exams 👍 If you find this lecture helpful, don’t forget to Like, Share, and Subscribe for more Physics lectures and problem-solving sessions. #Class12Physics #Electrostatics #ChargedDisc #ElectricField #PhysicsLecture #CBSEPhysics #NEETPhysics #JEEPhysics #ContinuousChargeDistribution #PhysicsByVinayRai