PHYS 0175 1st Edition Lecture 5Outline of Last Lecture II. Electric Field produced by point chargeIII. Electric Field LinesIV. Electric DipoleOutline of Current Lecture V. Electric DipoleVI. Field of Electric DipoleVII. Electric Dipole in an external electric fieldVIII.External Torque on electric dipoleIX. Dipole Potential EnergyX. Electric Field in Continuous Charge Distributiona. Rodb. Ringc. DiskCurrent LectureII. Electric Dipolea. Pair of point chargesb. Equal magnitudesc. Opposite signsd. Separated by distance de. Dipole momenti. p=q(charge)d(distance)ii. points from –q to positive qf. water moleculeIII. Field of Electric Dipolea. SI unite: kg m s^-2 A^-1b. E=P/(2pi*permittivity constant*z^3)c. As distance z increases, field will decrease much more dramatically than for other electric fields with distanced. Because from far away, two parts of dipole cancel each other outIV. Electric Dipole in an external electric fielda. Zero net force activeb. Nonzero net torqueV. External Torque on electric dipolei. Therefore, must have rotation/angular accelerationii. 0 translationiii. same direction(ccw or cw)iv. tau=pEsinPHIv. maximum torque1. PHI=pi/2 torque has maxvi. stable equilibrium1. PHI=0a. Smallest potential energyb. U=-pEvii. unstable equilibrium1. PHI=pia. Largest potential energy(U)VI. Dipole Potential Energya. W=-delta-Ub. U=-pEcosPHI=-p dot Ei. Dot productii. Must assume a uniform electric fieldVII. Electric Field in Continuous Charge Distributiona. Continuous must use integration techniquesi. Split charge distribution into volume elements dVii. Each element has charge dq=pdViii. Let r be distance of point p from dqiv. Find infinitesimal electric field dE generated by dq at point P using expression for point source1. dE=dq*k/r^2v. Integrate over the contributions of all volume elements, E=integralEb. P=charge densityi. dq/dVc. Rod d. Ringi. X and y components will cancelii. Net will b in z axis + or –e. Diski. Can think of as an onion layers of rings with radius
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