Lecture 3 Total Charge: Line, Surface and Volume Integrals Sections: 1.8, 1.9,.3 Homework: See homework file LECTURE 3 slide 1
y Line Elements 1 metric increment in position along a line -D SPACE rectangular CS polar CS a y dya y dl b a da a y a d ρa φ ρ a b dl ρdφa φ a ρ dφ dl = da + dya y dl = dρaρ + ρdφaφ LECTURE 3 slide
Line Elements line increment is a vector has direction: point of integration moves from point a to point b each of its components is a linear increment (in meters) RCS: d, dy, and d are linear by default dl = da + dya + da y CCS: dρ and d are linear, dϕ is not angular increment dϕ corresponds to linear increment ρdϕ dl = dρa + ρdφa + da dφ ρdφ ρ φ SCS: dr is linear, dθ and dϕ are not dl = dra + rdθa + r sinθdφa r θ φ dθ rdθ dφ rsinθdφ LECTURE 3 slide 3
SCS: dl = dra + rdθa + rsinθdφa r θ Line Elements 3 φ spherical CS dθ r θ dl rdθ dr a θ a r a φ φ dφ r sinθ rsinθdφ y LECTURE 3 slide 4
Line Integration: Charge on Lines 1 Q B = ρ dl A l the direction of integration does not matter: charge is scalar EASY SPECIAL CASES: CHARGE ON PRINCIPAL LINES straight line: choose RCS B ais along charged line Q = ρl ( ) d ϕ-line in CCS: circular φb charges in the =0 plane Q= ρl ( φ) ρ0d θ-line and ϕ-line in SCS Q A θb φb ρl ( θ) r0 dθ Q= ρl ( φ) r0sinθ0 θ φ = φ radius of arc φa radius of arc on parallel circle radius of arc on meridian circle A A LECTURE 3 slide 5 dφ
Side Note: Parallels and Meridians http://www.justtrails.com/nav-skills/navigation-skills-latitude-and-longitude/ LECTURE 3 slide 6
TRUE OR FALSE? Q1: A line charge of uniform density ρ l = 10 1 C/m is distributed along the line r = 1 m, π/ θ π, ϕ = 0. The total charge is 10 1 3.14 10 1 C Q = π Q: A line charge of uniform density ρ l = 10 1 C/m is distributed along the line r = 1 m, θ = 45, 0 ϕ 360. The total charge is Q = π / 10 1 4.443 10 1 C LECTURE 3 slide 7
Line Integration: Charge on Lines GENERAL CASE curved line: needs line equation in parametric form r( u) = u ( ) a + yu ( ) a + u ( ) a dl= dr = da + dyay + da y dl d dy d dl = d + dy + d = + + du du du du dl dr dr dl dr dr dr dr = = dl = du du du du du du du du du L AB u B = u A dr dr du du du r( u) dr y B u B dr dr Q = ρl ( u) du du du u A A LECTURE 3 slide 8
the surface element is defined by two line elements: ds= dl dl 1 the surface element is a vector Surface Elements dl ds dl 1 o its magnitude is the differential area it occupies o its direction is normal (perpendicular) to the differential area LECTURE 3 slide 9
Surface Elements in the RCS Principal Planes = const plane: ds= ± ddya y = const plane: ds= ± dda = const plane: ds= ± dyda d d y = a/ dy y = a/ dy d 0 y d = a/ = a/ a = a/ y = a/ LECTURE 3 slide 10
Surface Elements in the CCS Principal Planes ρ = const surface: ds= ± ρdφda φ = const plane: ds= ± dρda = const plane: ds= ± ρd ρdφa φ ρ = const a a dρ ρdφ d a φ d ρ φ = const d ρdφ a ρ ρ = const LECTURE 3 slide 11
Surface Elements in the SCS Principal Planes on r = const surface (sphere): ds= ± r sinθdθdφa on θ = const surface (cone): ds= ± rsinθdrdφaθ on φ = const surface (half-plane): ds= ± rdrdθaφ r LECTURE 3 slide 1
Surface Integration: Charge on Surfaces 1 We will limit ourselves to SPECIAL CASES charges on principal coordinate planes Q = ρ S s ds charge on a portion of a plane Q y = y 1 1 ρ (, y) ddy s y y 1 charge on a portion of a circular disk or a ring Q ρ φ = ρ φ 1 1 ρ( ρφρ, ) dφdρ s y ρ 1 1 φ φ 1 y ρ LECTURE 3 slide 13
Surface Integration: Charge on Surfaces charge on a portion of a cylinder Q ρ(, φρ ) dφd φ = φ 1 1 s charge on a portion of a sphere 0 ρ 0 φ φ 1 1 θ 1 0 r 0 θ y φ θ 1 1 0 Q ρ( θφ, ) r sinθdθdφ = φ θ s φ 1 φ LECTURE 3 slide 14
Surface Integration: Charge on Surfaces 3 charge on a cone φ r = ρs(, φ) sinθ0 φ 1 Q r r drd φ r 1 1 r φ φ r 1 θ 0 NOTE: If you set ρ s = 1 in the above formulas, you can compute the area of the respective surfaces. LECTURE 3 slide 15
TRUE OR FALSE? Q1: A surface charge of uniform density ρ s = 10 10 C/m is distributed on the cylindrical surface ρ = 1 m, π ϕ π, 1 0. The total charge is Q = π 10 10 3.14 10 10 C LECTURE 3 slide 16
Volume Elements 1 the volume element is defined by three line elements dv = ( dl dl ) dl 1 3 the volume element is a scalar dl dl dl 3 1 RCS: dv = ddyd LECTURE 3 slide 17
Volume Elements CCS: dv = ρdρdφd LECTURE 3 slide 18
SCS: dv = r sinθdrdθdφ Volume Elements 3 LECTURE 3 slide 19
Volume Integration: Volume Charges parallelogram Q y = y 1 1 1 ρ (, y, ) ddyd v cylindrical volume 1 Q ρ( ρφ,, ) ρd ρdφd φ ρ = φ ρ 1 1 1 v Q = ρ V v dv spherical volume φ θ r Q ρ( r, θφ, ) r sinθdrdθdφ = φ θ r 1 1 1 v NOTE: You can find the volume of the element by setting ρ v = 1. LECTURE 3 slide 0
TRUE OR FALSE? Q1: A volume charge of uniform density ρ s = 10 1 C/m is distributed in a volume defined by 0 r 1 m 0 θ 90 0 φ 180 The total charge is Q = π 3 10 1 C. Remider: Volume of sphere = 4 3 π r 3 LECTURE 3 slide 1
Volume Integration: Eample A light source shines onto a hemispherical dome of radius a = 5 m, and makes a round spot m in diameter, d = m. What is the volume of the light cone from the light source to the dome? Work in SCS. Assume the -ais is along the ais of the symmetrical conical light beam. π α a 3 a V = dv = r sinθdrdθdφ = π (1 cosα) 3 V φ= 0θ= 0r= 0a The sine of half the subtended angle α of the beam is d / sinα = = 0. cosα = 1 sin α = 0.9798 a 3 5 3 V = π 0.00 = 0.1, m 3 Homework: Find the area of the dome lit up by the beam. A = 3.173, m LECTURE 3 slide
You have learned how to find the total charge along a straight line or any other curved line find the total charge on any portion of the surface of a plane, disk, cylinder, sphere, cone find the total charge on any portion of a parallelogram, cylinder, sphere, cone use integration to find length, area and volume LECTURE 3 slide 3