Electric Flux And Gauss Theorem | Important Numericals With JEE Mains PYQs | Electrostatics | L-3

This video contains - 00:00 - Lecture Intro 00:15 - Area Vector 02:48 - Electric Flux 06:00- Component of a Vector 09:45 - Solid Angle 14:41 - Gauss Law 18:18 - Proof Of Gauss Theorem 22:51 - Continuous Charge Distribution 36:55 - Appilication Of Gauss Law 37:00 - Electric Field Intensity due to an infinitely long straight uniformly charged wire 47:46 To calculate Electric Field due to a uniformly charged infinite plane sheet 54:09 Electric Field due to a uniformly charged thin spherical shell 1:11:03 Q26. If the net electric flux through a closed surface is zero, then we can infer (A) no net charge is enclosed by the surface. (B) uniform electric field exists within the surface. (C) electric potential varies from point to point inside the surface. (D) charge is present inside the surface. 11:12:54 Draw the graph between E and r for the thin spherical shell where r is the radius of spherical shell? 1:14:50 Q28. A point charge is placed at the centre of a closed Gaussian spherical surface. How is the electric flux through gaussain surface affected when a. Sphere is replaced by a cylinder b. The charge is shifted from the centre to some other point inside the sphere c.Second charge is placed inside the sphere 1:18:29 Q29.A cylinder is placed at in a uniform electric field E with its axis parallel to field . Calculate the flux through the cylinder. 1:24:41 Three concentric metallic spherical shell of radii R,2R,3R are given charges Q1,Q2,Q3 respectively. It is found that the surface charge density on the outer surface of the shell are equal . Calculate the ratio of charge . 1:33:09 Q31 . A point charge +10µc is at a distance of 5cm directly above the centre of a square side 10cm . what is the magnitude of the electric flux through the square 1:37:16 Q32. Given a uniformly charge plane sheet of surface charge density σ=2×〖10〗^17 C/m^2. Find the electric field intensity at a point A, 5mm away from the sheet from the left side . 1:40:09 Q33. The electric field component are E_x=αx^(1/2) , E_(y )=E_z=0 in which α=800NC^(-1) m^(1/2). Calculate The flux through the cube The charge with in the cube assuming (l=0.1) 1:49:58 A pendulum bob of mass 80mg and carrying a charge 2×〖10〗^(-8) c is at rest in a horizontal uniform electric field 2×〖10〗^4 V/m. Find the tension in the thread of pendulum and the angle it makes with the vertical. 2:02:30 An infinite long wire is stretched horizontally 4m above the surface of the earth . It carries a charge 1µc/cm of its Length . Calculate its electric field at a point on the earth surface vertically below the wire. 2:07:30 Why do the electrostatic field lines not form close loop? 02:08:24 Consider two hollow concentric spheres S1 and S2 enclosing charges 2Q and 4Q respectively, as shown in the figure. i. Find out the ratio of the electric flux through them. ii. How will the electric flux through the sphere S1 change, if a medium of dielectric constant k is introduced in the space inside S1 in place of air? Deduce the necessary expression.