AljalalPhys 102 142 page 1 Chapter 23 Gauss

Aljalal-Phys. 102 -142 -page 1 Chapter 23 Gauss' Law 23 -1 Flux of an Electric Field 23 -2 Gauss' Law 23 -3 A Charged Isolated Conductor 23 -4 Applying Gauss' Law: Cylindrical Symmetry 23 -5 Applying Gauss' Law: Planar Symmetry 23 -6 Applying Gauss' Law: spherical Symmetry Objective

Aljalal-Phys. 102 -142 -page 2 23 -1 Flux of an Electric Field Another form of Coulomb's law Coulomb' Law Same as Gauss' Law Very useful in calculating the eclectic field from symmetrical shapes like spheres, cylinders, …

23 -1 Flux of an Electric Field Idea behind Gauss' law Gauss' Law Aljalal-Phys. 102 -142 -page 3 Gaussian surface If you know the electric field over any closed surface, you can calculate the net electric charge inside this closed surface

Aljalal-Phys. 102 -142 -page 4 23 -1 Flux of an Electric Field Illustration Gaussian surface imaginary sphere You can guess that the net charge inside the Gaussian surface is a positive charge. You can guess that the net charge inside the Gaussian surface is a negative charge.

23 -1 Flux of an Electric Field Gauss' law statement Aljalal-Phys. 102 -142 -page 5 Gauss' Law e 0 (Electric flux through any closed surface) = charge inside the surface

23 -1 Flux of an Electric Field Electric flux definition Aljalal-Phys. 102 -142 -page 6 Electric flux through a Gaussian surface Integration over a closed surface Area vector

23 -2 Flux of an Electric Field Gauss' law statement Aljalal-Phys. 102 -142 -page 7 Gauss' Law e 0 (Electric flux through any closed surface) = charge inside the surface

Aljalal-Phys. 102 -142 -page 8 23 -2 Flux of an Electric Field Area vector definition Area vector Magnitude : Area of a small surface on a Gaussian surface Normal to the surface Direction : and pointing to the outer side of the Gaussian closed surface Gaussian surface is any closed surface

Aljalal-Phys. 102 -142 -page 9 23 -1 Flux of an Electric Field Review - Dot product q Scalar quantity 900

23 -1 Flux of an Electric Field Electric flux depends on the angle between E and d. A 900 Aljalal-Phys. 102 -142 -page 10 450 Electric flux through the surface The electric flux through a surface is maximum when the electric field and the area vector are parallel.

23 -1 Flux of an Electric Field Electric flux and electric field lines Aljalal-Phys. 102 -142 -page 11 The magnitude of the electric field is proportional to the electric field lines per unit area perpendicular to the lines is proportional to the electric field lines passing through area The electric flux F through a Gaussian surface is proportional to the net number of electric field lines passing through that surface.

Aljalal-Phys. 102 -142 -page 12 23 -1 Flux of an Electric Field Example 1 Find the flux of the electric field through the closed surface of the cylinder. 900 Solution total flux = flux through surface a + flux through surface b + flux through surface c a c b

23 -1 Flux of an Electric Field Checkpoint 1 Aljalal-Phys. 102 -142 -page 13 y A Gaussian cube of face area A is immersed in a uniform electric field E pointing along the positive direction of z axis. A z Solution Flux through the front face = E A Flux through the rear face = - E A Flux through the top face = 0 Flux through the whole cube = 0 x

Aljalal-Phys. 102 -142 -page 14 23 -2 Flux of an Electric Field Example 2 y What is the electric flux through the right face? Solution cube z x =1 m x =3 m x Right face

Aljalal-Phys. 102 -142 -page 15 23 -1 Flux of an Electric Field Example 2 y left face What is the electric flux through the left face? Solution cube z x =1 m x =3 m x

Aljalal-Phys. 102 -142 -page 16 23 -1 Flux of an Electric Field Example 2 top face y What is the electric flux through the top face? Solution cube z x =1 m x =3 m x

23 -2 Gauss' Law Gauss' law statement Aljalal-Phys. 102 -142 -page 17 Gauss' Law e 0 (Electric flux through any closed surface) = charge inside the surface

23 -2 Gauss' Law Illustration Aljalal-Phys. 102 -142 -page 18 The electric field is outward for all points on this closed surface. The flux and charge inside are both positive. No Charge inside the surface. The electric field lines entering the surface also leaving it. The electric field is inward for all points on the surface. The flux and charge inside are both negative. Net charge inside this closed surface is zero. Number of the electric field lines entering the surface = number of lines leaving it.

Aljalal-Phys. 102 -142 -page 19 23 -2 Gauss' Law Checkpoint 2 What kind of net charge enclosed by these Gaussian cubes? Numbers indicate the flux on each face and arrows indicate the direction of the electric lines. 5 10 3 7 6 3 3 4 2 2 5 4 7 8 5 7 6 5 Solution No net charge Positive net charge Negative net charge

Aljalal-Phys. 102 -142 -page 20 23 -2 Gauss' Law For a point charge Gauss' law and Coulomb's law are the same. We will show this fact for the case of a point charge. Gauss' Law q Point charge Gaussian spherical surface of radius r Coulomb's Law

Aljalal-Phys. 102 -142 -page 21 23 -2 Gauss' Law Checkpoint 3 What is the relation between the electric flux through the three Gaussian surfaces? Small spherical surface q Solution All have the same flux through them. Cubical surface Large spherical surface

23 -2 Gauss' Law Example 3 Aljalal-Phys. 102 -142 -page 22 Solution From spherical symmetry and since the charge enclosed by the Gaussian sphere is positive, the electric field must be radially outward.

23 -2 Gauss' Law Example 3 Aljalal-Phys. 102 -142 -page 23 Solution From spherical symmetry and since the charge enclosed by the Gaussian sphere is negative, the electric field must be radially inward.

Aljalal-Phys. 102 -142 -page 24 23 -3 A Charged Isolated Conductor Electric field inside a conductor Conductor inside E = 0 -e Free electrons always present in any conductor If inside E≠ 0, free electrons move. Electrostatic Not electrostatic equilibrium. In electrostatic equilibrium, the electric field inside a conductor is zero.

Aljalal-Phys. 102 -142 -page 25 23 -3 A Charged Isolated Conductor Excess charge of a conductor In electrostatic equilibrium Gaussian surface inside the conductor Since the electric field inside the conductor = 0 Conductor inside E = 0 Flux through any Gaussian surface inside the conductor = 0 Net charge inside the conductor = 0 In electrostatic equilibrium, all excess charge on a conductor is entirely on the conductor's outer surface.

Aljalal-Phys. 102 -142 -page 26 23 -3 A Charged Isolated Conductor Net charge on a cavity within a conductor In electrostatic equilibrium inside Conductor cavity totally within the E = 0 conductor Flux though the closed surface = 0 No net charge inside the surface No net charge on a cavity wall which is totally within a conductor.

Review Gauss' Law Area vector Magnitude : Area of the surface Direction : Normal to the surface and pointing to the outer side of the Gaussian closed surface In electrostatic equilibrium, • The electric field inside a conductor is zero. • All excess charge on a conductor is entirely on the conductor's outer surface. • No net charge on a cavity wall which is totally within a conductor.

23 -3 A Charged Isolated Conductor Electric field is produced by the charge distribution Conductor Aljalal-Phys. 102 -142 -page 28 No conductor E = 0 Same charge distribution Same electric field Electric field is produced by the charge distribution.

23 -3 A Charged Isolated Conductor Electric field from non-uniform charge distribution Lower charge density Conductor sphere Uniform charge distribution Electric field is very easy to calculate. Aljalal-Phys. 102 -142 -page 29 Higher charge density Conductor charge distribution is not uniform In general, electric field is very difficult to calculate.

23 -3 A Charged Isolated Conductor Electric field near the surface Lower charge density Higher charge density Aljalal-Phys. 102 -142 -page 30 Choose a very small area on the conductor's surface such that the surface is almost flat and the charge distribution is almost uniform. Conductor charge distribution is not uniform In general, electric field is very difficult to calculate. But we can find the electric field just near the surface by using Gauss' law.

Aljalal-Phys. 102 -142 -page 31 23 -3 A Charged Isolated Conductor Electric field is perpendicular to the surface of a conductor In electrostatic equilibrium 900 Conductor In electrostatic equilibrium, electric field is perpendicular to the surface of conductors.

Aljalal-Phys. 102 -142 -page 32 23 -3 A Charged Isolated Conductor Proof - electric field is normal to the surface of a conductor If the electric field were not perpendicular to the surface of a conductor, free electrons would move. This would not be an electrostatic equilibrium. Free electrons -e Conductor always present on the surface of a conductor. There would be a component of E along the surface. In electrostatic equilibrium, electric field is perpendicular to the surface of conductors.

23 -3 A Charged Isolated Conductor A Gaussian surface through a surface of a conductor Aljalal-Phys. 102 -142 -page 33 Cross section Conductor a Gaussian surface Area of the cylinder cap = A

Aljalal-Phys. 102 -142 -page 34 23 -3 A Charged Isolated Conductor Magnitude of the electric field near the surface of a conductor Flux through the rear cap = 0, since E = 0 inside the conductor. Conductor Flux through the cylinder surface = 0, since the electric field is perpendicular to the area vector Flux through the front cap = E A, since just outside the conductor the electric field is paralell to the area vector. Here, A is the area of the cylinder cap. Total electric flux through the cylinder = E A e 0 F = qenc e 0 E A = qenc surface charge density

23 -3 A Charged Isolated Conductor Electric field just near the surface of a conductor Aljalal-Phys. 102 -142 -page 35 surface charge density Magnitude: Direction: Perpendicular to the conductor's surface Conductor Positively charged Conductor Negatively charged

Aljalal-Phys. 102 -142 -page 36 23 -3 A Charged Isolated Conductor Example 5 Neutral spherical q = -5 m. C metal shell q What is the induced charges on the inner surface? R R/2 Solution Metal = conductor. Inside the conductor, E = 0. Electric flux through the Gaussian surface = 0 e 0 F = qenc = 0 Gaussian Surface q= -5 m. C Net charge inside the Gaussian surface = 0 q + Charge on the inner surface of the shell = 0 Charge on the inner surface of the shell = - q = + 5 m. C - q = +5 m. C

Aljalal-Phys. 102 -142 -page 37 23 -3 A Charged Isolated Conductor Example 5 Neutral spherical metal shell q What is the induced charges on the outer surface? Solution +q The metal shell is neutral Net charge of the shell = 0 Charge on the outer surface + Charge on the inner surface = 0 Charge on the outer surface = - Charge on the inner surface Charge on the outer surface = - (-q) = q Charge on the outer surface = - 5 m. C - q

Aljalal-Phys. 102 -142 -page 38 23 -3 A Charged Isolated Conductor Example 5 Neutral spherical metal shell Are the charges induced on the inner surface uniformly distributed? Solution The point charge inside the shell repels electrons strongly from this area. More positive q + q q - q The point charge inside the shell repels electrons weakly from this area. less positive On the inner surface, charges are not uniformly distributed.

Aljalal-Phys. 102 -142 -page 39 23 -3 A Charged Isolated Conductor Example 5 Neutral spherical metal shell q Are the charges induced on the outer surface uniformly distributed? Solution + q - q Since in the conductor E = 0, the inside charges can not affect (apply force) on the outer charges. q Charges on the outer spherical surface distribute themselves to minimize the forces among themselves. On the outer surface, charges are uniformly distributed.

23 -4 Applying Gauss' Law: Cylindrical Symmetry Electric field points along the radial direction Aljalal-Phys. 102 -142 -page 40 Infinity 900 From symmetry, the electric field points along the radial direction Infinity Infinitely long cylindrical rod with a uniform linear charge density l. The linear charge density l is the charge per unit length.

23 -4 Applying Gauss' Law: Cylindrical Symmetry Electric field direction Aljalal-Phys. 102 -142 -page 41 Cross section of a cylindrical rod with a uniform linear charge density electric field lines Short rod Long rod Infinitely long rod For an infinitely long rod, the electric field points perfectly along the radial direction.

23 -4 Applying Gauss' Law: Cylindrical Symmetry Electric field outside an infinitely long rod Aljalal-Phys. 102 -142 -page 42 Infinitely long rod with a uniform linear charge density l r 900 Cylindrical Gaussian surface of radius r and coaxial with the rod h Flux through the upper and lower caps = 0 since the electric field is perpendicular to the area vector. Since the electric field is parallel to the area vector, flux through the cylinder surface = E (area of the cylinder surface) = E ( 2 p r h) Total electric flux through the Gaussian surface = E ( 2 p r h) e 0 F = qenc Linear charge density e 0 E (2 p r h) = qenc

23 -4 Applying Gauss' Law: Cylindrical Symmetry Electric field direction outside infinitely long rod Aljalal-Phys. 102 -142 -page 43 Magnitude: 900 outward in radial direction 900 inward in radial direction

23 -5 Applying Gauss' Law: Planar Symmetry Electric field direction Thin infinite nonconducting sheet Aljalal-Phys. 102 -142 -page 44 Cross section A Cylindrical Gaussian surface with cap area of A From symmetry, the electric field is perpendicular to the sheet

23 -5 Applying Gauss' Law: Planar Symmetry Electric field of a thin infinite nonconducting sheet Cross section Aljalal-Phys. 102 -142 -page 45 Flux through the cylinder surface = 0 since the electric field is perpendicular to the area vector. Flux through the right cap = Flux through the left cap = E A since the electric field is parallel to the area vector. Total electric flux through the cylinder = E A + E A = 2 E A e 0 F = qenc e 0 (2 E A) = qenc surface charge density

23 -5 Applying Gauss' Law: Planar Symmetry Two infinite nonconducting sheets The two sheets are parallel s 1 > s 2 Cross section Aljalal-Phys. 102 -142 -page 46 and close to each other Cross section Isolated nonconducting sheet s 1 Uniform surface charge density s 2 Uniform surface charge density s 1 s 2 Superposition In nonconducting sheets, charge cannot move.

23 -5 Applying Gauss' Law: Planar Symmetry Charge on a conducting sheet Aljalal-Phys. 102 -142 -page 47 No excess charge inside the conductor. The excess charge lies on the surface of the sheet. We define charge density for each side of the conductor. put charge Conducting Sheet s Nonconducting sheet s 2 sconductor = snonconductor. s

Aljalal-Phys. 102 -142 -page 48 23 -5 Applying Gauss' Law: Planar Symmetry Two infinite conducting sheets The two sheets are parallel and close to each other Cross section Isolated conducting sheet s 0 -s 0 Superposition of the electric fields Same magnitude In conducting sheets, charge can move.

Aljalal-Phys. 102 -142 -page 49 23 -5 Applying Gauss' Law: Planar Symmetry Surface charge densities of two infinite conducting sheets Two conducting sheets with the same surface charge density Cross section Flux = 0 since E =0 s =0 Flux = E A s =2 s 0 Flux = 0 since E =0 s =0 Flux =- E A s = - 2 s 0

Aljalal-Phys. 102 -142 -page 50 23 -5 Applying Gauss' Law: Planar Symmetry Example 6 Two nonconducting large sheets s(+) = 6. 8 m. C/m 2 s(-) = 4. 3 m. C/m 2 Find the electric field to the right, between and to the left of the two plates. Cross section Solution s(+) To the left To the right s(-)

Aljalal-Phys. 102 -142 -page 51 23 -6 Applying Gauss' Law: spherical Symmetry Electric field outside a uniformly charged spherical shell A thin, uniformly charged spherical shell with total charge q and of radius R Spherical Gaussian surface of radius r Outside the shell r > R r R Similar to the electric field from a point charge placed at the center of the sell. A shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell's charge were concentrated at its center.

23 -6 Applying Gauss' Law: spherical Symmetry Electric field inside a uniformly charged spherical shell Aljalal-Phys. 102 -142 -page 52 A thin, uniformly charged spherical shell with total charge q and of radius R inside the shell r < R R r If a charged particle is located inside a shell of uniform charge, there is no net electrostatic force on the particle from the shell.

Aljalal-Phys. 102 -142 -page 53 23 -6 Applying Gauss' Law: spherical Symmetry Electric field outside a uniformly charged sphere A sphere of radius R and with total charge q uniformly distributed throughout its volume Outside the sphere r > R r R Spherical Gaussian surface of radius r

Aljalal-Phys. 102 -142 -page 54 23 -6 Applying Gauss' Law: spherical Symmetry Electric field inside a uniformly charged sphere A sphere of radius R and with total charge q uniformly distributed throughout its volume inside the sphere r < R R r Spherical Gaussian surface of radius r

Aljalal-Phys. 102 -142 -page 55 23 -6 Applying Gauss' Law: spherical Symmetry Checkpoint 5 Large nonconducting sheets with identical positive uniform charge surface density Rank electric field at points, greatest first Solution 1 Electric field due the sheets is zero. You need only to consider the field from the sphere 3 and 4 tie then 2 then 1 2 d 4 3 d d s 0 Sphere with uniform positive charge density