Chapter 24 electric flux 24.1 Electric Flux 24.2 Gauss’s Law 24.3 Application of Gauss’s Law to Various Charge Distributions The Electric Field Due

Chapter 24 electric flux24.1 Electric

Flux24.2 Gauss’s Law24.3 Application of Gauss’s Law to Various Charge Distributions

The Electric Field Due to a Point Charge

A Spherically Symmetric Charge Distribution (An insulating solid sphere )

The Electric Field Due to a Thin Spherical ShellThe Electric Field Due to A Cylindrically Symmetric Charge Distributionthe electric field due to an infinite plane

24.4 Conductors in Electrostatic Equilibrium

1Norah Ali Al moneef

Norah Ali Almoneef2

Symbol Name Unit Charge per length C/m Charge per area C/m2 Charge per volume C/m3

We will use a “charge density”“charge density” to describe the distribution of charge.This charge density will be different depending on the geometry


DefinitionsSymmetry—The balanced structure of an object, the halves

of which are alike

Closed surface—A surface that divides space into an inside and outside region, so one can’t move from one region to another without crossing the surface

Gaussian surface—A hypothetical closed surface that has the same symmetry as the problem we are working on—note this is not a real surface it is just an mathematical one

Open and Closed Surfaces

A rectangle is an open surface — it does NOT contain a volume

A sphere is a closed surface — it DOES contain a volume


24-1 Electric Flux

• Like the flow of water, or light energy, we can think of the electric field as flowing through a surface (although in this case nothing is actually moving).

• We represent the flux of electric field as (greek letter phi),

The units for electric flux are Nm2/C.


The number of lines per unit area is proportional to the magnitude of the electric field.

This means that the total number of lines that penetrate a given area is proportional to the magnitude of the electric field times the area which is being penetrated. E x A

The product of the electric field (E) and surface area (A) which is perpendicular to the field is called the electric flux (ΦE).

Consider a uniform electric field, and a surface area through which the electric field is passing.

We define the angle of the given area, relative to the direction of the electric field, by a normal vector which is perpendicular to the surface.

24.1Electric Flux


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

A = A cos so E = EA = 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.

Because A is perpendicular to the surface, the amount of A parallel to the electric field is A cos .

Remember the dot product?

The electric flux passing through a surface is the number of electric field lines that pass through it.Because electric field lines are drawn arbitrarily, we quantify electric flux like this: E=EA, except that…

EA

E E A



• 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…

• so the flux of the electric field through an element of area A is

• When ˚, the flux is positive (out of the surface), and when ˚, the flux is negative.

• When we have a complicated surface, we can divide it up into tiny elemental areas: cos dAEAdEd

cos AEAE

iE i i

A 0i

lim E A

Ad . EE


Electric Flux


• We are going to be most interested in closed surfaces, in which case the outward direction becomes self-evident.

• We can ask, what is the electric flux out of such a closed surface? Just integrate over the closed surface:

• The symbol has a little circle to indicate that the integral is over a closed surface.

• The closed surface is called a gaussian gaussian surfacesurface, because such surfaces are used by Gauss’ Law, which states that:

Flux positive => out Flux negative => in

AdEdE


Spherical Gaussian surfaces around (a) positive and (b) negative point charge.


The flux is a maximum when the surface is perpendicular to the field Φ = EA

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

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

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

For a uniform electric field perpendicular to a rectangular surface, ΦE the electric flux is ΦE = EA (Nm^2/C)


Flux of Electric Field

1. Which of the following figures correctly shows a positive electric flux out of a surface element?

A. I.B. II.C. III.D. IV.E. I and III.

E

A

EA

E

A

E

A

I. II.

III. IV.


Example 1: flux through a cube of a uniform electric field

The field lines pass through two surfaces perpendicularly and are parallel to the other four surfaces

For side 1, ΦE = -El 2

For side 2, ΦE = El 2

For the other sides, ΦE = 0

Therefore, Φtotal = 014Norah Ali Al moneef

24-2Gauss’s LawGauss’s Law relates the electric flux through a closed surface with the charge Qin inside that surface.This is a useful tool for simply determining the electric field, but only for certain situations where the charge distribution is either rather simple or possesses a high degree of symmetry.

More general and elegant form of Coulomb’s law.The electric field by the distribution of charges

can be obtained using Coulomb’s law by summing (or integrating) over the charge distributions.

Gauss’ law, however, gives an additional insight into the nature of electrostatic field and a more general relationship between the charge and the field


16

Electric lines of flux and Derivation of Gauss’ Law using Coulombs law Consider a sphere drawn around a positive point charge.

Evaluate the net flux through the closed surface.

E d

A Net Flux =

E cosdA EdAFor a Point charge E=kq/r2

EdA kq /r2 dA

kq /r2 dA kq /r2(4r2)

4kq

4k 1/0 where 0 8.85x10 12 C2

Nm2

net qenc

0

Gauss’ Law


Gauss’ LawThe flux of electric field through a closed surface is proportional to the charge enclosed.

The Gaussian Surface and Gauss’s Law Closed surfaces of various shapes can surround

the charge Only S1 is spherical The flux through all other surfaces (S2 and S3)

are the same. These surfaces are all called the Gaussian

Surface. Gauss’s Law

The net flux through any closed surface surrounding a charge q is given by q/εo and is independent of the shape of that surface

The net electric flux through a closed surface that surrounds no charge is zero

Since the electric field due to many charges is the vector sum of the electric fields produced by the individual charges, the flux through any closed surface can be expressed as

• Gauss’s Law connects electric field with its source charge

0

AE

qdE

0

2121 A)EE(AE

...qq

d...dE


in

outq

Flux lines & FluxN N

and number of lines passing through each sphere is the same

In fact the number of flux lines passing through any surface surrounding this charge is the same1

2

s

even when a line passes in and out of the surface it crosses out once more than in

outin

out

012

QS

Just what we would expect because the number of field lines passing through each sphere is the same


Flux through a sphere from a point charge

210 ||4

1||

rE

Q

212

10

||4||4

1r

r

Q

0Q

r1

The electric field around a point charge

Thus the flux on a sphere is E × Area

AreaE

Cancelling we get


Now we change the radius of sphere

220 ||4

1||

rE

Q

222

202 ||4

||4

1r

r

Q

02

Q

r2

012

Q

Flux through a sphere from a point charge

210 ||4

1||

rE

Q

212

10

||4||4

1r

r

Q

0Q

r1

The electric field around a point charge

Thus the flux on a sphere is E ×Area

AreaE

Cancelling we get

The flux is the same as before


Principle of superposition:What is the flux from two charges?

s

Q1

Q2

0

2

0

1

QQ

S

0i

S

Q

In general

Gauss’s Law

For any

surface

Since the flux is related to the number of field lines passing through a surface the total flux is the total from each charge


22

The total flux through the below spherical surface is

+q+q

1. positive (net outward flux)

2. negative (net inward flux)

3. zero.

4. I don’t knowNo enclosed charge no net flux. Flux in on left cancelled by flux out on right


Norah Ali Al moneef 23

The net outward flux through the surface of a black box is 8 ×103 Nm2C–1.(a) What is the net charge inside the box?(b) If the net outward flux were zero, could you saythat there were no charges inside the box?

(a) Ф = 8 ×103 Nm2C–1

If the net charge inside is q, then,

(b) Not necessarily. We can only say that the net charge inside the box is zero.

Example

Cq

q

in

in

1071085.8108 81230

0

A surface is so constructed that, at all points on the surface, the E vector points inward. Therefore, it can be said that

A.the surface encloses a net positive charge.

B.the surface encloses a net negative charge.

C.the surface encloses no net charge.


Example:

The electric field E in Gauss’s Law is

A.only that part of the electric field due to the charges inside the surface.

B.only that part of the electric field due to the charges outside the surface.

C.the total electric field due to all the charges both inside and outside the surface.


Example:

The figure shows a surface enclosing the charges q and –q. The net flux through the surface surrounding the two charges is

A. q/0

B. 2q/0

C. –q/0

D. zero E. –2q/0


Example:

The figure shows a surface enclosing the charges 2q and –q. The net flux through the surface surrounding the two charges is

3

E.

zero D.

C.

2 B.

A.

0

0

0

0

q

q

ε

q

q


Example:

+q 2q

s

The figure shows a surface, S, with two charges q and –2q. The net flux through the surface is

0

0

0

0

3 E.

zero D.

C.

2 B.

A.

q

q

ε

q

q


Example:

These are two-dimensional cross sections through three-dimensional closed spheres and a cube. Rank order, from largest to smallest, the electric fluxes Фa to Фe through surfaces a to e.

1. Φa > Φc > Φb > Φd > Φe

2. Φb = Φe > Φa = Φc = Φd

3. Φe > Φd > Φb > Φc > Φa

4. Φb > Φa > Φc > Φe > Φd

5. Φd = Φe > Φc > Φa = Φb


Example:

Sample Problem

Four closed surfaces, S1 through S4, together with the charges -2Q, +Q, and –Q are sketched in the figure. Find the electric flux through each surface.

S1

S2

S3

S4

-

-Q

-2Q

+Q


a point charge q1 = 4.00 nC is located on the x-axis at x = 2.00 m, and a second point charge q2 = -6.00 nC is on the y-axis at y = 1.00 m. What is the total electric flux due to these two point charges through a spherical surface centered at the origin if the radius is (a) 0.500 m? (b) 1.50 m? (c) 2.50 m?

x

y

q1

q2

( ) 0Ea 0

6nC( ) Eb

0

2nC( ) Ec

2 m

1 m


Example:

At what angle will the flux be zero and what angle will it be a maximum.

1. π/4 ; π/22. π/2 ; 03. 0 ; π/2


Example:

Example

3 2cos 3.50 10 0.350 0.700 cos0 858 N m CE EA

An electric field with a magnitude of 3.50 kN/C is applied along the x axis. Calculate the electric flux through a rectangular plane 0.350 m wide and 0.700 m long assuming that

(a)the plane is parallel to the yz plane; (b)the plane is parallel to the xy plane; (c) the plane contains the y axis, and its normal

makes an angle of 40.0° with the x axis.

90.0 0E

3 23.50 10 0.350 0.700 cos40.0 657 N m CE


example

What is the net electricflux through the surface if Q1=q4=+3.1nC,q2=q5=-5.9nC,and q3=-3.1nC?

CmNqqqqenc /670 2

0

321

0

CmN

qqqqenc /670 2

0

321

0


ProblemWhat is the flux Φ of The electric field through This closed surface?

cba

AdEAdEAdEAEStep one:Step one:

Step two:Step two: EAdAEdAEAdE

a

180cos 0

EAdAEAdEc

0cos

0 90cos 0 dAEAdEb

00 EAEAStep three:Step three:


If no charge is enclosed within Gaussian surface – flux is zero!

Electric flux is proportional to the algebraic number of lines leavingthe surface, outgoing lines have positive sign, incoming - negative

Remember – electric field lines must start and must end on charges!


Example

The following charges are located inside a submarine: 5.00 μC, –9.00 μC, 27.0 μC, and –84.0 μC.

(a) Calculate the net electric flux through the hull of the submarine.

(b) Is the number of electric field lines leaving the submarine greater than, equal to, or less than the number entering it?

6 2 2in12 2 2

0

5.00 C 9.00 C 27.0 C 84.0 C6.89 10 N m C

8.85 10 C N mE

q


Consider a cube with each edge = 55cm. There is a 1.8 C chargeIn the center of the cube. Calculate the total flux exiting the cube.

CNmq

/1003.21085.8

108.1 2512

6

0

NOTE: INDEPENDENT OF THE SHAPE OF THE SURFACE!


Example:

Example

A point charge Q = 5.00 μC is located at the center of a cube of edge L = 0.100 m. In addition, six other identical point charges having q = –1.00 μC are positioned symmetrically around Q as shown in Figure P24.17. Determine the electric flux through one face of the cube.


Problem solving guide for Gauss’ law

Use the symmetry of the charge distribution to determine the pattern of the field lines.

Choose a Gaussian surface for which E is either parallel to or perpendicular to dA.

If E is parallel to dA, then the magnitude of E should be constant over this part of the surface. The integral then reduces to a sum over area elements.


24-3 Application of gauss s law to various charge distributionsGauss’s law is useful in determining electric fields when the charge distribution is characterized by a high degree of symmetry. The following examples demonstrate ways of choosing the Gaussian surface over which the surface integral given by Equation can be simplified and the electric field determined. In choosing the surface, we should always take advantage of the symmetry of the charge distribution so that we can remove E from the integral and solve for it. The goal in this type of calculation is to determine a surface that satisfies one or more of the following conditions:1. The value of the electric field can be argued by symmetry to be constant over the surface.2. The dot product in the Equation can be expressed as a simple algebraic product E dA because E and dA are parallel.3. The dot product in the Equation is zero because E and d A are perpendicular.4. The field can be argued to be zero over the surface.All four of these conditions are used in examples throughout the remainder of this chapter.

0

qin

EAdE


Gauss’ Law and Coulomb law

A spherical Gaussian surface centered on a point charge q

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

Coulomb’s Law

The field lines are directed radially outwards and are perpendicular to the surface at every point, so

24AE rEdAEEdAdAEd nE

0

inQ Ad .

EE

Example 24.4 The Electric Field Due to a Point ChargeStarting with Gauss’s law, calculate the electric field due to an isolated point charge q.


a

Since the charge distribution is spherically symmetric we select a spherical Gaussian surface of radius r > a centered on the charged sphere. Since the charged sphere has a positive charge, the field will be directed radially outward. On the Gaussian sphere E is always parallel to dA, and is constant.

Q

rE

dA

0 0

Right side: inQ Q

22 2

0 0

14 or

4 e

Q Q QE r E k

r r

0

inQ Ad .

E

) r4 ( AdAd 2EEE

An insulating solid sphere of radius a has a uniform volume charge density ρ and carries total charge Q.

(A) Find the magnitude of the E-field at a point outside the sphere

(B) Find the magnitude of the E-field at a point inside the sphere

Example 24.5 A Spherically Symmetric Charge Distribution


Example 3

a

Q

Find the electric field at a point inside the sphere.

Now we select a spherical Gaussian surface with radius r < a. Again the symmetry of the charge distribution allows us to simply evaluate the left side of Gauss’s law just as before.

r

The charge inside the Gaussian sphere is no longer Q. If we call the Gaussian sphere volume V’ then“V

3

2

0 0

44

3inQ r

E r

34

Right side: 3inQ V r

3

3 3230 00

4 1 but so

43 43 43

e

r Q Q QE r E r k r

a ar a

) r4 ( AdAd 2EEE

Volume charge density“: ρ = charge / unit volume is used to characterize the charge distribution


The electric field of a uniformly charged INSULATING sphere.

2

3

We found for ,

and for ,

e

e

Qr a E k

rk Q

r a E ra

inside

outsidesame as a point charge!

rrE ˆra4

Q=)(

3o


Example 24.6 The Electric Field Due to a Thin Spherical Shell

A thin spherical shell of radius a has a total charge Q distributed uniformly over its surface Find the electric field at points(A) outside and(B) inside the shell.


Let’s start with the Gaussian surface outside the sphere of charge, r > a

We know from symmetry arguments that the electric field will be radial outside the charged sphere If we rotate the sphere, the electric field cannot change

Spherical symmetryThus we can apply Gauss’ Law and get

… … so the electric field isso the electric field is

(Gauss) /

4Flux

0

2

q

rπEdAE

E 1

40

q

r2


Let’s let’s take the Gaussian surface inside the sphere of charge, r < a

We know that the enclosed charge is zero so

We find that the electric field is zero everywhere inside spherical shell of charge

Thus we obtain two results The electric field outside a spherical shell

of charge is the same as that of a point charge.

The electric field inside a spherical shell of charge is zero.

0E

0 Flux EAE


Let’s let’s take the Gaussian surface inside the sphere of charge, r < a

We know that the enclosed charge is zero so

We find that the electric field is zero everywhere inside spherical shell of charge

Thus we obtain two results The electric field outside a spherical shell

of charge is the same as that of a point charge.

The electric field inside a spherical shell of charge is zero.

0E

0 Flux EAE


Construct a “Gaussian Surface” that reflects the symmetry of the charge - cylindrical in this case, then evaluate

Find the E-field a distance r from a line of positive charge of infinite length and constant charge per unit length λ.


24-6 The Electric Field Due to A Cylindrically Symmetric Charge Distribution

Symmetry E field must be to line and can only depend on distance from line

q

L

E

20r

E

2kr

r̂

r

2

r2

L ) Lr 2 ( 0 AdAd

law s Gauss Q

Ad .

0

0

0

in

kE

EEE

E


Line of Charge

Example 24.8 A Plane of Charge•Find the electric field due to an infinite plane of positive charge with uniform surface charge density σ.

Assume that we have a thin, infinite non-conducting sheet of positive charge

The charge density in this case is the charge per unit area,

From symmetry, we can see that the electric field will be perpendicular to the surface of the sheet


Planar Symmetry (2)

To calculate the electric field using Gauss’ Law, we assume a Gaussian surface in the form of a right cylinder with cross sectional area A and height 2r, chosen to cut through the plane perpendicularly.

Because the electric field is perpendicular to the planeeverywhere, the electric field will be parallel to the walls of the cylinder and perpendicular to the ends of the cylinder.

Using Gauss’ Law we get

… so the electric field from an infinite non-conducting sheet with charge density

(Gauss) //

Flux

00 Aq

EAEAdAEE


ExampleA conducting spherical shell of inner radius a and outer radius b with a net charge -Q is centered on point charge +2Q. Use Gauss’s law to find the electric field everywhere, and to determine the charge distribution on the spherical shell.

a

b

-Q First find the field for 0 < r < a

This is the same as Ex. 2 and is the field due to apoint charge with charge +2Q.

2

2e

QE k

r

Now find the field for a < r < bThe field must be zero inside a conductor in equilibrium. Thus from Gauss’s law Qin is zero. There is a + 2Q from the point charge so we must have Qa = -2Q on the inner surface of the spherical shell. Since the net charge on the shell is -Q we can get the charge on the outer surface from Qnet = Qa + Qb.

Qb= Qnet - Qa = -Q - (-2Q) = + Q.

+2Q


ExampleFind the field for r > b

From the symmetry of the problem, the field in this region is radial and everywhere perpendicular to the spherical Gaussian surface. Furthermore, the field has the same value at every point on the Gaussian surface so the solution then proceeds exactly as in Ex. 2, but Qin=2Q-Q.

Gauss’s law now gives:

22 2

0 0 0 0

2 14 or

4in

e

Q Q Q Q Q QE r E k

r r

) r4 ( AdAd 2EEE


ExampleA spherical shell of radius 10 cm has a charge 2×10–6 C distributed uniformly over its surface. Find the electric field (a) Inside the shell(b) Just outside the shell(c) At a point 15 cm away from the centreq = 2 ×10–6 C, R = 0.1 m, r = 0.15 m

(a) Inside the shell, electric field is zero

–69

2 20

6 –1

1 q 2×10(b) E= =9×10 ×

4 ε R 0.1

=1.8×10 NC

–69

2 20

5 –1

1 q 2×10(c) E'= =9×10 ×

4 ε r 0.15

=8×10 NC


Example

An infinite line charge produces a field of 9×104 N C–1 at a distance of 2 cm. Calculate the linear charge density.

E = 9×104 N C–1, r = 2×10–2 m

4 –2–7 –1

9

1 9×10 ×2×10= × =10 Cm

9×10 2

0

1 λE=

2 ε r

0

Erλ=4 ε ×

2



+-

++-

-

-

+

++

- +

+-- -++++++

Conductor

Conductors.. In these materials, the charges ARE FREE TO MOVE.

24-4Conductors in Electrostatic Equilibrium

The electric field is zero everywhere inside the conductorAny net charge resides on the conductor’s surfaceThe electric field just outside a charged conductor is perpendicular to the conductor’s surface

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

By electrostatic equilibrium we mean a situation where there is no net motion of charge within the conductor


24-4Conductors in Electrostatic Equilibrium

Why is this so?

If there was a field in the conductor the charges would accelerate under the action of the field.

The electric field is zero everywhere inside the conductor

++

++

++

++

++

++

---------------------

Ein

E E

The charges in the conductor move creating an internal electric field that cancels the applied field on the inside of the conductor

Place a conducting slab in an external field, E.


When electric charges are at rest, the electric field within a conductor is zero.

The electric field is stronger where the surface is more sharply curved.

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


•Charge Resides on the Surface Choose a Gaussian surface inside but close

to the actual surface The electric field inside is zero There is no net flux through the gaussian

surface Because the gaussian surface can be as

close to the actual surface as desired, there can be no charge inside the surfaceSince no net charge can be inside the surface, any net charge must reside on the surfaceGauss’s law does not indicate the distribution of these charges, only that it must be on the surface of the conductor



Charges on Conductors

Field within conductor E=0


Charge on solid conductor resides on surface.

Charge in cavity makes a equal but opposite charge reside on inner surface of conductor.

E-Field’s Magnitude and Direction

Choose a cylinder as the Gaussian surface

The field must be perpendicular to the surface If there were a parallel component to

E, charges would experience a force and accelerate along the surface and it would not be in equilibrium

The net flux through the surface is through only the flat face outside the conductor The field here is perpendicular to the

surface Applying Gauss’s law

0

E

0A

EAE


Electric field = zero (electrostatic)

inside a solid conducting sphere

Under electrostatic conditions the electric

field inside a solid conducting sphere is

zero. Outside the sphere the electric field drops off

as 1 / r2, as though all the excess charge on the

sphere were concentrated at its

center.


Example Example • A spherical conducting shell has an excess

charge of +10 C. • A point charge of 15 C is located at center

of the sphere.• Use Gauss’ Law to calculate the charge on

inner and outer surface of sphere

-15 CR2

R1

• Since E = 0 inside the metal, flux through this surface = 0

• Gauss’ Law says total charge enclosed = 0 • Charge on inner surface = +15 C

a) Inner: +15 C; outer: 0(b) Inner: 0; outer: +10 C(c) Inner: +15 C; outer: -5 C

Since TOTAL charge on shell is +10 C,Charge on outer surface = +10 C 15 C = 5 C!

-15C

-5 C

+15C



A line charge (C/m) is placed along the axis of an uncharged conducting cylinder of inner radius ri = a, and outer radius ro = b as shown. What is the value of the charge

density o (C/m2) on the outer surface of the cylinder?

(a) (b) (c)

a

b

Example

The uniform surface charge density on a spherical copper shell is What is the electric field strength on the surface of the shell?

o

20

1 qE=

4 ε R

The electric field on the surface of a uniformly charged spherical conductor is given by,


A spherical charged conductor has a uniform surface charge density s . The electric field on its surface is E. If the radius of the sphere is doubled keeping the surface density of charge unchanged, what will be the electric field on the surface of the new sphere ?

Example

20

1 qE=

4 ε R

2

2o0

4 R

4 R

The electric field on the surface of a uniformly charged spherical conductor is given by,


Example

Consider a thin spherical shell of radius 14.0 cm with a total charge of 32.0 μC distributed uniformly on its surface. Find the electric field

(a) 10.0 cm and (b) 20.0 cm from the center of the charge distribution.

9 6

2 2

8.99 10 32.0 107.19 MN C

0.200ek Q

Er


ExampleA uniformly charged, straight filament 7.00 m in length has a total positive charge of 2.00 μC. An uncharged cardboard cylinder 2.00 cm in length and 10.0 cm in radius surrounds the filament at its center, with the filament as the axis of the cylinder. Using reasonable approximations, find (a) the electric field at the surface of the cylinder and (b) the total electric flux through the cylinder.

9 2 2 62 8.99 10 N m C 2.00 10 C 7.00 m20.100 m

ekE

r

51.4 kN C, radially outwardE

cos 2 cos0E EA E r

4 25.14 10 N C 2 0.100 m 0.0200 m 1.00 646 N m CE


ExampleA square plate of copper with 50.0-cm sides has no net charge and is

placed in a region of uniform electric field of 80.0 kN/C directed perpendicularly to the plate. Find

(a) the charge density of each face of the plate and (b) the total charge on each face.

4 12 7 28.00 10 8.85 10 7.08 10 C m

277.08 10 0.500 CQ A

71.77 10 C 177 nCQ


A B

A B

0

(σ +σ )=–

2ε

A B

0

(σ – σ )=

2ε

Two infinite parallel sheets of charge

1 2E = – E + EKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKK

1 2E = E – EKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKK

Due to uniform surface charge density, surface of the sheetEKKKKKKKKKKKKKK

Region II

Region I


A B

Two infinite parallel sheets of charge

Region III

A B

0

(σ +σ )=

2ε

1 2E = E + EKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKK



Two Parallel Nonconducting Sheets The situation is different if you bring two nonconducting sheets of

charge close to each other. In this case, the charges cannot move, so there is no shielding,

but now we can use the principle of superposition. In this case, the electric field on the left due to the positively

charged sheet is canceled by the electric field on the left of the negatively charged sheet, so the field there is zero.

Likewise, the electric field on the right due to the negatively charged sheet is canceled by the electric field on the right of the positively charged sheet. The result is much the same

as before, with the electric field in between being twice what it was previously.

The sheets have equal and opposite charge density

–

I II III

E 0 E 00

E

If

and

then

A

B –

In regions I and III, E = 0

In region II,

A Bas ( – ) 0

0 0

– (– )E

2

Special Case


ExampleTwo large thin metal plates with surface charge densities of

opposite signs but equal magnitude of 44.27 ×10–20 Cm–2

are placed parallel and close to each other. What is the field

(i) To the left of the plates?(ii) To the right of the plates?(iii) Between the plates?

Electric field exists only in the region between the plates. Therefore,

(i), (ii) E = 0

–20 –2

–208 –1

–120

(iii) σ =44.27×10 Cm ,

σ 44.27×10E= = =5×10 NC

ε 8.854×10


A uniform electric field of magnitude 1 N/C is pointing in the positive y direction. If the cube has sides of 1 meter, what is the flux through sides A, B, C?

A) 1, 0, 1B) 0, 0, 1C) 1, 0, 0D) 0, 0, 0E) 1, 1, 1

A

B

C


Example:

Consider three Gaussian surfaces. Surface 1 encloses a charge of +q, surface 2 encloses a charge of –q and surface 3 encloses both charges. Rank the 3 surfaces according to the flux, greatest first.A) 1, 2, 3B) 1, 3, 2C) 2, 1, 3D) 2, 3, 1E) 3, 2, 1

+q -q

1 2

3


Example:

Rank the following Gaussian surfaces by the amount of flux that passes through them, greatest first (q is at the center of each).

A) 1, 2, 3B) 1, 3, 2C) 2, 1, 3D) 3, 2, 1E) All tie

q q

1

2

q

3


Example:

Rank the following Gaussian surfaces by the strength of the field at the surface at the point direction below q (where the numbers are).

A) 1, 2, 3B) 1, 3, 2C) 2, 1, 3D) 3, 2, 1E) All tie

q q

1

2

q

3


Problem

What is the electric flux through the right the face, the left face,and the top face?

Right face:idAAd ˆ

jiE ˆ0.4ˆ0.3


CmN

dAdAxdA

jiAdEr

/36

0.90.30.30.3

ˆ0.4ˆ 0.3

2

CmNl /12 2Left face:

CmN

jdAjit

/16

ˆˆ0.4ˆ0.3

2

Top face:

A uniform electric field of magnitude 6000 N/C points upward in a charge-free region. What is the electric flux through the shaded side of the square pyramid shown if its base is 80 m on a side and it is 50 m high? .

All flux lines entering the base must leave the top, so the flux through the four sides is the fluxthrough the base:ФE = EA= (6000 N/C)(80 m)2= 38.4 MNm2/C.By symmetry, the flux through eachside is the same: ФE= 9.6 MNm2/C


Example:


Consider a closed triangular box resting within a horizontal electric field of magnitude E = 7.80 x104 N/C as shown in Figure. Calculate the electric flux through(a)the vertical rectangular surface, (b)(b) the slanted surface, (c) (c) the entire surface of the box

Example:



A uniform electric field aiˆ + bjˆ intersects a surface of areaA. What is the flux through this area if the surface lies(a) in the yz plane? (b) in the xz plane? (c) in the xy plane?

Example:


Consider a thin spherical shell of radius 14.0 cm with atotal charge of 32.0 μC distributed uniformly on itssurface. Find the electric field (a) 10.0 cm (b) 20.0 cm from the center of the charge distribution

Example:

Example - Charge in a Cube

Q=3.76 nC is at the center of a cube. What is the electric flux through one of the sides?

Gauss’ Law:

0/Q

Since a cube has 6 identical sides and the point charge is at the center

Q

C

Nm8.70 values)numerical thee(substitut

6

1

6

2

0face one

Q


Shown is an arrangement of five charged pieces of plastic (q1=q4=3nC, q2=q5=-5.9nC and q3=-3.1nC). A Gaussian surface S is indicated. What is the net electric flux through the surface?

A: =-6 x 10-9C/0= -678 Nm2/C

B: = x10-9C/0= -1356 Nm2/C

C: =0D: = x 10-9C/0= 328 Nm2/C

enclosed charge


Example:

ExampleTwo large thin metal plates with surface charge densities of

opposite signs but equal magnitude of 44.27 ×10–20 Cm–2

are placed parallel and close to each other. What is the field

(i) To the left of the plates?(ii) To the right of the plates?(iii) Between the plates?

Electric field exists only in the region between the plates. (i), (ii) E = 0

–20 –2

–208 –1

–120

(iii) σ =44.27×10 Cm ,

σ 44.27×10E= = =5×10 NC

ε 8.854×10


ExampleA spherical shell of radius 10 cm has a charge 2×10–6 C distributed uniformly over its surface. Find the electric field (a) Inside the shell(b) Just outside the shell(c) At a point 15 cm away from the centre

q = 2 ×10–6 C, R = 0.1 m, r = 0.15 m

(a) Inside the shell, electric field is zero

–69

2 20

6 –1

1 q 2×10(b) E= =9×10 ×

4 ε R 0.1

=1.8×10 NC

–69

2 20

5 –1

1 q 2×10(c) E'= =9×10 ×

4 ε r 0.15

=8×10 NC


Ex) What is the electric flux through this disk of radius r = 0.10 m if the uniform electric field has a magnitude E = 2.0x103 N/C?

3 2 0

3 2 2

cos (2.0 10 ) ( ) cos30

(2.0 10 ) (3.14 10 ) 0.866 54 /

E EA r

N m C


Example

A sphere of radius 8.0 cm carries a uniform volume charge density = 500 nC/m3. What is the electric field magnitude at r = 8.1 cm?

A.0.12 kN/CB.1.5 kN/CC.0.74 kN/CD.2.3 kN/CE.12 kN/C


Example

A spherical shell of radius 9.0 cm carries a uniform surface charge density = 9.0 nC/m2. The electric field magnitude at r = 4.0 cm is approximately

A.0.13 kN/C

B.1.0 kN/C C.0.32 kN/C

D.0.75 kN/C

E.zero 94Norah Ali Al moneef

Example

A spherical shell of radius 9.0 cm carries a uniform surface charge density = 9.0 nC/m2. The electric field magnitude at r = 9.1 cm is approximately

A.zeroB.1.0 kN/C C.0.65 kN/CD.0.32 kN/CE.0.13 kN/C


Example

An infinite plane of surface charge density = +8.00 nC/m2 lies in the yz plane at the origin, and a second infinite plane of surface charge density = –8.00 nC/m2 lies in a plane parallel to the yz plane at x = 4.00 m. The electric field magnitude at x = 3.50 m is

approximately A.226 N/C B.339 N/C C.904 N/C D.452 N/C E.zero


Example

An infinite plane of surface charge density = +8.00 nC/m2 lies in the yz plane at the origin, and a second infinite plane of surface charge density = –8.00 nC/m2 lies in a plane parallel to the yz plane at x =4.00 m. The electric field magnitude at x = 5.00 m is

approximately A.226 N/C B.339 N/C C.904 N/C D.452 N/C E.zero


Example

Qra

rb1

rb2

A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The electric field for ra r rb1 is

ero E.

ˆ2

D.

ˆ2

C.

ˆ B.

ˆ A.

2

2

z

rr

kQ

rr

kQ

rr

kQ

rr

kQ


Example

A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The electric field for rb1 r rb2 is

Qra

rb1

rb2ero E.

ˆ2

D.

ˆ2

C.

ˆ B.

ˆ A.

2

2

z

rr

kQ

rr

kQ

rr

kQ

rr

kQ


Example

A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The electric field for r rb1 is

Qra

rb1

rb2

ero E.

ˆ2

D.

ˆ2

C.

ˆ B.

ˆ A.

2

2

z

rr

kQ

rr

kQ

rr

kQ

rr

kQ


Example

Example

Consider a thin spherical shell of radius 14.0 cm with a total charge of 32.0 μC distributed uniformly on its surface. Find the electric field

(a)10.0 cm and (b)20.0 cm from the center of the charge

distribution.

9 6

2 2

8.99 10 32.0 107.19 MN C

0.200ek Q

Er


Gauss’ Law

· Gauss’ Law depends on the enclosed charge only

1. If there is a positive net flux there is a net positive charge enclosed

2. If there is a negative net flux there is a net negative charge enclosed

3. If there is a zero net flux there is no net charge enclosed

Gauss’ Law works in cases of symmetry

o

encqAdE

Summary:

GAUSS LAW – SPECIAL SYMMETRIES

SPHERICAL(point or sphere)

CYLINDRICAL(line or cylinder)

PLANAR(plane or sheet)

CHARGEDENSITY

Depends only on radial distance

from central point

Depends only onperpendicular distance from

line

Depends only on perpendicular distance from

plane

GAUSSIAN

SURFACE

Sphere centered at

point of symmetry

Cylinder centered at axis

of symmetry

Pillbox or cylinderwith axis

perpendicular to plane

ELECTRICFIELD E

E constant at surface

E ║A - cos = 1

E constant at curved surface

and E ║ AE ┴ A at end

surfacecos = 0

E constant at end surfaces and E ║ A

E ┴ A at curved

surfacecos = 0



Geometry

Charge Density

Gaussian surface

Electric field

Linear = q/L Cylindrical, with axis along line of charge

Sheet or Plane

= q/A Cylindrical, with axis along E.

Spherical

= q/V Spherical, with center on center of sphere

rE

02

0

E02

E

204 r

qE

r

R

qE

304

Line of Charge

Conducting

Nonconducting

r R

r < R


Documents

Chapter 24 electric flux 24.1 Electric Flux 24.2 Gauss’s Law 24.3 Application of Gauss’s Law to Various Charge Distributions The Electric Field Due