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Applications of Gauss Law

In cases of strong symmetry, Gauss's law may be

readily used to calculate E. Otherwise it is not

generally useful and integration over the charge

distribution is required. But when the symmetry

permits it, Gauss's law is the easiest way to go!

The KEY TO ITS APPLICATION is the choice of

Gaussian surface. Keep in mind that this is not a

surface of the charge distribution itself, but

rather an imaginary surface constructed for

application of Gauss's law.

- To make Gauss's law useful
- All sections of the gaussian surface should be

chosen so that they are either parallel or

perpendicular to E. For this we need to have

already determined the direction of E everywhere

on the surface by symmetry arguments. - E should be constant on each surface. Again,

this requires sufficient symmetry to "see" the

constancy of field strength. - And once again, make sure that the surface is

closed!

Gauss Law and Coulomb law

The angle ? between E and dA is zero at any

point on the surface, we can re-write Gauss Law

as

E has the has same value at all points on the

surface

E is can be moved out

Integral is the sum of surface area

A spherical Gaussian surface centered on a point

charge q

Coulombs Law

Cylindrical Symmetry

Evaluate the electric field that arises from an

infinite line of charge 1) Understand the

geometry Pick a point P for evaluation of E.

2) Understand the symmetry

The only variable on which E may depend is R, the

distance from the line of charge. There is no

angular dependence because the line is

cylindrically symmetric, i.e. it does not matter

at which angle about the line you view it. There

is no axial dependence because the line of charge

is infinitely long, i.e. there is no preferred

position along the line. Similar arguments

determine that the direction of E is directly

away from the line, perpendicular to it at all

locations. There may be no azimuthal or axial

components because the line is infinite and has

no extent in the transverse directions.

Cylindrical Symmetry

3) Construct the Gaussian surface

A cylindrical tube is chosen to match the

symmetry of the line of charge, centered about

it, with a radius equal to the distance between

the point of evaluation and the line of charge.

The electric field anywhere on this surface has

the same direction as the infinitesimal area

vector and has a constant value everywhere on the

surface. This surface is denoted by 2. Since a

gaussian surface must be closed, the tube is then

capped with flat endcaps, 1 and 3. The electric

field does not have the same value at all points

on the endcaps, but the field vector is

perpendicular to the area vector so there is no

flux through the endcaps.

Cylindrical Symmetry

4) Examine the Gaussian surface

The flux through the endcap 1 is zero since A1 is

perpendicular to E everywhere on this surface.

Similarly, the flux through surface 3 is zero.

Since the electric field is everywhere constant

on surface 2 and points in the same direction as

the surface vector at any point, the flux through

surface 2 will be the field magnitude E times the

area of the tube wall.

Cylindrical Symmetry

5) Evaluate the electric flux through the

Gaussian surface

6) Evaluate the charge enclosed by the Gaussian

surface

Cylindrical Symmetry

7) Apply Gauss's law for the result!

Source http//www.physics.udel.edu/watson/phys20

8/line-gauss/line-gauss1.html

For comparisonLine of Charge by Direct

Integration 1

Evaluate the electric field that arises from an

infinite line of charge.

Pick a point P for evaluation of E. The only

variable which matters is R, the distance from

the line of charge. There is no angular

dependence because the line is cylindrically

symmetric, i.e. it does not matter at which angle

about the line you view it. There is no axial

dependence because the line of charge is

infinitely long, i.e. there is no preferred

position along the line. The infinite line of

charge is certainly a physically-impossible

situation to set up (infinite amount of charge)

but one whose consideration is useful in

interpreting finite distributions, e.g. checking

the limiting behavior of solutions.

For comparisonLine of Charge by Direct

Integration 2

Span the charge distribution

Assuming a uniform charge distribution along a

line, the variable which appears to be the

easiest to use to span the distribution is the

variable x, which follows along the line from

negative infinity to positive infinity. Thus the

charge infinitesminal dq will be the product of

charge density and dx.

For comparisonLine of Charge by Direct

Integration 3

Evaluate the contribution from the infinitesimal

charge element

Considering a positive test charge at point P,

the direction of the electric field is shown. The

strength of the electric field contribution from

the infinitesimal charge shown is proportional to

dq and inversely proportional to the square of

the distance separating point P and the charge

element.

For comparisonLine of Charge by Direct

Integration 4

Exploit symmetry as appropriate

By examining the conjugate infinitesimal on the

other side of the origin, we see that the

horizontal components of the two contributions

will balance. We need only consider the vertical

contribution to the electric field by each

infinitesimal charge.

For comparisonLine of Charge by Direct

Integration 5

Set up the integral

In setting up the integral we wish to express all

factors entering the electric field contribution

as functions of the integration variable x. After

finding the magnitude of the infinitesimal

contribution to the field, we consider its

projection along the vertical axis. After the

integrand is determined, we evaluate the integral

between the limits of integration.

Substituting these terms, dEy becomes

Setting up the integrating over the charge

distributing

For comparisonLine of Charge by Direct

Integration 6

Solve the integral by trig substitution

Looking at the integrand, you may now realize

that trig substitution is required. All that

means is that we should have integrated over the

angle variable theta rather than the linear

variable x. The figure below shows the

relationship between the two variables for the

quantities involved

For comparisonLine of Charge by Direct

Integration 7

Expressing the integrand in terms of theta and

switching the limits of integration, the integral

may be solved and the result finally obtained!

A Charged Isolated Conductor

Gauss Law permits us to prove an important

theorem about conductors.

If excess charged is placed on an isolated

conductor, the charge will move to the surface of

the conductor. None of the excess charge will be

found within the body pf he conductor.

The electric field inside a conductor in static

equilibrium must be zero. Otherwise, the field

would exert forces on the conduction electrons

producing motion or currents which is not a

static equilibrium state. If we place a Gaussian

surface just inside the surface of the charged

conductor, the E field is zero for all points on

the Gaussian surface. Therefore the flux through

the Gaussian is zero and according to Gauss Law

the net charge enclosed the Gaussian surface must

also be zero. Therefore the excess charged must

be outside the Gaussian surface and must lie on

the actual surface of the conductor.

External Electric Field of a Conductor

Consider a section of the surface that is small

enough to to neglect any curvature and assume the

section is flat.

Embed a tiny cylindrical Gaussian surface with

one end cap full inside the conductor and the

other fully outside and cylinder is perpendicular

to the surface.

The electric field just outside the surface must

also be perpendicular to that surface otherwise

surface charges would be subject to motion.

External Electric Field of a Conductor

We now sum the flux through the Gaussian

surface. There is no flux through the internal

endcap. Why ?? There is no flux through the

curved surface. Why ?? The only flux through the

Gaussian surface is that through the external

endcap where E is perpendicular. Assuming a cap

area of A, the flux through the cap is EA. The

charge qenc enclosed by the Gaussian surface lies

on the conductors surface area A. If ? is the

charge per unit area , the qenc is equal to ? A.

Therefore Gauss Law becomes

?oEA ? A

E ? / ?o (conducting

surface)

Planar Symmetry

Two different views of a very large thin plastic

sheet, uniformly charged on one side with surface

density of ?. A closed cylindrical Gaussian

surface passes throught the sheet and is

perpendicular. From symmetry , E must be

perpedicular to the surface and endcaps. The

surface charge is positive so E emanates from

the surface. Since the field lines do not piece

the curved surface there is no flux.

What happens is we magically make the sheet a

conductor ?? Will the E field change ?

Planar Symmetry

Figure (a) show a cross-section of a thin,

infinite conducting plate with excess positive

charge. We know the this excess charge lies on

the surface of the plate. If there is no external

electric field to force the charge into some

particular distribution, it will spread out onto

the two surfaces with an uniform charge density

of ?1. From the previous slide we know this

charge sets up an electric field

which points

1

away from the plate. Figure (b) is a identical

plate with negative charge. In this case E points

inward.

Planar Symmetry

Suppose we move the two plates in the figures (a)

and (b) close together and parallel. Since the

plates are conductors, the excess charge on one

plate attracts the excess charge on the other and

all the excess charge moves into the inner faces

of the plates. The new surface charge density has

now doubled,

therefore the electric field between the plates is

The field points away from positive charged plate

and toward the negative plate. Since no excess is

left on the outer faces, the electric field

outside is zero.

Spherical Symmetry

Lets apply Gausss Law to a hollow charged sphere

What is electric field inside the hollow charged

sphere ?

x

Consider the point mark X in the figure located

on an interior equator By symmetry Ey0, what

about Ex ?? (Since X can be viewed at any angle,

the E-field must be in the radial direction.

Spherical Symmetry

x

x

Spherical Symmetry

Instead of deriving the solution through a long

integration Lets use Gauss Law. 1st choose an

Gaussian surface inside the shell that coincides

with the point x.

Since fields lines can not cross, the spherical

symmetry requires E field to be in the radially

direction and uniform on the Gaussian surface,

therefore Gauss Law

x

r

Spherical Symmetry

Outside the hollow sphere

Now lets consider the electric field a a point

outside the charged sphere. We define a Gaussian

surface around the sphere for which rgtR. Again

since fields lines can not cross, the spherical

symmetry requires E field to be in the radially

direction and uniform on the Gaussian surface,

therefore Gauss Law becomes

x

Same as point charge

Spherical Symmetry

Using Gauss Law we were able to prove two shell

theorems that were presented in the 1st lecture

without proof. A shell of uniform charge

attracts and repels a charged particle that is

outside the shell as if all shells charge were

concentrated at the center. If a charged particle

is located inside a shell of uniform charge,

there is no net electric force on the particle

from the shell.

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