PHYS 1444 Section 003 Lecture #5 Thursday Sep. 6, 2012 Dr. Andrew Brandt Chapter 22: - Electric Flux - Gauss Law - Gauss Law with many charges - What is Gauss Law good for? CH 23 Electrical Potential HW on ch 22 due weds at 11pm Thursday Sep. 6, 2012 PHYS 1444-03 Dr. Andrew Brandt 1 Electric Flux Lets imagine a surface of area A through which a uniform electric field E passes The electric flux is defined as E=EA, if the field is perpendicular to the surface E=EAcos, if the field makes an angle with r surface r the So the electric flux is defined as E E A. How would you define the electric flux in words?

Total number of field lines passing through the unit area perpendicular N E EA E to the field. Thursday Sep. 6, 2012 PHYS 1444-03 Dr. Andrew Brandt 2 Generalization of the Electric Flux Lets consider a surface of area A that has an irregular shape, and furthermore, that the field is not uniform. The surface can be divided up into infinitesimally small areas of Ai that can be considered flat. And the electric field through this area can be considered uniform since the area is r r very small. E Ei dA Then the electric flux through the entire n r r surface is E A E i open surface

i i 1 In the limit where Ai 0, the discrete r r summation becomes an integral. E Ei dA Thursday Sep. 6, 2012 PHYS 1444-03 Dr. Andrew Brandt closed surface 3 Generalization of the Electric Flux We define the direction of the area vector as pointing outward from the enclosed volume. For the line leaving the volume, /2, so cos>0. The flux is positive. For the line coming into the volume, /2, so cos<0. The flux is negative. If E>0, there is a net flux out of the volume. If E<0, there is flux into the volume. In the above figures, eachr fieldr that enters the volume also leaves the E volume, so E dA 0. The flux is non-zero only if one or more lines start or end inside the surface.

Thursday Sep. 6, 2012 PHYS 1444-03 Dr. Andrew Brandt 4 Generalization of the Electric Flux The field line starts or ends only on a charge. Sign of the net flux on the surface A1? Net outward flux (positive flux) How about A2? Net inward flux (negative flux) What is the flux in the bottom figure? There should be a net inward flux (negative flux) since the total charge inside the volume is negative. The flux that crosses an enclosed surface is proportional to the total charge inside the surface. This is the crux of Gauss law. Thursday Sep. 6, 2012 PHYS 1444-03 Dr. Andrew Brandt 5 Gauss Law The precise relation between flux and the enclosed charge is r r Qencl given by Gauss Law

E dA 0 0 is the permittivity of free space in the Coulombs law A few important points on Gauss Law Freedom to choose!! The integral is performed over the value of E on a closed surface of our choice in any given situation. Test of existence of electrical charge!! The charge Qencl is the net charge enclosed by the arbitrary closed surface of our choice. Universality of the law! It does NOT matter where or how much charge is distributed inside the surface. The charge outside the surface does not contribute to Qencl. Why? The charge outside the surface might impact field lines but not the total number of lines entering or leaving the surface Thursday Sep. 6, 2012 PHYS 1444-03 Dr. Andrew Brandt 6 Gauss Law q

q Lets consider the case in the above figure. What are the results of the closed integral of the gaussian surfaces A1 and A2? r r q E dA 0 r r q E dA 0 For A1 For A2 Thursday Sep. 6, 2012 PHYS 1444-03 Dr. Andrew Brandt 7 Coulombs Law from Gauss Law Lets consider a charge Q enclosed inside our imaginary Gaussian surface of sphere of radius r. Since we can choose any surface enclosing the charge, we choose the simplest possible one!

The surface is symmetric about the charge. What does this tell us about the field E? Must have the same magnitude at any point on the surface Points radially outward ( or inward) parallel to the surface vector dA. The Gaussian integral can be written as r r E dA EdA E dA E 4 r 2 Thursday Sep. 6, 2012 Qencl Q 0 0 PHYS 1444-03 Dr. Andrew Brandt

Solve for E Q E 4 0 r 2 Electric Field of 8 Coulombs Law Gauss Law from Coulombs Law Lets consider a single static point charge Q surrounded by an imaginary spherical surface. Coulombs law tells us that the electric field at a spherical surface is E 1 Q 4 0 r 2 Performing a closed integral over the surface, we obtain r r r 1 Q 1 Q r dA dA E dA 2 2 4 0 r 4 0 r 1 Q 1 Q

Q 2 dA 4 r 2 2 4 0 r 4 0 r 0 Thursday Sep. 6, 2012 PHYS 1444-03 Dr. Andrew Brandt Gauss Law 9

Gauss Law from Coulombs Law Irregular Surface Lets consider the same single static point charge Q surrounded by a symmetric spherical surface A1 and a randomly shaped surface A2. What is the difference in the number of field lines passing through the two surface due to the charge Q? None. What does this mean? The total number of field lines passing through the surface is the same no matter what the shape of therenclosed r surface r is.r So we can write: E dA E dA Q 0 A1 A2 What does this mean? The flux due to the given enclosed charge is the same no matter what the shape of r r Q Thursdaythe Sep.surface

6, 2012 enclosing it PHYS 1444-03 Dr. Andrew Brandt 10 E is. Gauss law, dA , is valid for any surface 0 Gauss Law w/ more than one charge Lets consider several charges inside a closed surface. For each charge, Qi, inside the chosen r closed surface, r Qi r What is Ei ? Ei dA The electric field produced by Qi alone! 0 Since electric fields can be added vectorially, following the superposition principle, ther total field r E is equal to the sum of the fields due to each charge E Ei . So r r E dA

r r Ei dA Q i 0 Qencl 0 What is Qencl? The total enclosed charge! The value of the flux depends on the charge enclosed in the surface!! Gauss law. Thursday Sep. 6, 2012 PHYS 1444-03 Dr. Andrew Brandt 11 So what is Gauss Law good for? Derivation of Gauss law from Coulombs law is only

valid for static electric charge. Electric field can also be produced by changing magnetic fields. Coulombs law cannot describe this field, but Gauss law is still valid Gauss law is more general than Coulombs law. Can be used to obtain electric field, forces, or charges Gauss Law: Any differences between the input and output flux of the electric field over any enclosed surface is due to the charge within that surface!!! Thursday Sep. 6, 2012 PHYS 1444-03 Dr. Andrew Brandt 12 21-7 Electric Field Calculations for Continuous Charge Distributions Conceptual Example 21-10: Charge at the center of a ring. Imagine a small positive charge placed at the center of a nonconducting ring carrying a uniformly distributed negative charge. Is the positive charge in equilibrium if it is displaced slightly from the center along the axis of the ring, and if so is it stable? What if the small charge is negative? Neglect gravity, as it is much smaller than the electrostatic forces. Solution: The positive charge is in stable equilibrium, as it is attracted uniformly by every part of the ring. The negative charge is also in equilibrium, but it is unstable; once it is displaced from its Thursday Sep. position, 6, 2012 1444-03 Dr. Andrew Brandt from the ring. 13

equilibrium it willPHYS accelerate away Example 22 2 Flux from Gauss Law: Consider the two Gaussian surfaces, A1 and A2, shown in the figure. The only charge present is the charge +Q at the center of surface A1. What is the net flux through each surface A1 and A2? The surface A1 encloses the charge +Q, so from Gauss law we obtain the total net flux For A2 the charge, +Q, is outside the surface, so the total net flux is 0. Thursday Sep. 6, 2012 r r Q E dA 0 r r 0 E dA 0 0 PHYS 1444-03 Dr. Andrew Brandt

14 Example 22 5 Long uniform line of charge: A very long straight wire possesses a uniform positive charge per unit length, . Calculate the electric field at points near but outside the wire, far from the ends. Which direction do you think the field due to the charge on the wire is? Radially outward from the wire, the direction of radial vector r. Due to cylindrical symmetry, the field is constant anywhere on the Gaussian surface of a cylinder that surrounds the wire. The end surfaces do not contribute to the flux at all. Why? Because the field vector E is perpendicular to the surface vector dA. From Gauss law Solving for E Thursday Sep. 6, 2012 r r Qencl l E dA E dA E 2 rl 0 0

E 2 0 r PHYS 1444-03 Dr. Andrew Brandt 15 Example 22-4: Solid sphere of charge. An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius r0. Determine the electric field (a) outside the sphere (r > r0) and (b) inside the sphere (r < r0). Solution: a. Outside the sphere, a gaussian surface encloses the total charge Q. Therefore, E = Q/(40r2). b. Within the sphere, a spherical gaussian surface encloses a fraction of the charge Qr3/r03 (the ratio of the volumes, as the charge density is constant). Integrating and solving for the field gives E = Qr/(40r03). Thursday Sep. 6, 2012 PHYS 1444-03 Dr. Andrew Brandt 16