Lecture 4Electric Field – Chapter 23Electric 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 chargeElectric Field (9)• Electric field, E, is the force per unit test charge in N/C• For a point charge so20rqqkF =r0qFErr=2rqkE =rElectric Field (10)• Direction of E = direction of F• E points towards (away from) a negative (positive) point charge• Superposition of electric fieldsnEEEErrrr+++= ...21Checkpoint #2 – Rank magnitude of net ECheckpoint #2 – Rank magnitude of net E•a)• Do this for the rest and find All EqualidqkdqkdqkExˆ532222=+=jdqkEyˆ52=22250dqkEEEyx=+=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• giving • where −+−== EEEEz22−+−=rqkrqkE2dzr −=+2dzr +=−Electric Field (13)• Substituting and rearranging gives• Assuming z>>d then expand using binomial theorem ignoring higher order terms d/z<<1+−−=−− 2222121zdzdzkqE+−−++= ...1...12zdzdzkqEElectric Field (14)• Approximate E field for a dipole is• Define electric dipole moment, p, aswhere direction of p is from the negative to positive end• E field along dipole axis at large distances (z>>d) is32zkqdE =qdp =r32zkpE =Electric Field (15)• What happens when a dipole is put in an electric field?• Net force, from uniform E, is zero •Butforce on charged ends produces a net torque about its center of massElectric Force (16)• Definition of torque• For dipole rewrite it as • Substitute F and d to get φτsinrFFr =×=rrrθθτsin)(sin FxdxF −+=Eprrr×=τElectric Field (17)• Torque acting on a dipole tends to rotate pinto the direction of E• Associate potential energy, U, with the orientation of an electric dipole in an Efield • Dipole has least U when p is lined up with EElectric Field (18)• Remember • Potential energy of a dipole • U is least (greatest) when p and E are in same (opposite) directions ∫∫=−=−=θθθθθτ9090sin dpEdWUEppEUrr•−=−=θcosCheckpoint #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 θτsinpE=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 densityLengthQ /=λAreaQ /=σVolumeQ /=ρElectric Field (20)• Ring of radius R and positive charge density λ•Use • Divide ring into diff. elements of charge so 2rqkE =dsdqλ=Electric Field (21)• Differential dE at P is• From trig• Look for symmetry–All ⊥ cancel and point upward22rdskrdqkdEλ==222Rzr+=Electric Field (22)• Parallel component dE is• Use trig to rewrite cos θ()θλθcoscos22RzdskdE+=()2/122cosRzzrz+==θElectric Field (23)• Substituting • Integrate around the ring()dsRzzkdE2/322cos+=λθ()∫∫+==rdsRzkzdEEπλθ202/322cosElectric Field (24)• Finally get • Replace λ with• Charge ring has E of ()()2/3222RzrkzE+=πλ()2/322RzkqzE+=()rqπλ2/=Electric Field (25)• Charge ring has E of • Check z >>R then• From far away ring looks like point charge()2/322RzkqzE+=2zkqE =Electric Field (26)• Also do this for charged disk• But now surface charge• Integrate to get()rdrAdqπσσ2==+−=22012RzzEεσElectric Field (27)• Charge disk of radius R• Let R→∞ then get• Acts as infinite sheet of a non-conductor with uniform charge
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