Most Important Selected Qs for JEE AdvancedPhysicsElectrostatics
Four identical point charges are fixed at the four corners of a square of a side length I. Another charged particle of mass m and charge +q is projected towards centre of square from a large distance along the line perpendicular to plane of square. The minimum value of initial velocity v₀ required to cross the square is? ( m =1 gm , l =4 2 ~m , Q =1 c , q =0.5 c )
Correct answer
3
Step-by-step solution
Particle will cross the square if it crosses the point of maximum potential between A and C -which is C itself. aligned & 1 2 m v₀^2=K+ 1 4 ₀ 4 Q^2 q (I / 2 ) & K 0 V₀^2=8 2 ( 1 4 ₀ ) ( Qq ml ) aligned