* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Document
Survey
Document related concepts
Introduction to gauge theory wikipedia , lookup
Speed of gravity wikipedia , lookup
History of electromagnetic theory wikipedia , lookup
Euclidean vector wikipedia , lookup
Four-vector wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Work (physics) wikipedia , lookup
Electromagnetism wikipedia , lookup
Maxwell's equations wikipedia , lookup
Centripetal force wikipedia , lookup
Circular dichroism wikipedia , lookup
Electric charge wikipedia , lookup
Field (physics) wikipedia , lookup
Transcript
Chapter 21 Electric Field and Coulomb’s Law (again) • • • Electric fields and forces (sec. 21.4) Electric field calculations (sec. 21.5) Vector addition (quick review) C 2012 J. Becker Learning Goals - we will learn: • How to use Coulomb’s Law (and vector addition) to calculate the force between electric charges. • How to calculate the electric field caused by discrete electric charges. • How to calculate the electric field caused by a continuous distribution of electric charge. Coulomb’s Law Coulomb’s Law lets us calculate the FORCE between two ELECTRIC CHARGES. 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. Figure 21.14 SYMMETRY! 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 is vector E is the electric field caused by q2. The direction of the E field is determined by the direction of the F, or the E field lines are directed away from positive q2 and toward -q2. The F on a charge q in an E field is F = q E and |E| = (k q2/r2) Fig. 21.15 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: F = qo E Therefore we can write for the electric field E = [ k Q / r2 ] E1 C See Lab #2 ET E2 See Fig. 21.23: Electric field at “C” set up by charges q1 and q1 Calculate E1, E2, and ETOTAL at point “C”: q = 12 nC A (an “electric dipole”) At “C” E1= 6.4 (10)3 N/C E2 = 6.4 (10)3 N/C ET = 4.9 (10)3 N/C in the +x-direction Need TABLE of ALL vector component VALUES. dq Fig. 21.24 Consider symmetry! Ey = 0 dEx = Ex o |dE| = k dq / r2 Xo cos a = xo / r dEx= dE cos a =[k dq /(xo2+a2)] [xo/(xo2+ a2)1/2] Ex = k xo dq /[xo2 + a2]3/2 where xo and a stay constant as we add all the dq’s ( dq = Q) in the integration: Ex = k xo Q/[xo2+a2]3/2 y Consider symmetry! Ey = 0 dq |dE| = k dq / r2 Xo Fig. 21.25 Electric field at P caused by a line of charge uniformly distributed along y-axis. |dE| = k dq / r2 cos a = xo/ r and and r = (xo2+ y2)1/2 cos a = dEx / dE dEx = dE cos a Ex = dEx = dE cos a Ex = [k dq /r2] [xo / r] Ex = [k dq /(xo2+y2)] [xo /(xo2+ y2)1/2] Linear charge density = l l = charge / length = Q / 2a = dq / dy dq = l dy Ex = [k dq /(xo2+y2)] [xo /(xo2+ y2)1/2] Ex = [k l dy /(xo2+y2)] [xo /(xo2+ y2)1/2] Ex = k lxo [dy /(xo2+y2)] [1 /(xo2+ y2)1/2] Ex = k lxo [dy /(xo2+y2) 3/2] Tabulated integral: (Integration variable “z”) dz / (c2+z2) 3/2 = z / c2 (c2+z2) 1/2 dy / (c2+y2) 3/2 = y / c2 (c2+y2) 1/2 dy / (Xo2+y2) 3/2 = y / Xo2 (Xo2+y2) 1/2 Ex = k lxo a 2+y2) 3/2] [dy /(x o -a Ex = k(Q/2a) Xo [y /Xo2 (Xo2+y2) 1/2 Ex = k (Q /2a) Xo [(a –(-a)) / ] a -a Xo2 (Xo2+a2) 1/2 Ex = k (Q /2a) Xo [2a / Xo2 (Xo2+a2) 1/2 ] Ex = k (Q / Xo) [1 / (Xo2+a2) 1/2 ] ] Tabulated integral: dz / (c-z) 2 = 1 / (c-z) l is uniform (= constant) +Q b Fig. 21.47 Calculate the electric field at the proton caused by the distributed charge +Q. Tabulated integrals: dz / (z2 + a2)3/2 = z / a2 (z2 + a2) ½ for calculation of Ex z dz / (z2 + a2)3/2 = -1 / (z2 + a2) ½ for calculation of Ey l is uniform (= constant) Fig. 21.48 Calculate the electric field at -q caused by +Q, and then the force on –q: F=qE 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=pxE ELECTRIC DIPOLE in E has potential energy: U=-p E ELECTRIC DIPOLE MOMENT is p = qd t=rxF t=pxE Fig. 21.32 Net force on an ELECTRIC DIPOLE is zero, but torque (t) is into the page. Review see www.physics.sjsu.edu/Becker/physics51 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.) A vector may be decomposed into its x- and y-components as shown: Ax A cos Ay A sin A Ax Ay 2 2 2 The scalar (or dot) product of two vectors is defined as A B AB cos Ax Bx Ay By Az Bz Note: The dot product of two vectors is a scalar quantity. 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: A B AB sin Right-hand rule for DIRECTION of vector cross product. PROFESSIONAL FORMAT