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 2009 J. F. Becker

INTRODUCTION: see Ch. 1 (Volume 1) Vectors (Review) Used extensively throughout the course INTRODUCTION C 2009 J. Becker

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.) Vectors - addition

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

The scalar (or dot) product of two vectors is defined as Note: The dot product of two vectors is a scalar quantity. Vectors – dot product C 2009 J. F. Becker

The MAGNITUDE of the vector product is given by: 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: Vectors – cross product Note: The dot product of two vectors is a scalar quantity. C 2009 J. F. Becker

Right-hand rule for DIRECTION of vector cross product. Vectors – right hand rule for cross product

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

THE FORCE ON q3 CAUSED BY q1 AND q2. Coulomb’s Law Coulomb’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.

21-9 Coulomb’s Law – vector problem 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.

Recall GRAVITATIONAL FIELD near Earth: F = G m1 m2/r2 = m1 (G m2/r2) = m1 g where the vector g = 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 2009 J. F. Becker

the electric field E = [ k Q / r2 ] 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 On board, calculate the E field due to a + and a - point charge… F = qo E the electric field E = [ k Q / r2 ]

Calculate E1, E2, and ETOTAL at points “A” & “C”: Electric field at“A” and “C” set up by charges q1 and q1 C Lab #1 Calculate E1, E2, and ETOTAL at points “A” & “C”: a) E1= 3.0 (10)4 N/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)

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

dEx = dE cos a =[k dq /xo2+a2][xo /(xo2+ a2)1/2] 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 xo ò 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

Tabulated integral: ò dz / (c-z) 2 = 1 / (c-z) Calculate the electric field at +q caused by the distributed charge +Q.

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

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

ò dz / (c2+z2) 3/2 = z / c2 (c2+z2) 1/2 Tabulated integral: (Integration variable “z”) ò dz / (c2+z2) 3/2 = z / c2 (c2+z2) 1/2 ò dy / (c2+y2) 3/2 = y / c2 (c2+y2) 1/2 ò dy / (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 2009 J. F. Becker

Tabulated integral: ò dz / (z2 + a2)3/2 = z / a2 (z2 + a2) 1/2 ò z dz / (z2 + a2)3/2 = -1 / (z2 + a2) 1/2 CALCULATE THE X- AND Y-COMPONENTS OF THE ELECTRIC FIELD Calculate the electric field at -q caused by +Q, and then the force on -q.

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 2009 J. F. Becker

2 r = d in figure in textbook. ELECTRIC DIPOLE MOMENT is p = qd t = r x F t = p x E 2 r = d in figure in textbook. Net force on an ELECTRIC DIPOLE is zero, but torque (t) is into the page.

See www.physics.edu/becker/physics51 Review OVERVIEW See www.physics.edu/becker/physics51 C 2009 J. F. Becker