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Electrostatics Lecture 15 – in this class we solve six advanced applications of electric field and superposition, perfect for CBSE Class 12 and JEE Main level. We carefully derive the field and force in different continuous charge configurations: Bent uniformly charged wire (two shapes) – Electric field at the centre for two wire configurations using symmetry and vector addition. Force between finite wire and infinite charged wire – Non‑uniform field of an infinite line, element method (dF = E,dQ), integration from (r_0) to (r_0 + l). Electron near an infinite charged sheet with a circular hole – Treat sheet with hole as (full sheet + disc of opposite charge). – Find (E(x)), then equation of motion (m v \frac{dv}{dx} = qE(x)), integrate to get speed as electron crosses the sheet. Electric field at the centre of a uniformly charged hollow hemisphere – Ring element on the hemisphere, symmetry to keep only axial components, integration over angle (\theta). – Important result: (E = \dfrac{\sigma}{4\varepsilon_0}) at the centre. Interaction force between a charged ring and a semi‑infinite charged thread on its axis – Field on the axis of a ring, element on the thread, integrate from (0) to (\infty) to get net force. Interaction between two perpendicular long charged rods – Element on one rod, field of the other, resolving components and integrating from (-\infty) to (\infty). – Beautiful result: net force is independent of separation (a), (F = \dfrac{\lambda_1 \lambda_2}{2\varepsilon_0}). All derivations are done step‑by‑step with clear justification of each formula, focusing on: Superposition of electric fields, Choosing smart elements (ring, line, disc), Using symmetry to simplify integrals, Connecting field (E(x)) to force and motion for charges. This lecture is ideal revision for: CBSE Class 12 Physics Electrostatics, JEE Main electrostatics problems on continuous charge distributions. Watch in sequence with the full Electrostatics playlist on “CBSE & JEE Physics | Dr Kedar Pathak” for complete theory + problems.