Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
History of quantum field theory wikipedia , lookup
Introduction to gauge theory wikipedia , lookup
History of electromagnetic theory wikipedia , lookup
Speed of gravity wikipedia , lookup
Circular dichroism wikipedia , lookup
Electromagnetism wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Maxwell's equations wikipedia , lookup
Lorentz force wikipedia , lookup
Field (physics) wikipedia , lookup
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 Electric Field (9) • Electric field, E, is the force per unit test charge in N/C r r F E = q0 • For a point charge r q0 q F =k 2 r so r q E = k 2 r Electric Field (10) • Direction of E = direction of F • E points towards (away from) a negative (positive) point charge • Superposition of electric fields r r r r E = E1 + E2 + ... + En Checkpoint #2 – Rank magnitude of net E Checkpoint #2 – Rank magnitude of net E • a) E x = k 2q + k 3q = k 5q iˆ d2 d2 d2 E= 5q ˆ Ey = k 2 j d 50 q E +E =k d2 2 x 2 y • Do this for the rest and find All Equal 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 (12) • E is on z-axis so E = E z = E+ − E− • giving • where q q E =k 2 −k 2 r+ r− d r+ = z − 2 d r− = z + 2 Electric Field (13) • Substituting and rearranging gives −2 −2 kq d d E = 2 1 − − 1 + z 2 z 2 z • Assuming z>>d then expand using binomial theorem ignoring higher order terms d/z<<1 kq d d E = 2 1 + + ... − 1 − + ... z z z Electric Field (14) • Approximate E field for a dipole is 2 kqd E = 3 z • 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 r p = qd 2kp E= 3 z 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) r r • Definition of torque τ = r × F = rF sin φ r • For dipole rewrite it as τ = xFsinθ +(d − x)Fsinθ • Substitute F and d to get r r τ = p×E r Electric Field (17) • Torque acting on a dipole tends to rotate p into the direction of E • Associate potential energy, U, with the orientation of an electric dipole in an E field • Dipole has least U when p is lined up with E Electric Field (18) • Remember θ θ 90 90 U = −W = − ∫ τ d θ = ∫ pE sin θ d θ • Potential energy of a dipole r r U = − pE cos θ = − p • E • U is least (greatest) when p and E are in same (opposite) directions Checkpoint #5 • Rank a) magnitude of torque and b) U , greatest to least τ = pE sin θ • a) Magnitudes are same U = − pE cosθ • U greatest at θ=180 • b) 1 & 3 tie, then 2 &4 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 λ = Q / Length • Surface charge density σ = Q / Area • Volume charge density ρ = Q / Volume Electric Field (20) • Ring of radius R and positive charge density λ • Use q E = k 2 r • Divide ring into diff. elements of charge so dq = λds Electric Field (21) • Differential dE at P is dq λ ds dE = k 2 = k 2 r r • From trig r =z +R 2 2 2 • Look for symmetry – All ⊥ cancel and point upward Electric Field (22) • Parallel component dE is dE cosθ = k (z λds 2 +R 2 ) cosθ • Use trig to rewrite cos θ z z cosθ = = 2 1/ 2 2 r (z + R ) Electric Field (23) • Substituting dE cosθ = k (z zλ 2 +R ) 2 3/ 2 ds • Integrate around the ring E = ∫ dE cosθ = kzλ 2πr ds ∫ (z + R ) 2 2 3/ 2 0 Electric Field (24) • Finally get E= • Replace λ with kzλ (2πr ) (z 2 +R λ = q / (2πr ) • Charge ring has E of E= (z kqz 2 ) 2 3/ 2 +R ) 2 3/ 2 Electric Field (25) • Charge ring has E of E= (z kqz 2 +R ) 2 3/ 2 • Check z >>R then kq E = 2 z • From far away ring looks like point charge Electric Field (26) • Also do this for charged disk • But now surface charge dq = σA = σ (2πrdr ) • Integrate to get σ z 1 − E= 2 2 2ε 0 z +R Electric Field (27) • Charge disk of radius R σ E = 2ε 0 1 − 2 2 z +R z • Let R→∞ then get σ E = 2ε 0 • Acts as infinite sheet of a nonconductor with uniform charge