Chapter 19 part 2: NCC SummerProf. Rizopoulos

Gauss’ Law

Gauss’ Law

Gauss’ Law can be used as an alternative procedure for calculating electric fields.

Gauss’ Law is based on the inverse-square behavior of the electric force between point charges.

It is convenient for calculating the electric field of highly symmetric charge distributions.

Gauss’ Law is important in understanding and verifying the properties of conductors in electrostatic equilibrium.

Electric Flux

Electric flux is the product of the magnitude of the electric field and the surface area, A, perpendicular to the field.

ΦE = EA

Units: N · m2 / C

Electric Flux

We have used electric field lines to visualize electric fields and indicate their strength.

We are now going to count the number of electric field lines passing through a surface, and use this count to determine the electric field. E

The electric flux passing through a surface is the number of electric field lines that pass through it.

we quantify electric flux as: E=EA,

If the surface is tilted, fewer lines cut the surface.

EA

E

The green lines miss!

E

A

The “amount of surface” perpendicular to the electric field is A cos .

AEffective = A cos E = EAEffective = EA cos .

We define A to be a vector having a magnitude equal to the area of the surface, in a direction normal to the surface.

Therefore, the amount of surface area effectively “cut through” by the electric field is A cos .

Electric Flux, General Area

The electric flux is proportional to the number of electric field lines penetrating some surface.

The field lines may make some angle θ with the perpendicular to the surface.

ΦE = EA cos θ

Electric Flux, Interpreting the Equation

The flux is a maximum when the surface is perpendicular to the field.

θ = 0°

The flux is zero when the surface is parallel to the field.

θ = 90°

If the field varies over the surface, Φ = EA cos θ is valid for only a small element of the area.

If the electric field is not uniform, or the surface is not flat…

divide the surface into infinitesimal surface elements and add the flux through each…

dAE

iE i iA 0 i

lim E A

E E dA

A

Remember, the direction of dA is normal to the surface.

a surface integral, therefore a double integral

Electric Flux

The surface integral means the integral must be evaluated over the surface in question

In general, the value of the flux will depend both on the field pattern and on the surface

Electric Flux, General

In the more general case, look at a small area element.

In general, this becomescosE i i i i iE A θ E A

0

surface

limi

E i iA

E

E A

d

E A

E E dA

If the surface is closed (completely encloses a volume)…

E

…we count lines going out as positive and lines going in as negative…

E E dA

dA

a surface integral, therefore a double integral

For a closed surface, dA is normal to the surface

and always points away from the inside.

Electric Flux, Closed Surface

Assume a closed surfaceThe vectors point in different directions. At each point, they are

perpendicular to the surface. By convention, they point

outward.

iA

At (1), the field lines are crossing the surface from theinside to the outside; θ < 90o, Φ is positive.At (2), the field lines graze surface; θ = 90o, Φ = 0At (3), the field lines are crossing the surface from the outside to the inside;180o > θ > 90o, Φ is negative.

19.8 Electric Flux

The net flux through the surface is proportional to the net number of lines leaving the surface

This net number of lines is the number of lines leaving the volume surrounding the surface minus the number entering the volume

The net flux through a closed surface is:

Consider a uniform electric field oriented in the x direction in empty space. A cube of edge length , is placed in the field. Find the net electric flux through the surface of the cube.

Example 19.9 Flux Through a Cube

Write the integrals for the net flux through faces 1 and 2:

Example 19.9 Flux Through a Cube

Find the flux through this face 1:

Find the flux through this face 2:

Find the net flux by adding the flux over all six faces:

Example 19.9 Flux Through a Cube

Gauss’s Law, Introduction

Gauss’s law is an expression of the general relationship between the net electric flux through a closed surface and the charge enclosed by the surface.

The closed surface is often called a gaussian surface.

Gauss’s law is of fundamental importance in the study of electric fields.

Gauss’ Law – General A positive point charge, q, is located at the center of a sphere of radius r

At all points on the surface:

The net flux is:

19.9 Gauss’s Law

19.9 Gauss’s Law

• For a spherical surface:

• or,

• The net flux through any closed surface surrounding a pint charge q is q/0

• It is independent of the position of the charge within the surface

• 0 = 8.85 x 10 -12 C2/Nm2

19.9 Gauss’s Law

In general, the net flux through any closed surface is:

Gauss’s law states:

qin is the net charge inside the surface,

and the electric field is at any point on the surface

The surface used is called a Gaussian surface

19.9 Gauss’s Law

Gauss’s Law is valid for the electric field of any system of charges or continuous distribution of charge

Although Gauss’s Law can, in theory, be solved to find E for any charge configuration, in practice it is limited to symmetric situations

Particularly spherical, cylindrical, or plane symmetry

Remember, the Gaussian surface is a surface you choose, it does not have to coincide with a real surface

Question: There are a few equations for electric flux. Which one do I need to use?

E E dA

E E dA

E E A

E EA cos

E EA

Answer: use the simplest (easiest!) one that works.

Flat surface, E A, E constant over surface. Easy!

Flat surface, E not A, E constant over surface.

Flat surface, E not A, E constant over surface.

Surface not flat, E not uniform. Avoid, if possible.

Closed surface.

If the surface is closed, you may be able to “break it up” into simple segments and still use E=E·A for each segment.

This is the definition of electric flux, so it is on your equation sheet.

The circle on the integral just reminds you to integrate over a closed surface.

Electric Flux Example: Calculate the electric flux through a cylinder with its axis parallel to the electric field direction.

E

Electric Flux Example: Calculate the electric flux through a cylinder with its axis parallel to the electric field direction.

E

I see three parts to the cylinder:

The left end cap.E

dA

Electric Flux Example: Calculate the electric flux through a cylinder with its axis parallel to the electric field direction.

E

I see three parts to the cylinder:

The tube.E

Electric Flux Example: Calculate the electric flux through a cylinder with its axis parallel to the electric field direction.

E

I see three parts to the cylinder:

The right end cap.E

dA

Let’s separately calculate the contribution of each part to the flux, then add to get the total flux.

E left tube rightleft tube right

E dA E dA E dA E dA

The right end cap.E

dA

The tube.E

E

dA

The left end cap.

left left leftleft left left

E dA E dA cos180 E dA

E

dA

The left end cap.

Every dA on the left end cap is antiparallel to E. The angle between the two vectors is 180

left left leftleft left

E dA E dA EA E is uniform, so

The tube.E

Let’s look down the axis of the tube.

E is pointing at you.

Every dA is radial (perpendicular to the tube surface).

dA

The angle between E and dA is 90.

dA

E

E

The angle between E and dA is 90.

dA

E

tube tube tubetube tube tube

E dA E dA cos90 0 dA 0

The tube contributes nothing to the flux!

right right rightright right right

E dA E dA cos0 E dA

Every dA on the right end cap is parallel to E. The angle between the two vectors is 0

E is uniform, so right right rightright right

E dA E dA EA

The right end cap.E

dA

The net (total) flux

E left tube rightleft tube right

E dA E dA E dA

E left rightEA 0 EA 0

The flux is zero! Every electric field line that goes in also goes out.

E

The net electric flux through a closed cylindrical surface is zero.

If there were a + charge inside the cylinder, there would be more lines going out than in.

If there were a - charge inside the cylinder, there would be more lines going in than out…

…which leads us to…

E+-

Gauss’ Law

enclosedE

o

qE dA

Gauss’ Law

We will find that Gauss’ law gives a simple way to calculate electric fields for charge distributions that exhibit a high degree of symmetry

OPEN ENDED LAB THU WEEK 2 day 4

Investigation OHMs LawSeries Parallel Circuits.

Derive an expression for the magnitude of E due to point charge q at a distance r from the charge.

The electric field everywhere points away from the charge.

+q

E

If you go any distance r away from +q, the electric field is always directed “out” and has the same magnitude as the electric field at any other r.What is the symmetry of the electric field?

“spherical”

To apply Gauss’ Law, we really want to pick a surface for which we can easily evaluate

+q

Eenclosed

o

qE dA

E dA.

That means we want to everywhere be either parallel or perpendicular to the surface.

E

So let’s draw a Gaussian sphere of radius r, enclosing and centered on +q. “Centered on” makes it easy to evaluate

+q

Eenclosed

o

qE dA

E dA. r

Everywhere on the sphere, are parallel and E is constant so

E dA

and

2sphereE dA E dA E dA E A E 4 r

dA

The charge enclosed by my Gaussian sphere is q, so

+q

Eenclosed

o

qE dA

r

2

o

qE dA E 4 r

2

o

qE 4 r

2o

qE =4 r

The direction of is shown in the diagram.Or you can say is “radially out.”

E

dA

E

+q

E

r

2o

qE = , away from +q4 r

Field at a Distance from a Line of Charge (example 19.11)

Calculate E at a distance r from an infinite line charge with a charge density

Field at a Distance from a Line of Charge

Select a cylindrical charge distribution . The cylinder has a radius of r

and a length of ℓ.is constant in magnitude and

perpendicular to the surface at every point on the curved part of the surface.Use Gauss’s law to find the field.

E

in

2

22

Eo

o

eo

qd EdAε

λE πrε

λ λE kπε r r

E A

Field Due to a Line of Charge, cont.

The end view confirms the field is perpendicular to the curved surface.

The field through the ends of the cylinder is 0 since the field is parallel to these surfaces.

WEEK #3 Day 1

Calculate the electric field inside and outside an insulating solid sphere of uniform charge density 2 x 10 -7 C/m3 with a radius of 10 m.

Field Due to a Spherically Symmetric Charge Distribution

Select a sphere as the gaussian surface.

For r >a

in

2 24

Eo

eo

qd EdAε

Q QE kπε r r

E A

Spherically Symmetric, cont.Select a sphere as the gaussian surface, r < a.

qin < Q

qin = V = (4/3πr3)

Spherically Symmetric Distribution, final

Inside the sphere, E varies linearly with r

E → 0 as r → 0

The field outside the sphere is equivalent to that of a point charge located at the center of the sphere.

Properties of a Conductor in Electrostatic Equilibrium

When there is no net motion of charge within a conductor, the conductor is said to be in electrostatic equilibrium.The electric field is zero everywhere inside the conductor. Whether the conductor is solid or hollow

If the conductor is isolated and carries a charge, the charge resides on its surface.The electric field at a point just outside a charged conductor is perpendicular to the surface and has a magnitude of σ/εo.

is the surface charge density at that point.On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature is the smallest.

Charge tends to accumulate at points of high curvature on the surface (sharp Points)

HW#5:

31 page 652

37, 40 pg 653