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The Electric Field 26.1 I) Electric Field Models Review A) Fields point away from positive, toward negative source charges B) Field strength is proportional to source charge, inversely proportional to distance from source charge C) Net electric field is the vector sum of the fields due to each charge (superposition) 1 π II) πΈβ = 4ππ π2 πΜ 0 26.2 I) The Electric Field of Multiple Point Charges Superposition A) (πΈπππ‘ )π₯ = (πΈ1 )π₯ + (πΈ2 )π₯ + β― = β(πΈπ )π₯ B) (πΈπππ‘ )π¦ = (πΈ1 )π¦ + (πΈ2 )π¦ + β― = β(πΈπ )π¦ C) (πΈπππ‘ )π§ = (πΈ1 )π§ + (πΈ2 )π§ + β― = β(πΈπ )π§ D) Problem Solving Strategy for Electric Field due to multiple point charge II) The electric Field of a Dipole A) Derivation of Equation 2π B) πΈβππππππ = π π3 , where π = (ππ , from the negative to the positive charge) along the dipole axis π C) πΈβππππππ = βπ π 3 along the bisecting plane III) Picturing the Electric Field A) Drawing the Field 26.3 I) The Electric Field of a Continuous Charge Distribution Charge Density A) Linear π= π πΏ B) Surface π= π π΄ C) Example 26.3 II) Infinite Line of Charge A) πΈrod = π 2|π| 1 π β1+4π 2 βπΏ2 B) But if πΏ β β then πΈline = π 26.4 I) 2|π| π The Electric Field of Rings, Disks, Planes and Spheres Example 26.4 derives (πΈππππ ) π§ A) Different than the field generated by a line of charge or rod, all the segments of charge are at the same position relative to the point in question along the z-axis B) Consequently, no summation of distances is needed, nor summation of segments of charge II) A Disk of Charge A) π = π π΄ = π ππ 2 B) Method includes using segments of charge in the shape of rings, since the πΈππππ is already derived C) A bit of calculusβ¦ D) (πΈdisk )π§ = π 2π0 [1 β π§ βπ§ 2 +π 2 ] III) A Plane of Charge A) Plane could be a disk of infinite radius B) if π β β in the equation (πΈdisk )π§ = π 2π0 [1 β π§ βπ§ 2 +π 2 ] then πΈplane = π 2π0 IV) A Sphere of Charge A) If π is the radius of the sphere and π is the distance from the center of the sphere π B) πΈβsphere = 2 πΜ 4ππ0 π 26.5 I) The Parallel-Plate Capacitor Two parallel plates, essential two planes of charge, comprise a parallel plate capacitor π πΈβcapacitor = πΈβ+ + πΈββ = ( , from positive to negative) π0 π πΈβcapacitor = πΈβ+ + πΈββ = ( , from positive to negative) π0 π΄ II) Uniform electric fields 26.6 Motion of a Charged Particle in an Electric Field I) Recall that electric fields exert forces onto charges within the electric field A) πΉon π = ππΈβ B) π = πΉon π π = π πΈβ π II) Motion in a Uniform Field A) A charged particle in a uniform electric field will move with constant acceleration B) π = 26.7 I) πΉon π π = π πΈβ π Motion of a Dipole in an Electric Field Factors influencing polarization force A) The charge polarizes a neutral object, creating the dipole B) The charge then exerts an attractive force on one end, that is larger than the repulsive force on the other end II) Dipoles in a Uniform Field A) πΉnet = πΉ+ + πΉβ = 0, but the forces do create unbalanced torques 1 2 B) π = 2 × ππΉ+ = 2( π sin π)(ππΈ) C) π = π × πΈβ III) Dipoles in a Nonuniform Field A) Net force on a dipole is toward the strongest field B) A dipole will experience a net force with any charged object