Gauss’s Law

Physics 2 Gauss’s Law Lecture 2  Gauss’s Law  Conductors in Electrostatic Equilibrium 1Ngac An Bang, Faculty of Physics, HUS Gauss’s Law Gauss’s Law Ngac An Bang, Faculty of Physics, HUS 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) 2 Gauss’s Law Gauss’s Law Introduction A very useful computational technique  Two of Maxwell’s Equations  Gauss’s Law –The Equation 0 in surfaceClosed d  qAEE    Ngac An Bang, Faculty of Physics, HUS 3 The total “flux” of field lines penetrating any of these surfaces is the same and depends only on the amount of charge inside Gauss’s Law Gauss’s Law Introduction Ngac An Bang, Faculty of Physics, HUS nAA    4 Air flow rate (flux) through A vAAv   cosvAAv   Gauss’s Law Electric flux nAA    Gauss’s Law Ngac An Bang, Faculty of Physics, HUS 5 • 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. EAEAAEE  00cos  'cos EAEAAEE    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 • The electric flux through an area element is • The total flux through the entire surface can be obtained by summing over all the area elements Gauss’s Law Ngac An Bang, Faculty of Physics, HUS 6 iii nAA   iiiiiE AEAE cos      S i i iAE AdEAE i  10 lim 0 90 0 E  0 90 0 E  0 90180 00  E  Gauss’s Law Electric flux Gauss’s Law Ngac An Bang, Faculty of Physics, HUS 7 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 Gauss’s Law Gauss’s law Gauss’s Law Ngac An Bang, Faculty of Physics, HUS 8 For a closed surface the normal vector is chosen to point in the outward normal direction. outpointsif0 EE  inpointsif0 EE  Gauss’s Law Gauss’s law Gauss’s Law outpointsif0 EE  inpointsif0 EE  Ngac An Bang, Faculty of Physics, HUS 9 Gauss’s Law Gauss’s Law Gauss’s law Ngac An Bang, Faculty of Physics, HUS 10 The net electric flux through any closed surface S is where qin represents the net charge inside the surface and E represents the electric field at any point on the surface S. 0surfaceclosed  in S E qAdE    Gauss’s law Gauss’s Law Gauss’s Law Ngac An Bang, Faculty of Physics, HUS 11 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 . 0321  qAdEAdEAdE SSS E    The number of electric field lines entering the surface equals the number leaving the surface. The net flux is zero. 0 0surfaceclosed   inSE qAdE  Gauss’s law Gauss’s Law Gauss’s Law Ngac An Bang, Faculty of Physics, HUS 12 Gauss’s Law and Coulomb’s Law Gauss’s Law 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 Applying Gauss’s law, we have  By symmetry, E is constant over the surface S, thus  The electric field E at any point on surface S is of the form AEAE dd  0 d  qAEAdE SS E    Ngac An Bang, Faculty of Physics, HUS 13 0 24dd   q rEAEAE SS E   2 04 1 r qE   Gauss’s Law Applying Gauss’s Law Gauss’s Law 1. Identify the symmetry associated with the charge distribution. 2. Identify regions in which to calculate E field. 3. Determine the direction of the electric field, and a “Gaussian surface” on which the magnitude of the electric field is constant over portions of the surface. 4. Calculate 5. Calculate q, charge enclosed by surface S. 6. Apply Gauss’s Law to calculate E: Ngac An Bang, Faculty of Physics, HUS 14  S E AdE  0 qAdE S E    Gauss’s Law Some Applications Planar Symmetry Consider an infinitely large non-conducting plane with uniform surface charge density σ. Determine the electric field everywhere in space. Ngac An Bang, Faculty of Physics, HUS 15 1. An infinitely large plane possesses a planar symmetry. 2. The electric field must point perpendicularly away from the plane. The magnitude of the electric field is constant on planes parallel to the non-conducting plane. 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 (4) The amount of charge enclosed by the Gaussian surface is q= σA. (5) Applying Gauss’s law gives EAAEAE AdEAdEAdEAdE E SSSS E 202211 332211 321     Gauss’s Law Ngac An Bang, Faculty of Physics, HUS 160 2  AEA  02 E 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. Gauss’s Law Ngac An Bang, Faculty of Physics, HUS 17 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 (4) The amount of charge enclosed by the Gaussian surface is q=λl. (5) Applying Gauss’s law gives rlErlEAE AdEAdEAdEAdE s E SSSS E  22d00 333 332211 3 321      Gauss’s Law Ngac An Bang, Faculty of Physics, HUS 180 2    l rlE  r E 02  Some Applications Cylindrical Symmetry r E 02  Gauss’s Law Ngac An Bang, Faculty of Physics, HUS 19 j lyy qE  44 1 22 000    For very long rod, l ∞, then j y j l q y j ly qE  0000 2 00 22/4 1 44 1     Some Applications Spherical Symmetry Gauss’s Law 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 • The electric field must be radially symmetric and directed outward. 24 a Q    Ngac An Bang, Faculty of Physics, HUS 20 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 0 0   inSE qAdE  arE  ,0 Some Applications Spherical Symmetry Gauss’s Law 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 00 2 2 0 44     Qa rE qAdE E in S E     Ngac An Bang, Faculty of Physics, HUS 21 2 0 2 0 2 4 1 r Q r aE    Note that the field outside the sphere is the same as if all the charges were concentrated at the center of the sphere. Spherical Symmetry Gauss’s Law Some Applications 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 Ngac An Bang, Faculty of Physics, HUS 22 0 2 0 4    Q rE qAdE E in S E     2 04 1 r QE   Note that the field outside the sphere is the same as if all the charges were concentrated at the center of the sphere. Point charge Spherical Symmetry Gauss’s Law Some Applications 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 Ngac An Bang, Faculty of Physics, HUS 23 0 in S E qAdE    24 rEAdE S E    The total charge qin enclosed by the Gaussian sphere can be calculated as 3 4dd 3 ' VV '' rVVVqin     r a Qr a QrE   3 00 3 0 43 3 43    Spherical Symmetry Gauss’s Law Some Applications An insulating solid sphere of radius a has a uniform volume charge density ρ and carries a total positive charge Q. The magnitude of the electric field at a point outside the sphere The magnitude of the electric field at a point inside the sphere 2 04 1 r QE   r a QE 3 04  Ngac An Bang, Faculty of Physics, HUS 24 Charged Isolated Conductor Conductors in Electrostatic Equilibrium Gauss’s Law 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. Ngac An Bang, Faculty of Physics, HUS 25 When there is no net motion of charge within a conductor, the conductor is in electrostatic equilibrium. Charged Isolated Conductor Conductors in Electrostatic Equilibrium Gauss’s Law As we shall see, a conductor in electrostatic equilibrium has the following properties: 1. The electric field is zero everywhere inside the conductor. 2. If an isolated conductor carries a charge, the charge resides on its surface. 3. 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. 4. On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature of the surface is smallest. Ngac An Bang, Faculty of Physics, HUS 26 Charged Isolated Conductor Conductors in Electrostatic Equilibrium Gauss’s Law 1. 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. Ngac An Bang, Faculty of Physics, HUS 27 2. If an isolated conductor carries a charge, the charge resides on its surface. The electric field is zero everywhere inside the conductor, thus 0 0   inSE qAdE  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. Charged Isolated Conductor Conductors in Electrostatic Equilibrium Gauss’s Law 3. 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. nt EEE   • 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 • If Et ≠ 0, the free charges would move along the surface; in such a case, the conductor would not be in equilibrium. Ngac An Bang, Faculty of Physics, HUS 28 0ddddd   dacdbcab sEsEsEsEsE   00d '  xExElEsE nnt In the limit where both Δx and Δx’→ 0, we have EtΔl= 0. Δl is finite 0tE  Charged Isolated Conductor Conductors in Electrostatic Equilibrium Gauss’s Law 3. 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. nt EEE   • The tangential component of E is zero on the surface of a conductor. • The normal component of E has a magnitude σ/ε0 Ngac An Bang, Faculty of Physics, HUS 29 0tE  0 0    AAEEA qAdE nE in S E     0  nEE 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. Gauss’s Law 30 See you all next week!