Lecture 7 Total Surface and Volume Charges

Transcription

1 Lecture 7 Total Surface and Volume Charges Sections: 2.3 Homework: See homework file

2 the surface element is defined by two line elements: ds= dl dl 1 2 the surface element is a vector Surface Elements dl ds 2 dl 1 o its magnitude is its differential area o its direction is normal (perpendicular) to the differential area LECTURE 7 slide 2

3 Surface Elements in the RCS Principal Planes = const plane: ds= ± dxdya y = const plane: ds= ± dxda x = const plane: ds= ± dyda d dx y x x= a/2 dy y = a/2 dy d 0 y dx x x= a/2 = a/2 a = a/2 LECTURE 7 slide 3 y = a/2

4 ρ = φ = const surface: ds = ± ρdφda const plane: ds= ± dρda = cont s Surface Elements in the CCS Principal Planes plane: ds φ ρ = ± ρdρdφa = const a a d ρ ρdφ d a φ d ρ φ = const d ρdφ a ρ ρ = const LECTURE 7 slide 4

5 Surface Elements in the SCS Principal Planes on r = const surface (sphere): 2 ds= ± r sinθdθdφa on θ = const surface (cone): ds= ± rsinθdrdφaθ on φ = const surface (half-plane): ds= ± rdrdθaφ r LECTURE 7 slide 5

6 Surface Charge on Principal Planes 1 Q = ρ ( uvdsuv, ) (, ) s S( uv, ) We limit this lecture to charges on principal coordinate planes y 2 charge on a portion of an RCS plane -plane: Q y x 2 2 = y x 1 1 ρ ( x, y) dxdy s y 1 y x x1 x2 charge on a portion of a circular disk or a ring in CCS φ 2 -plane: Q ρ φ 2 2 = ρ φ 1 1 ρ( ρφρ, ) dφdρ s ρ 1 ρ 2 φ 1 y LECTURE 7 slide 6 x

7 Surface Charge on Principal Planes 2 Can we define a surface charge on a disk or a ring in SCS? LECTURE 7 slide 7

8 Surface Charge on Principal Planes 3 charge on a portion of a cylinder in CCS 2 φ2 Q ρ(, φρ ) dφd = φ 1 1 s charge on a portion of a sphere in SCS 0 ρ 0 φ φ θ 1 0 r 0 θ 2 y φ θ Q ρ( θφ, ) r sinθdθdφ = φ θ s x φ 1 φ 2 LECTURE 7 slide 8

9 Surface Integration: Charge on Surfaces 4 charge on a cone in SCS φ2 r2 = ρs(, φ) sinθ0 φ 1 Q r r drd φ r 1 1 r 2 φ φ 2 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 7 slide 9

10 TRUE OR FALSE? Q1: A surface charge of uniform density ρ s = C/m 2 is distributed on the cylindrical surface ρ = 1 m, π ϕ 2π, 1 0. The total charge is Q = π C LECTURE 7 slide 10

11 Surface Charge Example A surface is defined by r 0 = 2 m, 30 θ 50, 20 ϕ 60. (a) Find the area of the surface. (b) Find the total charge on this surface if the charge density ρ s = C/m 2 is distributed uniformly. LECTURE 7 slide 11

12 Volume Elements 1 the volume element is defined by three line elements dv = ( dl dl ) dl the volume element is a scalar dl 2 dl dl 3 1 RCS: dv = dxdyd LECTURE 7 slide 12

13 Volume Elements 2 CCS: dv = ρdρdφd LECTURE 7 slide 13

14 SCS: 2 dv = r sinθdrdθdφ Volume Elements 3 LECTURE 7 slide 14

15 Volume Integration: Volume Charges parallelogram Q y x = y x ρ ( x, y, ) dxdyd v cylindrical volume Q ρ( ρφ,, ) ρd ρdφd φ ρ = φ ρ v Q = ρ V v dv spherical volume φ θ r Q ρ( r, θφ, ) r sinθdrdθdφ = φ θ r v 2 NOTE: You can find the volume of the element by setting ρ v = 1. LECTURE 7 slide 15

16 TRUE OR FALSE? Q1: A volume charge of uniform density ρ v = C/m 3 is distributed in a volume defined by 0 r 1 m 0 θ 90 0 φ 180 The total charge is Q = π C. Remider: Volume of sphere = 4 3 π r 3 LECTURE 7 slide 16

17 Volume Integration: Example A light source shines onto a hemispherical dome of radius a = 5 m, and makes a round spot 2 m in diameter (d = 2 m). What is the volume of the light cone from the light source to the dome? LECTURE 7 slide 17

18 You have learned how to find the total charge on any portion of the surface of a plane, cylinder, sphere, or cone find the total charge on any portion of a parallelogram, cylinder, sphere, cone use integration to find surface area and volume LECTURE 7 slide 18