Electric Field: Electric Field Lines, Point Charges, Superposition, and Electric Dipoles, Study notes of Geology

Download Electric Field: Electric Field Lines, Point Charges, Superposition, and Electric Dipoles and more Study notes Geology in PDF only on Docsity! Lecture 4 Electric Field – Chapter 23 Electric Field (8) • Electric field lines: – Point away from positive and towards negative – Tangent to the field line is the direction of the E field at that point – # lines is proportional to magnitude of the charge Checkpoint #2 — Rank magnitude of net E 3 ¥ Checkpoint #2 – Rank magnitude of net E • a) • Do this for the rest and find All Equal i d qk d qk d qkEx ˆ 532 222 =+= jd qkEy ˆ 5 2= 2 22 50 d qkEEE yx =+= Electric Field (11) • Electric dipole – 2 equal magnitude, opposite charged particles separated by distance d • What’s the electric field at point P due to the dipole? Electric Field (14) • Approximate E field for a dipole is • Define electric dipole moment, p, as where direction of p is from the negative to positive end • E field along dipole axis at large distances (z>>d) is 3 2 z kqdE = qdp =r 3 2 z kpE = Electric Field (15) • What happens when a dipole is put in an electric field? • Net force, from uniform E, is zero • But force on charged ends produces a net torque about its center of mass Electric Force (16) • Definition of torque • For dipole rewrite it as • Substitute F and d to get φτ sinrFFr =×= rrr θθτ sin)(sin FxdxF −+= Ep rrr ×=τ Checkpoint #5 • Rank a) magnitude of torque and b) U , greatest to least • a) Magnitudes are same • U greatest at θ=180 • b) 1 & 3 tie, then 2 &4 θτ sinpE= θcospEU −= Electric Field (19) • E field from a continuous line of charge • Use calculus and a charge density instead of total charge, Q • Linear charge density • Surface charge density • Volume charge density LengthQ /=λ AreaQ /=σ VolumeQ /=ρ Electric Field (20) • Ring of radius R and positive charge density λ • Use • Divide ring into diff. elements of charge so 2r qkE = dsdq λ= Electric Field (23) • Substituting • Integrate around the ring ( ) dsRz zkdE 2/322cos + = λθ ( ) ∫∫ +== r ds Rz kzdEE πλθ 2 0 2/322 cos Electric Field (24) • Finally get • Replace λ with • Charge ring has E of ( ) ( ) 2/322 2 Rz rkzE + = πλ ( ) 2/322 Rz kqzE + = ( )rq πλ 2/= Electric Field (25) • Charge ring has E of • Check z >>R then • From far away ring looks like point charge ( ) 2/322 Rz kqzE + = 2z kqE =