Physics 2 lecture 2 gauss s law

Physics Lecture Gauss’s Law  Gauss’s Law  Conductors in Electrostatic Equilibrium Ngac An Bang, Faculty of Physics, HUS Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Gauss’s Law In a table-top plasma ball, the colorful lines emanating from the sphere give evidence of strong electric fields Using Gauss’s law, we show in this chapter that the electric field surrounding a charged sphere is identical to that of a point charge (Getty Images) Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Gauss’s Law Introduction  A very useful computational technique  Two of Maxwell’s Equations  Gauss’s Law –The Equation   q in E  EdA   0 Closed surface The total “flux” of field lines penetrating any of these surfaces is the same and depends only on the amount of charge inside Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Gauss’s Law   A  An Introduction Air flow rate (flux) through A     v  A  vA     v  A  vA cos  Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Gauss’s Law Electric flux   E  E A  EA cos 0  EA   A  An   E  E A  EA cos   EA ' • We showed that the strength of an electric field is proportional to the number of field lines per area • Electric flux ΦE is proportional to the number of electric field lines penetrating some surface Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Gauss’s Law Electric flux In general, a surface S can be curved and the electric field E may vary over the surface • In order to compute the electric flux, we divide the surface into a large number of infinitesimal area elements    Ai   Ai ni • The electric flux through an area element is    E  E i  Ai  E i  Ai cos  i • The total flux through the entire surface can be obtained by summing over all the area elements    E  lim  E i  Ai    Ai  i 1    E dA S 180    90   90   90  E   E   E  Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Gauss’s Law Electric flux A rectangle is an open surface —it does NOT contain a volume A sphere is a closed surface —it DOES contain a volume We are interested in closed surface Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Gauss’s Law Gauss’s law For a closed surface the normal vector is chosen to point in the outward normal direction   E  if E points out   E  if E points in Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Gauss’s Law Gauss’s law   E  if E points out   E  if E points in Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Gauss’s Law Gauss’s law The net electric flux through any closed surface S is   q in E  EdA   0 closed surface S where qin represents the net charge inside the surface and E represents the electric field at any point on the surface S 10 Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Some Applications Planar Symmetry (3) The Gaussian surface consists of three parts: a two ends S1and S2 plus the curved side wall S3 The flux through the Gaussian surface is    E   E d A  S        E1dA1   E dA2   E dA3 S1 S2  E  E1 A1  E A2   EA S3 (4) The amount of charge enclosed by the Gaussian surface is q= σA (5) Applying Gauss’s law gives EA  A 0 E  2 16 Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Some Applications Cylindrical Symmetry An infinitely long rod of negligible radius has a uniform positive charge density λ Find the electric field a distance r from the rod (1) An infinitely long rod possesses cylindrical symmetry (2) The electric field must be point radially away from the symmetry axis of the rod The magnitude of the electric field is constant on cylindrical surfaces of radius r Therefore, we choose a coaxial cylinder as our Gaussian surface 17 Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Some Applications Cylindrical Symmetry (3) The Gaussian surface consists of three parts: a two ends S1and S2 plus the curved side wall S3 The flux through the Gaussian surface is    E   E d A  S        E1dA1   E dA2   E dA3 S1 S2 S3  E    E  d A3  E 2rl  E 2rl s3 (4) The amount of charge enclosed by the Gaussian surface is q=λl (5) Applying Gauss’s law gives E 2rl  l 0 E  2 r 18 Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Some Applications Cylindrical Symmetry E  E   2 r  y q y 02  l  j For very long rod, l  ∞, then  E    q q   j  j  j  y l 4  y l / 2  y 19 Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Some Applications Spherical Symmetry A thin spherical shell of radius a has a charge +Q evenly distributed over its surface Find the electric field both inside and outside the shell • The charge distribution is spherically symmetric, with a surface charge density Q   4a • The electric field must be radially symmetric and directed outward Case 1: Inside the shell r < a • We choose our Gaussian surface to be a sphere of radius r, with r < a • Applying Gauss’s law   q in  E   E d A  0 0 S E  0, r  a 20 Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Some Applications Spherical Symmetry Case 2: outside the shell r > a • We choose our Gaussian surface to be a sphere of radius r, with r > a • Applying Gauss’s law   q in  E   E d A  0 S   a Q  E  E 4r   0 0 a Q E   r 4 r Note that the field outside the sphere is the same as if all the charges were concentrated at the center of the sphere 21 Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Some Applications Spherical Symmetry An insulating solid sphere of radius a has a uniform volume charge density ρ and carries a total positive charge Q (a) Calculate the magnitude of the electric field at a point outside the sphere Solution • The charge distribution is spherically symmetric, we again select a spherical gaussian surface of radius r > a, concentric with the sphere, as shown in Figure a • Applying Gauss’s law   q in  E   E d A  0 S Q  E  E 4r  0 Q E 4 r Point charge Note that the field outside the sphere is the same as if all the charges were concentrated at the center of the sphere 22 Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Some Applications Spherical Symmetry An insulating solid sphere of radius a has a uniform volume charge density ρ and carries a total positive charge Q (b) Find the magnitude of the electric field at a point inside the sphere Solution • We select a spherical gaussian surface having radius r < a, concentric with the insulating sphere (Fig b) • Applying Gauss’s law   q in  E   E d A  0 S    E   E d A  E 4r The total charge qin enclosed by the Gaussian sphere can be calculated as S    r  Q r Q E   r 3 4a 3 4 a 3 4r q in    d V    d V   V   V' V' ' 23 Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Some Applications Spherical Symmetry An insulating solid sphere of radius a has a uniform volume charge density ρ and carries a total positive charge Q Q The magnitude of the electric field at a point outside the sphere E 4 r The magnitude of the electric field at a point inside the sphere E Q r 4 a 24 Gauss’s Law Charged Isolated Conductor Ngac An Bang, Faculty of Physics, HUS Conductors in Electrostatic Equilibrium An insulator such as glass or paper is a material in which electrons are attached to some particular atoms and cannot move freely  A good electrical conductor contains charges (electrons) that are not bound to any atom and therefore are free to move about within the material When there is no net motion of charge within a conductor, the conductor is in electrostatic equilibrium 25 Gauss’s Law Charged Isolated Conductor Ngac An Bang, Faculty of Physics, HUS Conductors in Electrostatic Equilibrium As we shall see, a conductor in electrostatic equilibrium has the following properties: 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 of the conductor and has a magnitude σ/ε0 , where σ is the surface charge density at that point On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature of the surface is smallest 26 Gauss’s Law Charged Isolated Conductor Ngac An Bang, Faculty of Physics, HUS Conductors in Electrostatic Equilibrium The electric field is zero everywhere inside the conductor If the field was not zero, free charges in the conductor would accelerate under the action of the field Thus, the existence of electrostatic equilibrium is consistent only with a zero field in the conductor If an isolated conductor carries a charge, the charge resides on its surface The electric field is zero everywhere inside the conductor, thus   q in  E   E d A  0 0 S There can be no net charge inside the gaussian surface (which is arbitrarily close to the conductor’s surface), any exess charge on the conductor must reside on its surface 27 Gauss’s Law Charged Isolated Conductor Ngac An Bang, Faculty of Physics, HUS Conductors in Electrostatic Equilibrium The electric field just outside a charged conductor is perpendicular to the surface of the conductor and has a magnitude σ/ε0, where σ is the surface charge density at that point    E  Et  E n • The tangential component of E is zero on the surface of a conductor • Let’s consider a path integral of Eds around a closed path shown in the figure        Eds   Eds   Eds  ab bc    Eds  cd    Eds  da   ' E d s  E  l  E  x   E  x 0 t n n  In the limit where both Δx and Δx’→ 0, we have EtΔl= Δl is finite • If Et ≠ 0, the free charges would move along the surface; in such a case, the conductor would not be in equilibrium  Et  28 Gauss’s Law Charged Isolated Conductor Ngac An Bang, Faculty of Physics, HUS Conductors in Electrostatic Equilibrium The electric field just outside a charged conductor is perpendicular to the surface of the conductor and has a magnitude σ/ε0, where σ is the surface charge density at that point    E  Et  E n • The tangential component of E is zero on the surface of a conductor  Et  • The normal component of E has a magnitude σ/ε0   q in  E   E d A  0 S A  E  EA  E n A  0  E  En  0 29 Gauss’s Law Ngac An Bang, Faculty of Physics, HUS That’s enough for today  Please redo all the example problems given in your textbook  Feel free to contact me via email See you all next week! 30 [...]... flux is zero   q in E  EdA  0  0 closed surface S 11 Ngac An Bang, Faculty of Physics, HUS Gauss s Law Gauss s law Gauss s Law 12 Gauss s Law Gauss s Law and Coulomb s Law Ngac An Bang, Faculty of Physics, HUS  Around q we have drawn a spherical surface S of radius r centered at q  At any point on the surface S, E is parallel to the vector dA, thus   EdA  EdA  Applying Gauss s law, we... parts: a two ends S1 and S2 plus the curved side wall S3 The flux through the Gaussian surface is    E   E d A  S        E1dA1   E 2 dA2   E 3 dA3 S1 S2  E  E1 A1  E 2 A2  0  2 EA S3 (4) The amount of charge enclosed by the Gaussian surface is q= σA (5) Applying Gauss s law gives 2 EA  A 0 E  2 0 16 Ngac An Bang, Faculty of Physics, HUS Gauss s Law Some Applications Cylindrical... cylinder as our Gaussian surface 17 Ngac An Bang, Faculty of Physics, HUS Gauss s Law Some Applications Cylindrical Symmetry (3) The Gaussian surface consists of three parts: a two ends S1 and S2 plus the curved side wall S3 The flux through the Gaussian surface is    E   E d A  S        E1dA1   E 2 dA2   E 3 dA3 S1 S2 S3  E  0  0  E 3  d A3  E 3 2 rl  E 2 rl s3 (4) The... Faculty of Physics, HUS Gauss s Law Gauss s law Gauss s Law Consider several closed surfaces surrounding a charge q, as shown in figure above The number of field lines through S1 is equal to the number of field lines through the surfaces S2 and S3    E   E d A  S1    E dA  S2   q  S E dA   0 3 The number of electric field lines entering the surface equals the number leaving the surface... enclosed by the Gaussian surface is q=λl (5) Applying Gauss s law gives E 2 rl  l 0 E  2  0 r 18 Ngac An Bang, Faculty of Physics, HUS Gauss s Law Some Applications Cylindrical Symmetry E  E   2  0 r 1 4  0 y 0 q y 02  l 2 4  j For very long rod, l  ∞, then  E    1 q 1 q   j  j  j 2 4  0 y 0 l 4 4  0 y 0 l / 2 2  0 y 0 19 Ngac An Bang, Faculty of Physics, HUS Gauss s Law Some... to be a sphere of radius r, with r < a • Applying Gauss s law   q in  E   E d A  0 0 S E  0, r  a 20 Ngac An Bang, Faculty of Physics, HUS Gauss s Law Some Applications Spherical Symmetry Case 2: outside the shell r > a • We choose our Gaussian surface to be a sphere of radius r, with r > a • Applying Gauss s law   q in  E   E d A  0 S 2  4  a Q  E  E 4r 2   0 0 a 2 1 Q E...  E   E d A  S  E dA  S q 0  By symmetry, E is constant over the surface S, thus q  E   E d A  E  d A  E 4r  0 S S 2  The electric field E at any point on surface S is of the form 1 q E 4 0 r 2 13 Ngac An Bang, Faculty of Physics, HUS Gauss s Law Gauss s Law Applying Gauss s Law 1 Identify the symmetry associated with the charge distribution 2 Identify regions in which to calculate... the sphere 22 Ngac An Bang, Faculty of Physics, HUS Gauss s Law Some Applications Spherical Symmetry An insulating solid sphere of radius a has a uniform volume charge density ρ and carries a total positive charge Q (b) Find the magnitude of the electric field at a point inside the sphere Solution • We select a spherical gaussian surface having radius r < a, concentric with the insulating sphere (Fig... Applying Gauss s law   q in  E   E d A  0 S    E   E d A  E 4r 2 The total charge qin enclosed by the Gaussian sphere can be calculated as S    r  Q r Q E   r 3 0 4a 3 3 0 4 0 a 3 3 4r 3 q in    d V    d V   V   3 V' V' ' 23 Ngac An Bang, Faculty of Physics, HUS Gauss s Law Some Applications Spherical Symmetry An insulating solid sphere of radius a has a uniform... 2 4 0 r 2 Note that the field outside the sphere is the same as if all the charges were concentrated at the center of the sphere 21 Ngac An Bang, Faculty of Physics, HUS Gauss s Law Some Applications Spherical Symmetry An insulating solid sphere of radius a has a uniform volume charge density ρ and carries a total positive charge Q (a) Calculate the magnitude of the electric field at a point outside ...  0 closed surface S 11 Ngac An Bang, Faculty of Physics, HUS Gauss s Law Gauss s law Gauss s Law 12 Gauss s Law Gauss s Law and Coulomb s Law Ngac An Bang, Faculty of Physics, HUS  Around... of Physics, HUS Gauss s Law Some Applications Planar Symmetry (3) The Gaussian surface consists of three parts: a two ends S1 and S2 plus the curved side wall S3 The flux through the Gaussian surface... volume A sphere is a closed surface —it DOES contain a volume We are interested in closed surface Ngac An Bang, Faculty of Physics, HUS Gauss s Law Gauss s Law Gauss s law For a closed surface the