Physics 122B Electricity and Magnetism Lecture 6 Flux and Gauss’s Law April 06, 2007 Martin Savage

Lecture 6 Announcements Lecture Homework #2 has been posted on the Tycho system. It is due at 10 PM on Wednesday, April 11. On Friday, April 13, we will have Midterm 1, covering Chapters 25-27. You may bring one page of notes. Also bring a Scantron sheet and a scientific calculator with good batteries. A discussion forum for this class now exists…use it as you please…discussions related to physics only !!

Chapter 26 Summary (1) 12/27/2018 Physics 122B - Lecture 6

Chapter 26 Summary (2) 12/27/2018 Physics 122B - Lecture 6

Types of Symmetry (again) Symmetry Principle: The electric field must have the same symmetries as the charge distribution that produced it. Full rotational symmetry: any rotation about any axis does not change object: Sphere Translational symmetry: translation along a line does not change object. Cylindrical symmetry: any rotation about one axis does not change object: Cylinder Reflectional symmetry: mirror image reflection about some plane is same as object. A (yes) A E (no) E 720 Partial rotational symmetry: 720 rotation about one axis does not change object: Star 12/27/2018 Physics 122B - Lecture 6

Symmetry of a Long Cylinder z A long cylinder has the following symmetries: Translational symmetry along the z axis; Full rotational symmetry about the z axis; Reflectional symmetry in any plane containing the z axis; Reflectional symmetry in any plane ^ the z axis. 12/27/2018 Physics 122B - Lecture 6

Rejected Possibilities (1) Question: Could the E field have z-axis components? Symmetry Answer: Consider the 4th symmetry: (reflection in a plane ^ z) The field changes on this reflection, but the charged object does not. Therefore, the E-field of the long cylinder cannot have z-axis components 12/27/2018 Physics 122B - Lecture 6

Rejected Possibilities (2) Question: Could the E-field have q-axis components? Symmetry Answer: Consider the 3rd symmetry: (reflection plane contains z) r z q The field changes on this reflection, but the charged object does not. Therefore, the E-field cannot have q-axis components. r Conclusion: In cylindrical coordinates, the electric field must be radial (r), with no components in the q or z directions. 12/27/2018 Physics 122B - Lecture 6

What Good is Symmetry? Symmetry allows us to rule out many conceivable field shapes that are found to be incompatible with the symmetries of the charge distribution. Knowing what cannot happen is often as useful as knowing what must happen. By using symmetry arguments, we can avoid lengthy mathematical calculations that will ultimately give a zero result. Reasoning on the basis of symmetry can be a powerful tool in physics, chemistry, and engineering. It is a tool worth mastering. 12/27/2018 Physics 122B - Lecture 6

Symmetries of Three Geometries Translation along plane Full rotation about ^ axis. Reflection about charged plane or about ^ plane Translation along z Full rotation about z Reflection about plane including or ^ z. Full rotation about any axis through center. Reflection about any plane including center. 12/27/2018 Physics 122B - Lecture 6

Charge Conjugation Symmetry There is an additional symmetry associated with electric charges. If we change all positive charges to negative and all negative charges to positive without changing the magnitudes of the charges, the field must reverse direction, but must not change in any other way. Charge Conjugation 12/27/2018 Physics 122B - Lecture 6

Question 2 A uniformly charged rod has positive charge Q and finite length L. The rod is symmetric under rotations about the z axis and under reflection in any plane containing the z axis. It does not have translational symmetry or symmetry on reflection about any plane ^ z, except the plane through the center of the rod. Which of the field shapes below matches the symmetries of the rod? (a) (c) (b) (d) 12/27/2018 Physics 122B - Lecture 6

The Porcupine Theory of Flux Imagine that we have some porcupines with very long quills. Careful examination shows each porcupine has exactly 30 quills. Now suppose that we hide a number of these porcupines in a loose-weave burlap bag. Assuming that the quills of one beast can completely penetrate another beast (i.e they are transparent to quills, or ghosts) ….. How many porcupines are in the bag? Answer: Count the number of quills coming out through the cloth. Then, by dividing the number of quills by 30, we can tell how many porcupines are in the bag. + + + + This is a good analogy to Gauss’s Law. The quill-count is the electric flux, the porcupines are electric charges, and the bag is a “Gaussian surface” surrounding the charges. Each charge Q has a “quill count” of Q/e0. 12/27/2018 Physics 122B - Lecture 6

The Concept of Flux The basic idea here is that the qualitative idea of field lines can be quantified by defining electric flux as a conserved quantity that is “delivered” by electric charge. Charge creates flux, and observing flux allows one to deduce the presence of charge. If an electric field is observed to be coming out of all walls of an opaque box as in (a), we can conclude that the box contains a net positive charge. If an electric field is observed to be going into all walls of an opaque box as in (b), we can conclude that the box contains a net negative charge. If equal amounts of electric field are observed to be entering and exiting the box as in (c), we can conclude that the box contains no net charge. 12/27/2018 Physics 122B - Lecture 6

Gaussian Surfaces A “Gaussian surface” is a closed virtual surface (you get to choose it as you wish) that surrounds a region of space that may contain a quantity of charge. Gaussian surfaces are most useful when the symmetry of the Gaussian surface matches the symmetry of the electric field that passes through the surface (the same symmetry as the charge distribution that produces it), and when the field lines are perpendicular to the surfaces that they cross. 12/27/2018 Physics 122B - Lecture 6

Matching the Field with the Gaussian Surface We would like to choose a closed Gaussian surface to assess the flux so that: The electric field & surface are ^. The electric field is constant or zero over each surface element. To do this, we must match the symmetries of the field and the closed surface. 12/27/2018 Physics 122B - Lecture 6

Question 1 The box contains: (a) A single positive charge; (b) A single negative charge; (c) No charge; (d) A net positive charge; (e) A net negative charge. 12/27/2018 Physics 122B - Lecture 6

The Basic Definition of Flux Imagine that you are holding a loop of wire of area A in front of a fan, and you want to know the volume per second, F, of air with velocity v flowing through the loop. When the loop is perpendicular to the air flow (a), F will be a maximum, and when the plane of the loop is parallel to the air flow (b), F=0. Then F = vA cosq, where q is the angle between v and the unit vector n normal to the plane of the loop, i.e., F = v^A. By analogy, the electric flux is: ^ 12/27/2018 Physics 122B - Lecture 6

The Area Vector We can make the expression for the flux a vector equation by defining an area vector: Here, n is the unit vector perpendicular to the plane of the area. If the surface is closed, n points away from the interior. We can use this definition of n for curved surfaces as well as flat ones by defining n as perpendicular to the local surface. With this definition, the electric flux Fe for a constant E field is: ^ ^ ^ ^ 12/27/2018 Physics 122B - Lecture 6

The Electric Flux of a Nonuniform Electric Field For a non-uniform field, the flux can be calculated by breaking up the area A of interest into small surface elements dA and integrating the E·dA flux increments over the surface. 12/27/2018 Physics 122B - Lecture 6

Flux through a Curved Surface For a non-uniform field plus a curved surface, the same approach works. Break up the area A into small surface elements dA and integrate the E·dA flux increments over the surface. When E is ^ to dA the flux contributions are zero. When E is || to dA the flux contributions are maximum. 12/27/2018 Physics 122B - Lecture 6

Flux Surface Integral For closed Gaussian surfaces (i.e. encloses a non-zero volume) we write the flux integral as: Here, the “loop” on the integral sign indicates a surface integral over a closed surface. Since a simple non-intersecting closed surface has a definite inside and outside, we take the vector direction of dA as always pointing out of the volume (and normal to the surface of course). This integral is related to the amount of charge enclosed by the surface !! 12/27/2018 Physics 122B - Lecture 6

Example: Flux through a Closed Cylinder A cylindrically symmetric charge distribution has created an electric field E(r)=E0(r/r0)2r, where E0 and r0 are constants and the unit vector r lies in the xy plane. Calculate the electric flux through a closed cylinder of length L and radius R that is centered along the z axis. ^ ^ The electric field is directed radially outward from the z axis, so it will be perpendicular to the area vectors for the “end caps” of the cylinder. It will be parallel to the normal vectors of the curved sides of the cylinder and constant in magnitude there. Therefore: 12/27/2018 Physics 122B - Lecture 6

Question 2 The total electric flux through the box is: (a) 0 Nm2/C; (b) 1 Nm2/C; (c) 2 Nm2/C; (d) 4 Nm2/C; (e) 6 Nm2/C. 12/27/2018 Physics 122B - Lecture 6

End of Lecture 6 Lecture Homework #2 has been posted on the Tycho system. It is due at 10 PM on Wednesday, April 11. On Friday, April 13, we will have Midterm 1, covering Chapters 25-27. You may bring one page of notes. Also bring a Scantron sheet and a scientific calculator with good batteries. A discussion forum for this class now exists…use it as you please…discussions related to physics only !!