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 numberof lines passing through a unitarea 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.

EAE

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.

AE cosEAE

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

iiiii AE AE cos

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.

surfaceii

AE d

i

AEAE0

lim

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- dAEd nE AE

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

EAE

224 r

r

qkeE

qkeE 4

oE

q

oek 4

1

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.

oE

q

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.

o

inE

qd

AE

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 σ is 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

o

inE

qEdA

o

AEA

o

E

24.4

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 24.6Example 24.10