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Chapter 21 Electric Field and Coulomb’s Law (again)

Chapter 21 Electric Field and Coulomb’s Law (again). Coulomb’s Law (sec. 21.3) Electric fields & forces (sec. 21.4 & -6) Vector addition. C 2007 J. F. Becker. INTRODUCTION: see Ch. 1 (Volume 1). Vectors (Review) Used extensively throughout the course. C 2007 J. Becker.

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Chapter 21 Electric Field and Coulomb’s Law (again)

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  1. Chapter 21 Electric Field and Coulomb’s Law (again) • Coulomb’s Law (sec. 21.3) • Electric fields & forces (sec. 21.4 & -6) • Vector addition C 2007 J. F. Becker

  2. INTRODUCTION: see Ch. 1 (Volume 1) Vectors (Review) Used extensively throughout the course C 2007 J. Becker

  3. Vectors are quantities that have both magnitude and direction. An example of a vector quantity is velocity. A velocity has both magnitude (speed) and direction, say 60 miles per hour in a DIRECTION due west. (A scalar quantity is different; it has only magnitude – mass, time, temperature, etc.)

  4. A vector may be composed of its x- and y-components as shown.

  5. The scalar (or dot) product of two vectors is defined as Note: The dot product of two vectors is a scalar quantity. C 2007 J. F. Becker

  6. The vector (or cross) product of two vectors is a vector where the direction of the vector product is given by the right-hand rule. The MAGNITUDE of the vector product is given by: Note: The dot product of two vectors is a scalar quantity. C 2007 J. F. Becker

  7. Right-hand rule for DIRECTION of vector crossproduct.

  8. Coulomb’s Law Coulomb’s Law lets us calculate the FORCE between two ELECTRIC CHARGES.

  9. Coulomb’s LawCoulomb’s Law lets us calculate the force between MANY charges. We calculate the forces one at a time and ADD them AS VECTORS. (This is called “superposition.”) THE FORCE ON q3 CAUSED BY q1 AND q2.

  10. Coulomb’s Law -forces between two charges 21-9 Coulomb’s Law – vector problem Net force on charge Q is the vector sum of the forces by the other two charges.

  11. Recall GRAVITATIONAL FIELD near Earth:F = G m1 m2/r2 = m1 (G m2/r2) = m1g where the vectorg = 9.8 m/s2 in the downward direction, and F = m g. ELECTRIC FIELD is obtained in a similar way:F = k q1 q2/r2 = q1 (k q2/r2) = q1 (E) where the vector E is the electric field caused by q2. The direction of the E field is determined by the direction of the F, or as you noticed in lab #1, the E field lines are directed away from a positive q2 and toward a -q2.The F on a charge q in an E field is F = q E and |E| = (k q2/r2) C 2007 J. F. Becker

  12. A charged body creates an electric field.Coulomb force of repulsion between two charged bodies at A and B, (having charges Q and qo respectively) has magnitude: F = k |Q qo |/r2 = qo [ k Q/r2 ]where we have factored out the small charge qo. We can write the force in terms of an electric field E: Therefore we can write for F = qoE the electric fieldE = [ k Q / r2 ]

  13. Electric field at“A” and “C” set up by charges q1 and q1 C Lab #1 Calculate E1, E2, and ETOTALat points “A” & “C”: a) E1= 3.0 (10)4N/C E2 = 6.8 (10)4 N/C EA= 9.8 (10)4 N/C c) E1= 6.4 (10)3 N/C E2 = 6.4 (10)3 N/C EC = 4.9 (10)3 N/C in the +x-direction q= 12 nC A (an electric dipole)

  14. y Consider symmetry! Ey= 0 Xo Electric field at Pcaused by a line of charge along the y-axis.

  15. Consider symmetry! Ey= 0 dq |dE| = k dq / r2 o Xo cos a =xo / r dEx = dE cos a =[k dq /xo2+a2][xo /(xo2+ a2)1/2] Ex = k xoò dq /[xo2 + a2]3/2 where xo is constant as we add all the dq’s (=Q) in the integration: Ex = k xo Q/[xo2+a2]3/2

  16. Tabulated integral: òdz / (c-z) 2 = 1 / (c-z) d b Calculate the electric fieldat +qcaused by the distributed charge +Q.

  17. y Consider symmetry! Ey= 0 Xo Electric field at Pcaused by a line of charge along the y-axis.

  18. y Consider symmetry! Ey= 0 Xo Tabulated integral: òdz / (c2+z2) 3/2 = z / c2 (c2+z2) 1/2 Electric field at Pcaused by a line of charge along the y-axis.

  19. Tabulated integral: (Integration variable “z”) òdz / (c2+z2) 3/2 = z / c2 (c2+z2) 1/2 ò dy / (c2+y2) 3/2 = y / c2 (c2+y2) 1/2 òdy / (Xo2+y2) 3/2 = y / Xo2 (Xo2+y2) 1/2 Our integral=k (Q/2a)Xo 2[y /Xo2 (Xo2+y2) 1/2 ]0a Ex = k (Q /2a)Xo 2[(a –0) / Xo2 (Xo2+a2) 1/2 ] Ex = k (Q /2a)Xo 2 [a / Xo2 (Xo2+a2) 1/2 ] Ex = k (Q / Xo)[1 / (Xo2+a2) 1/2 ] Notation change C 2007 J. F. Becker

  20. Tabulated integral: òdz / (z2 + a2)3/2 = z / a2 (z2 + a2) 1/2òzdz / (z2 + a2)3/2 = -1 / (z2 + a2) 1/2 Calculate the electric field at -q caused by +Q, and then the force on -q.

  21. An ELECTRIC DIPOLE consists of a +q and –q separated by a distance d. ELECTRIC DIPOLE MOMENT is p = q d ELECTRIC DIPOLE in E experiences a torque:t = p x E ELECTRIC DIPOLE in E has potential energy:U = - p E C 2007 J. F. Becker

  22. ELECTRIC DIPOLE MOMENT is p = qd t = r x F t = p x E Net force on an ELECTRIC DIPOLE is zero, but torque (t) is into the page.

  23. Review See www.physics.edu/becker/physics51 C 2007 J. F. Becker

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