Applications of Gauss’ Law : Applications of Gauss’ Law In cases of strong symmetry, Gauss's law may be readily used to calculate E . Otherwise it is not generally useful and integration over the charge distribution is required. But when the symmetry permits it, Gauss's law is the easiest way to go! The KEY TO ITS APPLICATION is the choice of Gaussian surface . Keep in mind that this is not a surface of the charge distribution itself, but rather an imaginary surface constructed for application of Gauss's law. To make Gauss's law useful: All sections of the gaussian surface should be chosen so that they are either parallel or perpendicular to E . For this we need to have already determined the direction of E everywhere on the surface by symmetry arguments. | E | should be constant on each surface. Again, this requires sufficient symmetry to "see" the constancy of field strength. And once again, make sure that the surface is closed!
Gauss’ Law and Coulomb law: Gauss’ Law and Coulomb law A spherical Gaussian surface centered on a point charge q The angle between E and dA is zero at any point on the surface, we can re-write Gauss’ Law as E has the has same value at all points on the surface E is can be moved out Integral is the sum of surface area Coulomb’s Law
Cylindrical Symmetry: Cylindrical Symmetry Evaluate the electric field that arises from an infinite line of charge 1) Understand the geometry Pick a point P for evaluation of E . The only variable on which E may depend is R, the distance from the line of charge. There is no angular dependence because the line is cylindrically symmetric, i.e. it does not matter at which angle about the line you view it. There is no axial dependence because the line of charge is infinitely long, i.e. there is no preferred position along the line. Similar arguments determine that the direction of E is directly away from the line, perpendicular to it at all locations. There may be no azimuthal or axial components because the line is infinite and has no extent in the transverse directions. 2) Understand the symmetry
Cylindrical Symmetry: Cylindrical Symmetry 3) Construct the Gaussian surface A cylindrical tube is chosen to match the symmetry of the line of charge, centered about it, with a radius equal to the distance between the point of evaluation and the line of charge. The electric field anywhere on this surface has the same direction as the infinitesimal area vector and has a constant value everywhere on the surface. This surface is denoted by 2. Since a gaussian surface must be closed, the tube is then capped with flat endcaps, 1 and 3. The electric field does not have the same value at all points on the endcaps, but the field vector is perpendicular to the area vector so there is no flux through the endcaps.
Cylindrical Symmetry: Cylindrical Symmetry The flux through the endcap 1 is zero since A 1 is perpendicular to E everywhere on this surface. Similarly, the flux through surface 3 is zero. Since the electric field is everywhere constant on surface 2 and points in the same direction as the surface vector at any point, the flux through surface 2 will be the field magnitude E times the area of the tube wall. 4) Examine the Gaussian surface
Cylindrical Symmetry: Cylindrical Symmetry 5) Evaluate the electric flux through the Gaussian surface 6) Evaluate the charge enclosed by the Gaussian surface
Cylindrical Symmetry: Cylindrical Symmetry 7) Apply Gauss's law for the result! Source: http://www.physics.udel.edu/~watson/phys208/line-gauss/line-gauss1.html
For comparison Line of Charge by Direct Integration 1: For comparison Line of Charge by Direct Integration 1 Evaluate the electric field that arises from an infinite line of charge. Pick a point P for evaluation of E . The only variable which matters is R, the distance from the line of charge. There is no angular dependence because the line is cylindrically symmetric, i.e. it does not matter at which angle about the line you view it. There is no axial dependence because the line of charge is infinitely long, i.e. there is no preferred position along the line. The infinite line of charge is certainly a physically-impossible situation to set up (infinite amount of charge) but one whose consideration is useful in interpreting finite distributions, e.g. checking the limiting behavior of solutions.
For comparison Line of Charge by Direct Integration 2: Span the charge distribution Assuming a uniform charge distribution along a line, the variable which appears to be the easiest to use to span the distribution is the variable x , which follows along the line from negative infinity to positive infinity. Thus the charge infinitesminal dq will be the product of charge density and dx. For comparison Line of Charge by Direct Integration 2
For comparison Line of Charge by Direct Integration 3: For comparison Line of Charge by Direct Integration 3 Evaluate the contribution from the infinitesimal charge element Considering a positive test charge at point P, the direction of the electric field is shown. The strength of the electric field contribution from the infinitesimal charge shown is proportional to dq and inversely proportional to the square of the distance separating point P and the charge element.
For comparison Line of Charge by Direct Integration 4: For comparison Line of Charge by Direct Integration 4 Exploit symmetry as appropriate By examining the conjugate infinitesimal on the other side of the origin, we see that the horizontal components of the two contributions will balance. We need only consider the vertical contribution to the electric field by each infinitesimal charge.
For comparison Line of Charge by Direct Integration 5: For comparison Line of Charge by Direct Integration 5 In setting up the integral we wish to express all factors entering the electric field contribution as functions of the integration variable x . After finding the magnitude of the infinitesimal contribution to the field, we consider its projection along the vertical axis. After the integrand is determined, we evaluate the integral between the limits of integration. Substituting these terms, dEy becomes Setting up the integrating over the charge distributing Set up the integral
For comparison Line of Charge by Direct Integration 6: For comparison Line of Charge by Direct Integration 6 Solve the integral by trig substitution Looking at the integrand, you may now realize that trig substitution is required. All that means is that we should have integrated over the angle variable theta rather than the linear variable x . The figure below shows the relationship between the two variables for the quantities involved
For comparison Line of Charge by Direct Integration 7: For comparison Line of Charge by Direct Integration 7 Expressing the integrand in terms of theta and switching the limits of integration, the integral may be solved and the result finally obtained!
A Charged Isolated Conductor: A Charged Isolated Conductor Gauss’ Law permits us to prove an important theorem about conductors. If excess charged is placed on an isolated conductor, the charge will move to the surface of the conductor. None of the excess charge will be found within the body pf he conductor. The electric field inside a conductor in static equilibrium must be zero. Otherwise, the field would exert forces on the conduction electrons producing motion or currents which is not a static equilibrium state. If we place a Gaussian surface just inside the surface of the charged conductor, the E field is zero for all points on the Gaussian surface. Therefore the flux through the Gaussian is zero and according to Gauss’ Law the net charge enclosed the Gaussian surface must also be zero. Therefore the excess charged must be outside the Gaussian surface and must lie on the actual surface of the conductor.
External Electric Field of a Conductor: External Electric Field of a Conductor Consider a section of the surface that is small enough to to neglect any curvature and assume the section is flat. Embed a tiny cylindrical Gaussian surface with one end cap full inside the conductor and the other fully outside and cylinder is perpendicular to the surface . The electric field just outside the surface must also be perpendicular to that surface otherwise surface charges would be subject to motion.
External Electric Field of a Conductor: External Electric Field of a Conductor We now sum the flux through the Gaussian surface. There is no flux through the internal endcap. Why ?? There is no flux through the curved surface. Why ?? The only flux through the Gaussian surface is that through the external endcap where E is perpendicular. Assuming a cap area of A, the flux through the cap is EA. The charge q enc enclosed by the Gaussian surface lies on the conductor’s surface area A. If is the charge per unit area , the q enc is equal to A. Therefore Gauss’ Law becomes o EA = A E = / o (conducting surface)
Planar Symmetry: Planar Symmetry Two different views of a very large thin plastic sheet, uniformly charged on one side with surface density of . A closed cylindrical Gaussian surface passes throught the sheet and is perpendicular. From symmetry , E must be perpedicular to the surface and endcaps. The surface charge is positive so E emanates from the surface. Since the field lines do not piece the curved surface there is no flux. What happens is we magically make the sheet a conductor ?? Will the E field change ?
Planar Symmetry: Planar Symmetry Figure (a) show a cross-section of a thin, infinite conducting plate with excess positive charge. We know the this excess charge lies on the surface of the plate. If there is no external electric field to force the charge into some particular distribution, it will spread out onto the two surfaces with an uniform charge density of 1 . From the previous slide we know this charge sets up an electric field which points 1 away from the plate. Figure (b) is a identical plate with negative charge. In this case E points inward.
Planar Symmetry: Planar Symmetry Suppose we move the two plates in the figures (a) and (b) close together and parallel. Since the plates are conductors, the excess charge on one plate attracts the excess charge on the other and all the excess charge moves into the inner faces of the plates. The new surface charge density has now doubled, therefore the electric field between the plates is The field points away from positive charged plate and toward the negative plate. Since no excess is left on the outer faces, the electric field outside is zero.
Spherical Symmetry: Spherical Symmetry Let’s apply Gauss’s Law to a hollow charged sphere What is electric field inside the hollow charged sphere ? Consider the point mark X in the figure located on an interior equator By symmetry Ey=0, what about Ex ?? (Since X can be viewed at any angle, the E-field must be in the radial direction. x
Spherical Symmetry: Spherical Symmetry x x
Spherical Symmetry: Spherical Symmetry Instead of deriving the solution through a long integration Let’s use Gauss’ Law. 1 st choose an Gaussian surface inside the shell that coincides with the point x. x Since fields lines can not cross, the spherical symmetry requires E field to be in the radially direction and uniform on the Gaussian surface, therefore Gauss’ Law r
Slide 24: Spherical Symmetry Outside the hollow sphere Now let’s consider the electric field a a point outside the charged sphere. We define a Gaussian surface around the sphere for which r>R. Again since fields lines can not cross, the spherical symmetry requires E field to be in the radially direction and uniform on the Gaussian surface, therefore Gauss’ Law becomes x Same as point charge
Spherical Symmetry: Spherical Symmetry Using Gauss’ Law we were able to prove two shell theorems that were presented in the 1 st lecture without proof. A shell of uniform charge attracts and repels a charged particle that is outside the shell as if all shell’s charge were concentrated at the center. If a charged particle is located inside a shell of uniform charge, there is no net electric force on the particle from the shell.