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Chapter 24. Gauss’s Law. Intro. Gauss’s Law is an alternative method for determining electric fields. While it stem’s from Coulomb’s law, Gauss’s law is more convenient for highly symmetric charge distributions and allows qualitative reasoning to simply complicated scenarios. .
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Chapter 24 Gauss’s Law
Intro • Gauss’s Law is an alternative method for determining electric fields. • While it stem’s from Coulomb’s law, Gauss’s law is more convenient for highly symmetric charge distributions and allows qualitative reasoning to simply complicated scenarios.
24.1 Electric Flux • Ch 23 dealt with electric field lines qualitatively. • Consider a uniform electric field passing through a surface of area A • Remember that the number of lines passing through a unit area is proportional to the magnitude of the E-field.
24.1 • We can then draw the conclusion that the total number of electric field lines penetrating the surface is proportional to the product EA • The product of E and perpendicular A will be called Electric Flux • Electric Flux (N.m2/C) is proportional to the number of electric field lines penetrating some surface.
24.1 • Example 24.1 • If the surface is not perpendicular to the E-field, it will have less flux than ΦE = EA • As the area is rotated away from perpendicular, fewer lines will pass through it.
24.1 • The electric flux through a given area will be the dot product of the E and A • A is a vector? • vector A points in a direction normal to the surface area. (See Diagram)
24.1 • This all assumes a uniform field, although in more general situations, the field can vary over a given surface. • Our definition of flux now only applies to small elements of the surface area (ΔA) • The electric flux through each element is therefore
24.1 • The total flux through the entire surface will be the sum of flux through each element as the elements become infinitely small. • A surface integral, evaluated over the surface in question. Generally depends on the field pattern and the surface.
24.1 • Often we are interested in evaluating flux through a closed surface.
24.1 • Elements where the electric field lines enter the surface experience negative flux (90o<θ<180o) • Elements where the electric field skims the surface experience zero flux (θ = 90o) • Elements where the electric field leaves the surface experience positive flux (θ < 90o)
24.1 • The net flux on a closed surface is equal to the number of lines leaving the surface minus the number of lines entering the surface. • More lines leaving, net ΦE is positive • More lines entering, net ΦE is negative • Net Flux-
24.1 • The symbol indicates integration over a closed surface. • En is the component of the E field that is normal to the surface. • If the entire field is normal to each point of the surface, and constant in magnitude the calculation is very straightforward as in Ex. 24.1
24.1 • Quick quizzes p. 742 • Example 24.2
24.2 Gauss’s Law • Gauss’s Law gives the general relationship between the net electric flux through a closed (Gaussian) surface and the charge enclosed by the surface. • Again we will look at a positive charge q at the center of a sphere of radius r.
24.2 • We found that flux • Simplifying Since
24.2 • We can use to verify our results from example 24.1 • Consider multiple Gaussian surfaces • The flux through each of the surfaces should be the same.
24.2 • The net flux through any closed surface surrounding a point charge is given by q/εo and is independent to the shape of that surface.
24.2 • Now consider a point charge located outside the closed arbitrary surface. • Every field line that enters the surface also leaves the surface. • The net electric flux through a closed surface that surrounds no charge is zero.
24.2 • Gauss’s Law- The Net Electric Flux through any closed surface is • While qin represents just the enclosed charges E represents the TOTAL E-field through an element of area, from both inside charge and outside charge.
24.2 • Gauss’s Law can be used to solve for E, for a system of charges or a continous distribution of charge. • It is however limited to situations of high symmetry. By choosing the shape of the gaussian surface carefully, the integral in the equation becomes very simple. • Quick Quizzes p. 745 • Example 24.3
24.3 Applications of Gauss’s Law • Applied to charge distributions of high symmetry. • A gaussian surface is not a real surface, but rather one that is chosen that will simplify the integral by- • Making the value of E constant over the entire surface • The dot product has no angle because E and dA are parallel
24.3 • The dot product is zero because E and dA are perpendicular. • The field can be argued to be zero over the entire surface. • Examples 24.4-24.8
24.4 Conductors In Electrostatic Equilibrium • A conductor has charges (e-) that are unbound to atoms/molecules, therefore free to move. • Electrostatic Equilibrium- there is no net motion of the charge within the conductor • A conductor in equilibrium has the following four properties.
24.4 • 1) The electric field is zero everywhere inside the conductor. • Consider a conducting slab. • If there is a field inside the conductor, charges present would experience an electric force and therefore acceleration, which would not be considered equilibrium
24.4 • 2) If an isolated conductor carries a charge, the charge resides on its surface. • Charged Conductor • No E-Field inside • Gaussian Surface as near as possible to actual surface, qin = 0 • The charge must reside on the surface.
24.4 • The electric field just outside a charged conductor is perpendicular to the surface and has magnitude σ/εo where σis the surface charge density at that point. • If there is any component of E that is parallel to the surface of the object, free electrons would accelerate, indicating no equilibrium
24.4 • We can Determine the E field just outside the surface by choosing the cylindrical gaussian surface whose two flat sides are just inside/outside of the charged surface. • Since the E-field inside is zero from property #1
24.4 • 4) On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature is smallest. • The 4th property will be verified in Ch 25. Quick quiz 24.6 Example 24.10