Transcript Gauss` law
Gauss’s Law Summer 2008 Gauss’ law In this chapter we will introduce the following new concepts: The flux (symbol Φ ) of the electric field. Symmetry Gauss’ law We will then apply Gauss’ law and determine the electric field generated by: An infinite, uniformly charged insulating plane. An infinite, uniformly charged insulating rod. A uniformly charged spherical shell. A uniform spherical charge distribution. We will also apply Gauss’ law to determine the electric field inside and outside charged conductors. (23-1) Last Time: Definition-Electric Field Lines Field Lines Electric Field For a charged plane Parallel plate capacitor s/2e0 s/2e0 E=0 s/2e0 s/2e0 s/2e0 s/2e0 E=s/e0 We will use this a lot! E=0 NEW RULES Imagine a region of space where the ELECTRIC FIELD LINES HAVE BEEN DRAWN. The electric field at a point in this region is TANGENT to the Electric Field lines that have been drawn. If you construct a small rectangle normal to the field lines, the Electric Field is proportional to the number of field lines that cross the small area. The DENSITY of the lines. We won’t use this much n̂ n̂ So far … The electric field exiting a closed surface seems to be related to the charge inside. But … what does “exiting a closed surface mean”? How do we really talk about “the electric field exiting” a surface? How do we define such a concept? CAN we define such a concept? Mr. Gauss answered the question with.. The “normal component” of the ELECTRIC FIELD E n E n = E cos( )n E n = E cos( ) n = E n En DEFINITION FLUX E n En Flux = E n A = (E n)A = E A cos( ) “Element” of Flux of a vector E leaving a surface d = E dA = E NORMAL A also d = E dA = E ndA For a CLOSED surface: n is a unit OUTWARD pointing vector. Definition of TOTAL FLUX through a surface = d surface Total Flux of the Electric Field LEAVING a surface is = E n outdA Visualizing Flux flux = E ndA = n is the OUTWARD pointing unit normal. Remember flux = E ndA = E n = E n cos( ) n E A Component of E perpendicular to surface. This is the flux passing through the surface and n is the OUTWARD pointing unit normal vector! Example Cube in a UNIFORM Electric Field Flux is EL2 E Flux is -EL2 L area Note sign E is parallel to four of the surfaces of the cube so the flux is zero across these because E is perpendicular to A and the dot product is zero. Total Flux leaving the cube is zero Simple Example 1 r q = E ndA = dA 2 4e0 r Sphere 1 q q = 2 4e0 r 1 q dA = 4e0 r 2 A q q 2 = 4r = 2 4e0 r e0 1 Gauss’ Law Flux is total EXITING the Surface. n is the OUTWARD pointing unit normal. qenclosed = E ndA = e0 q is the total charge ENCLOSED by the Gaussian Surface. qenclosed = dA E n e0 Line of Charge Q L charge Q = = length L Materials Conductors Electrons are free to move. In equilibrium, all charges are a rest. If they are at rest, they aren’t moving! If they aren’t moving, there is no net force on them. If there is no net force on them, the electric field must be zero. THE ELECTRIC FIELD INSIDE A CONDUCTOR IS ZERO! (23-6) e F v E The electric field inside a conductor We shall prove the the electric field inside a conductor vanishes Consider the conductor shown in the figure to the left. It is an experimental fact that such an object contains negatively charged electrons which are free to move inside the conductor. Lets assume for a moment that the electric field is not equal to zero. In such a case an non-vanishing force F = eE is exerted by the field on each electron. This force would result in a non-zero velocity v and the moving electrons would constitute an electric current. We will see in subserquent chapters that electric currents manifest themselves in a variety of ways: a) they heat the conductor b) they generate magnetic fields around the concuctor No such effects have ever been observed. Thus the original assumption that there exists a non-zero electric field inside the conductor. Thus we conclude that : The electrostatic electric field E inside a conductor is equal to zero Isolated Conductor Electric Field is ZERO in the interior of a conductor. Gauss’ law on surface shown Also says that the enclosed Charge must be ZERO. All charge on a Conductor must reside on The SURFACE. Charged Conductors Charge Must reside on the SURFACE - E=0 - - s Very SMALL Gaussian Surface E sA EA = e0 or s E= e0 Charged Isolated Conductor The ELECTRIC FIELD is normal to the surface outside of the conductor. s The field is given by: E= e0 Inside of the isolated conductor, the Electric field is ZERO. If the electric field had a component parallel to the surface, there would be a current flow!!! Isolated (Charged) Conductor with a HOLE in it. Q E dA = 0 = e n 0 Because E=0 everywhere inside the conductor. There is no charge on the cavity walls. All the excess charge q remains on the outer surface of the conductor Symmetry: We say that an object is symmetric under a particular mathematical operation (e.g. rotation, translation, …) if to an observer the object looks the same before and after the operation. Note: Symmetry is a primitive notion and as such is very powerful Rotational symmetry featureless sphere rotation (23-10)axis observer Example of spherical symmetry Consider a featureless beach ball that can be rotated about a vertical axis that passes through its center. The observer closes his eyes and we rotate the sphere. Then, when the observer opens his eyes, he cannot tell whether the sphere has been rotated or not. We conclude that the sphere has rotational symmetry about the rotation axis. Line of Charge q E dA = e n 0 h E 2rh = e0 E= 2e0 r From SYMMETRY E is Radial and Outward Infinite Sheet of Charge +s h E cylinder sA EA + EA = e0 s E= 2e 0 We got this same result from that ugly integration! E= s 2e o Electric field generated by a thin, infinite, non - conducting uniformly charged sheet. We assume that the sheet has a positive charge of surface density s . From symmetry, the electric field vector E is perpenducular to the sheet and has a constant magnitude. Furthermore it points away from the sheet. We choose a cylindrical Gaussian surface S with the caps of area A on either side of the sheet as shown in the figure. We devide S into three sections: S1 is the cap to the right of the sheet. S2 is the curved surface of the cylinder. S3 is the cap to the left of the sheet. The net flux through S is = 1 + 2 + 3 n̂2 S3 n̂3 1 = 2 = EA cos 0 = EA. 3 = 0 S2 n̂1 S1 ( =90) = 2 EA. From Gauss's law we have: = 2 EA = sA eo E= s 2e o qenc eo = sA eo (23-15) Ei = S A 2s 1 eo Eo = 0 A' S' The electric field generated by two parallel conducting infinite planes charged with surface densities s 1 and -s 1. In fig.a and b we shown the two plates isolated so that one does not influence the charge distribution of the other. The charge speads out equally on both faces of each sheet. When the two plates are moved close to each other as shown in fig.c, then the charges one one plate attract those on the other. As a result the charges move on the inner faces of each plate. To find the field Ei between the plates we apply Gauss' law for the cylindrical surface S which has caps of area A. q 2s A 2s The net flux = Ei A = enc = 1 Ei = 1 To find the field Eo outside eo eo eo the plates we apply Gauss' law for the cylindrical surface S which has caps of area A. q s s1 The net flux = Eo A = enc = 1 = 0 Eo = 0 (23-16) eo eo n̂2 Eo The electric field generated by a spherical shell of charge q and radius R. Inside the shell : Consider a Gaussian surface S1 which is a sphere with radius r R and whose center coincides with that of the charge shell. q The electric field flux = 4 r 2 Ei = enc = 0 eo Thus Ei = 0 n̂1 Ei = 0 Eo = q 4e o r 2 (23-17) Ei Outside the shell : Consider a Gaussian surface S2 which is a sphere with radius r R and whose center coincides with that of the charge shell. The electric field flux = 4 r 2 Eo = Thus Eo = qenc eo = q eo q 4e o r 2 Note : Outside the shell the electric field is the same as if all the charge of the shell were concentrated at the shell center. Insulators In an insulator all of the charge is bound. None of the charge can move. We can therefore have charge anywhere in the volume and it can’t “flow” anywhere so it stays there. You can therefore have a charge density inside an insulator. You can also have an ELECTRIC FIELD in an insulator as well. Recipe for applying Gauss’ Law 1. Make a sketch of the charge distribution 2. Identify the symmetry of the distribution and its effect on the electric field 3. Gauss’ law is true for any closed surface S. Choose one that makes the calculation of the flux as easy as possible. 4. Use Gauss’ law to determine the electric field vector qenc = eo (23-13) A uniformly charged cylinder. rR s R s (r 2 h) E (2rh ) = e0 r E= 2e 0 rR (R 2 h) E (2rh ) = = e0 e0 q R 2 E= 2e 0 r