Chapter 22. Dr. Armen Kocharian. Gauss s Law Lecture 4

Chapter 22 Dr. Armen Kocharian Gauss s Law Lecture 4

Field Due to a Plane of Charge E must be perpendicular to the plane and must have the same magnitude at all points equidistant from the plane Choose a small cylinder whose axis is perpendicular to the plane for the gaussian surface qin σa Φ E = 2EA = = ε ε 2EA o σa = E = ε o o σ 2ε o

Field Due to a Plane of Charge, cont E is parallel to the curved surface and there is no contribution to the surface area from this curved part of the cylinder The flux through each end of the cylinder is EA and so the total flux is 2EA

Field Due to a Plane of Charge, final The total charge in the surface is σa Applying Gauss s law σa Φ E = 2EA = and E = ε σ 2ε Note, this does not depend on r Therefore, the field is uniform everywhere o o

Electrostatic Equilibrium When there is no net motion of charge within a conductor, the conductor is said to be in electrostatic equilibrium

Properties of a Conductor in Electrostatic Equilibrium The electric field is zero everywhere inside the conductor If an isolated conductor carries a charge, the charge resides on its surface The electric field just outside a charged conductor is perpendicular to the surface and has a magnitude of σ/ε o On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature is the smallest

Property 1: E inside = 0 Consider a conducting slab in an external field E If the field inside the conductor were not zero, free electrons in the conductor would experience an electrical force These electrons would accelerate These electrons would not be in equilibrium Therefore, there cannot be a field inside the conductor

Property 1: E inside = 0, cont. Before the external field is applied, free electrons are distributed throughout the conductor When the external field is applied, the electrons redistribute until the magnitude of the internal field equals the magnitude of the external field There is a net field of zero inside the conductor This redistribution takes about 10-15 s and can be considered instantaneous

Property 2: Charge Resides on the Surface Choose a gaussian surface inside but close to the actual surface The electric field inside is zero (prop. 1) There is no net flux through the gaussian surface Because the gaussian surface can be as close to the actual surface as desired, there can be no charge inside the surface

Property 2: Charge Resides on the Surface, cont Since no net charge can be inside the surface, any net charge must reside on the surface Gauss s law does not indicate the distribution of these charges, only that it must be on the surface of the conductor

Property 3: Field s Magnitude and Direction Choose a cylinder as the gaussian surface The field must be perpendicular to the surface If there were a parallel component to E, charges would experience a force and accelerate along the surface and it would not be in equilibrium

Property 3: Field s Magnitude and Direction, cont. The net flux through the gaussian surface is through only the flat face outside the conductor The field here is perpendicular to the surface Applying Gauss s law σa Φ E = EA = and E = ε o σ ε o

Conductors in Equilibrium, example The field lines are perpendicular to both conductors There are no field lines inside the cylinder

Example Find the field in regions 1, 2, 3 and 4 Φ = E EA Find in region 1 Field in region 2 ( 2 ) qin EA 1 = E1 4 π r =, ε q in 1 = 0 ( ) E = 0 r a 0 2 2 in = 2Q ( 2 4π ) E A= E r = q 2Q 2kQ E = = a r b e 2 2 2 4πε0r r q ε in 0 ( )

Example Find the field in regions 1, 2, 3 and 4 Φ = E EA Field in region 3 Field in region 4 ( 2 ) qin EA 3 = E3 4 π r =, ε q in 3 = 0 ( ) E = 0 b r c 0 ( 2 ) qin EA 4 = E4 4 π r =, ε ( ) q = 2Q+ Q = Q in Q kq E = = r c e 4 2 2 4πε0r r 0 ( )

Example Electrical field is zero inside conductors E1 = 0( r a) E3 = 0( b r c) Field exist outside of conductors 2kQ E2 = e ( a r b) 2 r Q kq e E4 = = r c 2 2 4πε r r 0 ( )

Derivation of Gauss s Law We will use a solid angle, Ω A spherical surface of radius r contains an area element ΔA The solid angle subtended at the center of the sphere Δ is defined to be Ω= 2 A r

Some Notes About Solid Angles A= 4πr 2 and r 2 have the same units, so Ω is a dimensionless ratio We give the name steradian to this dimensionless ratio The total solid angle subtended by a sphere is 4π steradians

Derivation of Gauss s Law, cont. Consider a point charge, q, surrounded by a closed surface of arbitrary shape The total flux through this surface can be found by evaluating E. ΔA for each small area element and summing over all the elements

Derivation of Gauss s Law, final The flux through each element is ΔAcosθ Φ E = E Δ A = ( E cosθ) Δ A= keq 2 r Relating to the solid angle ΔΩ = ΔAcosθ 2 r where this is the solid angle subtended by ΔA The total flux is dacosθ q r ε Φ E = kq e = kq 2 e dω = o

Derivation of Gauss s Law, cont. (a) qin =+ 3Q Q = + 2Q q in > 0 The charge distribution is spherically symmetric and. directed radially outward. E kq 2k Q = = r c e in e 2 2 r r

Insulating sphere 3Q within conducting shell -Q. Prob. 55 (d) q in = 0 b< r< c Since all points within this region are located. inside conducting material, E=0. Φ E = E d A = 0 qin = 0 Φ E = 0

Insulating sphere 3Q within conducting shell Q, cont. (f) qin = + 3 Q E kq e in 3ke Q = = a r< b 2 2. r r Directed radially outward.

Insulating sphere 3Q within conducting shell Q, cont. (h) q V r Q 3 + 3Q 4 3 r in = ρ = 3 4 3 = + 3 3 3 a π π a kq k r r E = = + Q = kq r r a a 3 e in e 3 3 2 2 3 e 3 Directed radially outward..

Problem (a) The inner charge is + q. (b) on the inner surface of the conductor, q where its surface density is: Inner surface density σ a = q 4π a 2 The outer surface carries charge Q. Outer surface density + q σ b = Q + q 4π b 2

Derivation of Gauss s Law Notes This result matches the earlier result for the equation for Gauss s law This derivation was independent of the shape of the closed surface This derivation was independent of the position of the charge within the surface

Problem Find the flux through the circular cap Φ = E EA 2π r= 2πRsin θ ds= R dθ θ 2 2 θ 2 2π ( 2π sinθ) θ 2π sinθ θ 2π ( cos θ) 2π ( 1 cos θ) A= rds= R Rd = R d = R = R 0 0 θ Φ 1 2 E = Q Q 2 R ( 1 cos ) ( 1 cos ) 4 π θ = 2 θ π 2 0 R 0 0

Electric Flux Field lines penetrating an area A perpendicular to the field The product of EA is the flux, Φ In general: Φ E = E A sin θ

Assessment Test Chapter 24

A uniformly charged rod has a finite length L. The rod is symmetric under rotations about the axis and under reflection in any plane containing the axis. It is not symmetric under translations or under reflections in a plane perpendicular to the axis other than the plane that bisects the rod. Which field shape or shapes match the symmetry of the rod? A. a and d B. c and e C. b only D. e only E. none of the above

This box contains A. a net positive charge. B. no net charge. C. a net negative charge. D. a positive charge. E. a negative charge.

The total electric flux through this box is A. 0 Nm 2 /C. B. 1 Nm 2 /C. C. 2 Nm 2 /C. D. 4 Nm 2 /C. E. 6 Nm 2 /C.

These are two-dimensional cross sections through threedimensional closed spheres and a cube. Rank order, from largest to smallest, the electric fluxes a to e through surfaces a to e. A. Φ a > Φ c > Φ b > Φ d > Φ e B. Φ b = Φ e > Φ a = Φ c = Φ d C. Φ e > Φ d > Φ b > Φ c > Φ a D. Φ b > Φ a > Φ c > Φ e > Φ d E. Φ d = Φ e > Φ c > Φ a = Φ b

These are two-dimensional cross sections through threedimensional closed spheres and a cube. Rank order, from largest to smallest, the electric fluxes a to e through surfaces a to e.

Which Gaussian surface would allow you to use Gauss s law to determine the electric field outside a uniformly charged cube? A. A sphere whose center coincides with the center of the charged cube. B. A cube whose center coincides with the center of the charged cube and which has parallel faces. C. Either A or B. D. Neither A nor B.

Chapter 24 Reading Quiz

The amount of electric field passing through a surface is called A. Electric flux. B. Gauss s Law. C. Electricity. D. Charge surface density. E. None of the above.

Gauss s law is useful for calculating electric fields that are A. due to point charges. B. uniform. C. symmetric. D. due to continuous charges.

Gauss s law applies to A. lines. B. flat surfaces. C. spheres only. D. closed surfaces.

The electric field inside a conductor in electrostatic equilibrium is A. uniform. B. zero. C. radial. D. symmetric.