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1 CHAPTER 2 GAUSS’S LAW CHAPTER 2 GAUSS’S LAW 2.1 THE FLUX OF A VECTOR FIELD 2.2 GAUSS’S LAW 2.3 APPLICATIONS OF GAUSS’S LAW 2.3.1 The Charge Distribution with Spherical Symmetry 2.3.2 The Charge Distribution With A Cylindrical Symmetry 2.3.3 The Charge Distribution With A Planar Symmetry CHAPTER SUMMARY EXERCISES AND PROBLEMS

Gauss' Law

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This is a lecture note about Gauss' law and how to apply with electromagnetic problems.

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1

CHAPTER 2

GAUSS’S LAW

CHAPTER 2 GAUSS’S LAW

2.1 THE FLUX OF A VECTOR FIELD 2.2 GAUSS’S LAW 2.3 APPLICATIONS OF GAUSS’S LAW

2.3.1 The Charge Distribution with Spherical Symmetry 2.3.2 The Charge Distribution With A Cylindrical Symmetry 2.3.3 The Charge Distribution With A Planar Symmetry

CHAPTER SUMMARY EXERCISES AND PROBLEMS

2

2.1 THE FLUX OF A VECTOR FIELD In this chapter we introduce a property of an electric field called flux, which is similar to the flow of fluid through a surface. The calculation of flux of electric field leads to Gauss’s law. To illustrate the idea of flux, it is useful to look at the volume flow of a fluid per unit time (cubic meter per second in SI units) through a given area. Imagine a pipe through which a fluid flows. Let the velocity of the flow be described by the velocity vector field lines as shown in Figure 2-1. Since flow velocity may be different at different places, it is a vector function of space coordinates just as the electric field. For the sake of simplicity, here we consider only the case of a steady flow described by a constant velocity field. Now, construct a wire frame of area A, hold it at a particular place in the fluid oriented in a definite way, and observe the volume of fluid flowing through it per unit time. The volume of the fluid passing through the wire frame per unit time is equal to the flux of velocity field though the frame.

Figure 2-1: Illustrations of flux of a vector field vr

From the figure, it is clear that the orientation of the area of the wire frame relative to the velocity vector plays an important role in determining the flux. In order to take this into account, we define an area vector of a flat surface A

r whose magnitude is the area, and

whose direction is given by a unit vector normal to the flat surface.

!

Area vector r A of a flat surface

!

Magnitude = Area

!

Direction = Normal to the surface Equation 2-1

You may be wondering which way would the normal point, i.e., should it point up or point down for a horizontal area. If a surface is not closed, e.g. the area of the wire frame here, the normal direction is ambiguous up to a sign; it may point in either direction perpendicular

(c) Flux = v A cos θ

= Av

rr•

(b) Flux = 0

(b)

A

r

(c)

(a) (a) Flux = vA

A

r θ

vr

A

r

vr

A cos θ = Area

(π/2)-θ

3

to the surface. If a surface is closed, i.e. it encloses a volume, such as the surface of a cubical box, then the normal unit vector will be taken to point from the inside of the box to the outside. In Figure (1c), notice that when the wire frame is tilted by an angle, then, only the fluid passing through the projected area shown will pass through the area of the wire frame. The projected area in question is A cos θ , if θ is the angle the area vector makes with the velocity vector. The volume of the fluid flow per unit time will then be v A cos θ, or,

!

r v "

r A . Hence,

the flux of the velocity field through the surface of the wire frame

!

"vis given as:

!

"v=

!

r v "

r A (uniform

!

r v ) Equation 2-2

Clearly,

!

"v is the volume of the fluid flowing through the area per unit time.

We define the flux of electric field in a similar way. Consider a region of space where the electric field is constant, e.g. between two uniformly charged infinite parallel plates, then the situation would be akin to the uniform fluid flow discussed above, and you can think E

r in

place of vr

in Fig. 1 above. By, analogy, then, we would get the flux of electric field through a flat area

E! :

E

! = AE

rr! (uniform E

r) Equation 2-3

In the case of the fluid flow, the flux of the velocity field corresponds to a tangible physical quantity, namely, the volume of fluid passing the area per unit time. Here, if you think of the electric field at a point as number of field lines per unit cross-sectional area, then flux of electric field will correspond to “the number of electric field lines passing through the area”, which is a very abstract idea. Because of the abstractness inherent in description of electric phenomena, it is often very hard for a person to get used to these ideas. We encourage you to learn the language presented here as it will open the door for more in-depth studies later. In our studies below, we would be interested in more general cases, where electric field may be non-uniform, and the surface may not be flat. In order to generalize the definition of flux given in Equation 2-3 above, let us consider an arbitrarily shaped, but smooth, closed surface S in space where electric field is non-homogeneous as shown in Figure 2-2.

4

Figure 2-2: A surface in a region of non-uniform electric field We can approximate any smooth non-flat surface by a collection of tiny approximately flat surfaces (Figure 2-3). If we divide the surface S into small patches, then we notice that, as the patches become small in area, they can be approximated by flat surfaces as long as there are no kinks in the surface and the surface S is smooth. Figure 2-3: A surface is divided up in patches to find the flux. (a) Patches (b) Normal to a few patches shown – by convention the normal to a closed surface points from inside to outside. To keep track of patches, we can number them, from 1 through N, with N being a large number. Now, the area vector for each patch can be defined as the area of the patch pointed in the direction of an outward normal. Let us denote the area vector for the ith patch by

iA

r! . We have used the symbol Δ to remind us that the area is of an arbitrarily

small patch. Since, we expect the electric field to be non-uniform, we will have different directions and/or different magnitudes of the electric field at the sites of different patches. Let us denote by

iE

rthe electric field at the location of the i-th patch. Since the patch is

arbitrarily small, we can use the average of the electric field over the area of the i-th patch. Let

Surface S

5

!

r E

i= Average electric field over the ith patch.

Hence, we can use our answer for the uniform field for flux through the i-th patch

i! .

!

"i=

r E

i# \$

r A

iith patch( ) Equation 2-4

Although electric field is a vector, its flux is a scalar number, i.e., a positive or a negative real number. The flux through each of the individual patches can be constructed in this manner, and then added up to give us an estimate of the flux through the entire surface S.

!

" =r E

i# \$

r A

i

i=1

N

% N - patch estimate( ) Equation 2-5

The estimate gets better as we decrease the size of the patches, thereby increasing the number of patches needed to cover the surface. In the limit of infinitesimally small patches, the sum becomes a surface integral.

!

" =r E #d

r A

Surface S

\$\$ Equation 2-6

If the surface is closed, another symbol is used to denote the result.

!! "=#

SSurfaceClosed

Equation 2-7

Thus, to find the flux of a given vector field through a surface, first divide up the surface into small patches, then find the area vector for each patch, compute the dot product of the vector field with the area vector of each patch, and finally sum over the results from all patches. The calculation of flux through arbitrary surfaces, and for arbitrary vector fields can be very complicated, but that kind of calculation is hardly needed or even useful in physics. Fortunately, most useful cases have some symmetry that simplifies the calculations considerably as you will in the rest of the chapter.

6

EXAMPLE 2-1 Flux of a constant electric field # 1

Let there be a given constant electric field in the z direction

!

Eo

ˆ k . We want to find electric flux through a rectangle (a!b) in the xy plane. Solution: The electric field lines pass through the area perpendicularly. Therefore, the flux will have magnitude of (Electric Field) times (Area), i.e.

!

E0ab . Since the area is

not part of a closed surface, there is no inside or outside, therefore, the sign of flux will be undetermined, and the flux through the given area is either

!

+E0ab or

!

"E0ab .

These results can also be obtained by carrying out the integration. An infintesimal area element in the xy-plane is given as:

!

dr A = dx dy (± ˆ k ) (1)

(We can have + and – as both directions are possible for an open surface. ) Therefore, the electric flux is:

!

" = (E0ˆ k ) # (± ˆ k dx dy)\$\$ = ±E

0ab (2)

EXAMPLE 2-2 Flux of a constant field #2

Given a constant electric field in the z direction,

!

E0

ˆ k , find the electric flux through the rectangle (a!b) in the xz plane. Solution: Here the electric filed lines do not cross the area in question. Hence, the electric flux through the area will be zero. Algebraically, this result is due to the dot product of two vectors at right angle be zero as seen in the following calculations. The area element in the xz-plane is:

!

dr A = dx dz(± ˆ j ) (1)

The electric flux of the given electric field pointed along the z-axis is then:

!

" = ± (E0

ˆ k ) # ( ˆ j ) dx dz = 0\$\$ (2) since the dot product of two different unit vectors is zero.

b

a

x

y

z

7

EXAMPLE 2-3 Flux through a slanted area

Given a constant electric field in the z direction

!

E0

ˆ k find the electric flux through the surface (PQRS) inside the cube as shown in the Figure. Solution: The flux is easy to work out here if you think conceptually in terms of electric field lines. Since the electric field lines are parallel to the z-axis, the field lines that pass through the slanted area also pass through the bottom of the box, i.e. PQTU. Therefore, the flux through the slanted area is equal to the flux through PQTU, which is found easily to be

!

±E0a2 .

Of course, we can also find the flux through the slated area from a direct calculation of the dot product of an area element on the slated surface and the electric field. The direction of the area element of the slanted surface is found by taking the cross-product of two arbitrary vectors that lie on that surface.

!

Direction of dr A = ±

PQ

" # "

PQ\$

PS

" # "

PS

%

&

' ' '

(

)

* * *

= ± (1,0,0) \$1

2(0,1,1)

+

, - .

/ 0 = ±

1

2ˆ j + ˆ k ( ) (1)

Therefore flux through the area is

!

" = ± E0ˆ k #

1

2(ˆ i \$ ˆ j )dA = ±

E0

2dA%%%% (2)

Hence,

!

" = ±E0a2 (3)

which is the same result obtained much more easily by the physical argument using electric field lines.

z

y

x Q

P

R

S

T

U

8

EXAMPLE 2-4 Flux of a Non-homogeneous Field

Consider an electric field of a point charge q situated at the origin. Find the electric field of the point charge through a circular area of radius R about the z-axis. The center of the circle coincides with the z-axis and is a distance h from the origin. Solution: Note that the electric field of a point charge is inhomogeneous, meaning different at different point in space. In a Cartesian coordinate system, the field is

given by the following

!

r E =

q

4"#0

(xˆ i + yˆ j + z ˆ k )

x2 + y

2 + z2( )

3 / 2. The

use of electric field lines for an easy calculation of the electric flux is limited to constant fields and situations with symmetry. Here we do not have any simplifying features. Now, the electric flux cannot be obtained by simply multiplying the area with the electric field. Therefore, we must resort to the calculation based on the definition using the surface integral. First we note that the surface is parallel to the xy-plane. Therefore the area element vector will be directed along the z-axis and will be written as:

!

dr A = dx dy (± ˆ k ) (1)

Taking the dot product with the electric field, and imposing z = h we find the flux through an element of the area to be:

!

d" =q

4#\$0

±h dx dy

x2 + y

2 + h2( )3 / 2

(2)

We need to integrate over the area to find the total flux. Since the given area is circular in shape, the limits on the integral will be more easily placed if we use polar coordinates instead of Cartesian coordinates. The conversion between the two system of coordinates is:

!

x = rcos"; y = rsin"

r = x2

+ y2; " = tan

#1 y

x

\$

% & '

( ) (3)

The area elements in the two systems are related as follows.

!

dx dy = rdrd" (4)

c

z

y

x

S

9

The limits of integration are r = 0 to R and θ = 0 to 2π. The flux through a surface element written in polar coordinates becomes:

!

d" =q

4#\$0

±h dx dy

x2 + y

2 + h2( )3 / 2

=q

4#\$0

±h rdr d%

r2 + h2( )

3 / 2 (5)

Integrating over theta is trivial and yields just 2π. The integral of r is found by substitution method: Let u = r2, and do the integral over u.

!

" =±#hq

4#\$0

du

u + h2( )3 / 2

=0

R2

%±2#hq

4#\$0

1

h&

1

h2 + R2

'

( )

*

+ , (6)

2.2 GAUSS’S LAW Let us calculate the electric flux through a spherical surface around a charge q. Finding flux in this case is rather easy. On each patch, the electric field has the same magnitude q/(4 πε0r

2), and the direction of electric field is parallel to the area vector of the patch (Figure 2-4).

Figure 2-4: Gauss’s law applied to a point charge q. Therefore, we obtain the following for the flux of the electric field:

0

2

2

0

2

0

41

4

1

4 !"

"!"!

qr

r

qA

r

q

patchesE =# \$=%=& Equation 2-8

A remarkable fact about Equation 2-8 is that the flux is independent of the size of the spherical surface. This can be directly attributed to the fact that electric field of a point

A

r!

q

E

r

10

charge decreases as 1/r2 with distance, which just cancels the r2 rate of increase of the surface area. Gauss’s law generalizes the result obtained above to the case of any number of charges and any location inside the closed surface. According to Gauss’s law, the flux of electric field

!

r E

through any CLOSED surface is equal to the charge enclosed (qenc) divided by the permittivity of free space (ε0).

!! ="

SurfaceClosed

rr

Equation 2-9

Charge enclosed (qenc) is the sum total of all charges in the volume bounded by the closed surface. If the charges are discrete point charges, then we just sum over them, and if instead they are described by a continuous charge distribution, then we would need to integrate appropriately to get the total charge that resides inside the enclosed volume. We can also understand Gauss’s law graphically in terms of electric field lines as illustrated in Figure 5. Recall that the flux of electric field can also be thought to be equal to “the number of field lines” crossing the surface. Since, the field lines crossing a surface have directions, two field lines crossing the surface in opposite directions should cancel each other. Hence, we adopt the following convention for counting the number of filed lines that cross a given surface. At each instance, when a field line crosses the surface from inside to the outside, it will contribute plus one to the flux, and when a field line crosses the surface from outside to the inside, it will contribute minus one to the flux. For example, a field line that enters a closed surface at one point (count = -1), and then later exits it at another (count =+1), will contribute zero to the flux. With this understanding, you can immediately see that if no charges are included within a closed surface, then the electric flux through it must be zero since as many lines cross the surface in one direction as in the other direction (Figure 2-5(a)). The same will happen if the charges of equal and opposite sign are included inside the closed surface, so that the total charge included is zero (Figure 2-5(b)). All surfaces that include the same amount of charge have the same number of field lines crossing it, regardless of the shape or size of the surface, as long as the surfaces enclose the same amount of charge (Figure 2-5(c)).

11

Figure 2-5: Understanding flux in terms of field lines. (a) No charges included -> Flux through the closed surface =0. (b) Net charge included =0, hence flux through the closed surface is zero. Same number of lines enters it as exit it. (c) Shape and size of the surfaces that enclose a charge does not matter – all surfaces enclosing the same charge will have the same flux. You might be wondering if other 1/r2 force laws, most notably the Newton’s law of gravitational force, would also have a Gauss’s law. The answer is yes. See Figure 2-6. There we define a gravitational field g

r as the gravitational force on a test object per unit of its

mass m, in analogy with the electric force per unit charge for electric field. This gives the gravitational field g

r at a distance r of an isolated point mass M located at origin to be equal

to GM/r 2 in magnitude, and pointed radially towards the mass. As the patch area vector is 1800 from the direction of gravitational field g

r, the flux will be a negative number.

(a)

(c) S1

S2

(b)

12

Figure 2-6: Gauss’s law for Newton’s gravitational force Therefore, we get the following for the flux of the gravitational field.

GMrrr

GMA

r

GM

patches

!! 4422

22"=# \$"=%"=& Equation 2-10

Again, the flux is independent of the size of the spherical surface. Equations 2-8 and 2-10 demonstrate that flux of all vector fields that drop off as 1/r2 from the source will be independent of the size of the spherical surface.

2.3 APPLICATIONS OF GAUSS’S LAW Gauss’s law is very helpful in determining electric field of certain highly symmetric charge distributions. Whenever Gauss’s law is useful, it can make a complicated problem very simple. Therefore, it is worthwhile for you to learn the basic arguments necessary that makes a problem accessible through Gauss’s law. Basically, there are only three types of problems that have the requisite symmetry so that Gauss’s law can be fruitfully used to make predictions about their electric field. They are as follows.

• A charge distribution with spherical symmetry • A charge distribution with cylindrical symmetry • A charge distribution with planar symmetry

In each case, the symmetry of the charge distribution dictates a particular dependence of electric field on space coordinates. A closed surface in space, called the Gaussian surface, is then imagined that can make use of this additional information when setting up the surface integral. This leads to the flux becoming equal to the electric field on any point on the surface times the appropriate area. When you use this flux in Gauss’s law, you get an algebraic equation for the magnitude of electric field, which can be easily solved for the electric field. I shall illustrate the method for each of the three types of symmetries of the charge distribution.

A

r!

M

gr

13

2.3.1 The Charge Distribution with Spherical Symmetry A charge distribution has a spherical symmetry if the density of charge depends only on the distance from a center, and not on the direction in space. For instance, if a sphere of radius R is uniformly charged with charge density ρ0 then the distribution has a spherical symmetry (Figure 2-7(a)). On the other hand, if a sphere of radius R is charged so that top half of the sphere has uniform charge density of +ρ0 and the bottom half a uniform charge density of -ρ0 then it does not have a spherical symmetry because the charge density depends on the direction ( Figure 2-7(b)). So, just because a charged object is a sphere does not mean that the charge distribution has a spherical symmetry! Figure 2-7: Illustration of spherically symmetric and non-symmetric situations. We have spherical symmetry only when charge density does not depend on direction. In Figure 2-7(c), a sphere with four different charge densities is shown. Although this is a situation where charge density is not uniform, but since the charge density depends only on the distance from the center and not on the direction, it has a spherical symmetry. In all spherically symmetric cases, the electric field at any point must be radially directed, either towards the center or away from the center, because there are no preferred directions in the charge distribution. Therefore, using spherical coordinates we can write down the expected form of the electric field at a point a distance r from the center.

rrEEsymmetrySpherical ˆ)(: =r

Equation 2-11

Hence, the electric field flux through a spherical surface (Figure 2-8) having the same center as the charge distribution will simply be:

24)( rrEE

!"=# Equation 2-12

ρ 0 +ρ 0

-ρ 0

(a) Spherically Symmetric (b) Not spherically symmetric (c) Spherically Symmetric

14

Figure 2-8: Gaussian surface for a spherical symmetric charge distribution is a spherical surface with radius r . According to Gauss’s law, the flux through the imagined spherical surface will be equal to the total charge enclosed divided by the permittivity of vacuum.

!

4" r2E =qenc

#0

Equation 2-13

Hence, the electric field at a distance r from the center of a spherically symmetric charge distribution has the following magnitude.

!

E(r) =1

4"#0

qenc

r2

(SPHERICAL CASE) Equation 2-14

where qenc depends on whether r<R or r>R. If the radial distance r takes you outside the charge distribution, then charge enclosed will be the total charge. But, if the radial distance r goes to a point inside the charge distribution, charge enclosed will be the charge inside a sphere of radius r.

!

qenc =The entire charge if r > R

Only the charges in the sphere of radius r if r < R

" # \$

Equation 2-15

Note that electric field outside a spherically symmetric charge distribution is identical with that of a point charge at the center!

R

r

15

EXAMPLE 2-5 Uniformly Charged Sphere

Consider a sphere of radius R (meters) with uniform charge density ρ0 (C/m3). Find the electric field at a point outside the sphere and at a point inside the sphere. Solution: Here the charge distribution has a spherical symmetry since the charge density does not depend on the direction, we can use Equation 2-14 above. The answer can be written down immediately, giving the magnitude of the electric field as follows.

2

04)(

r

qrE tot

out!"

= , since qenc = total charge = qtot = (4/3) πR3 ρ0 (1)

0

0

3)(

!

" rrE

in= , since qenc = charge inside Gaussian surface = (4/3) πr3 ρ0 (2)

It is interesting to note that the electric field increases inside the material as you go out since the amount of charge by the imagined spherical surface increases with the volume enclosed which goes as r3 while the surface area increases only as r2. The electric field outside the sphere decreases as you go away from it since the included charge is same as long as you are outside the material. I have drawn a sketch of the electric field magnitude as a function of r in the Figure to give you a better feel for how the electric field of a uniformly charged sphere varies in space. Note that the direction of the electric field is radially outward if ρ0 is positive, and inward, i.e. towards the center, if ρ0 is negative. Figure Electric field of a uniformly charged non-conducting sphere increases inside the sphere to a maximum at the surface, and then decreases as 1/r2.

E

R r

0

0

3!

" R constant/r2

16

EXAMPLE 2-6 Non-Uniformly Charged Sphere

Consider a non-conducting sphere of radius R (meter) with a non-uniform charge density that varies with distance from its center as given by the following function. n

rar =)(! (coulomb per cubic meter) (1) Using Gauss’s law find electric field at a point outside the sphere, and at a point inside the sphere. Solution. Again, since the charge density has only a radial distance dependence and no dependence on the direction, we have a spherically symmetric situation. Therefore, we know that the answer will be as given in Equation 2-14. We just need to find the enclosed charge qenc, which will be different for the two cases. Since the charge density is not constant here, we need to integrate the charge density over the volume of the sphere. For the spherically symmetric charge density, think of dividing the sphere into thin spherical shells. One such shell between r and r +dr is shown in Figure. Its volume will be area times thickness =

!

4"r2dr . Multiplying the volume with the density gives the charge in the shell as drr

24!" # .

Figure Spherical symmetry with non-uniform charge distribution We must integrate this from r =0 to R to get the total charge in the sphere.

!

qenc = " dV =### = a rn( ) 4\$r2dr( )

0

R

# =4\$aRn+3

n + 3 (2)

r+dr

r

R

17

Hence electric field at a point outside the sphere is radially directed, and has the following magnitude.

!

Eout(r) =

aRn+3

(n + 3)"0r2

(3)

The charge enclosed will be only within a sphere of radius r.

!

qenc =4" a r

n+3

n + 3 (4)

Hence, the electric field at a point inside has the following magnitude.

0

1

)3()(

!+=

+

n

rarE

n

in (5)

2.3.2 The Charge Distribution With A Cylindrical Symmetry A charge distribution has a cylindrical symmetry if the following two conditions are satisfied: (1) the charge is distributed in a cylindrical shape extending to infinity along its axis, and (2) the charge density depends only on the distance s from the axis (Figure 2-9). For instance, a uniform charge density λ0 on a straight wire has a cylindrical symmetry, and so does an infinite cylinder of radius R and infinite length with a constant charge density ρ0. An infinite cylinder that has a charge density +ρ0 for z>0, and -ρ0 for the z<0 does not have a cylindrical symmetry, neither does an infinite cylinder which has charge density +ρ0 for 0≤θ≤π and -ρ0 for π<θ<2π. On the other hand, a system with concentric cylindrical shells each with uniform charge densities, albeit different in different shell, does have a cylindrical symmetry. The best way to tell whether a system has a cylindrical symmetry is to first make sure that you have an infinite cylinder, and then you look at the cross-section map of the charge density. If the charge density is a function only of the radial coordinate, then the system has a cylindrical symmetry. Figure 2-9: To tell if you have a cylindrical symmetry, look at the cross-section of an infinite cylinder. If the charge density does not depend on the polar angle on the cross-section, then you have a cylindrical symmetry.

(a) Cylindrically symmetric (b) Not cylindrically symmetric (c) Cylindrically symmetric

ρ 0

+ρ 0

-ρ 0

18

In all cylindrical symmetry cases, the electric field at any point must be directed perpendicular to the axis of the cylinder, and can only depend on the distance from the axis.

ssEEsymmetrylCylindrica ˆ)(: =r

Equation 2-16

Here s is the distance from the axis, and s is a unit vector directed perpendicular from the axis as shown in Figure 2-10. Figure 2-10: Electric field in a cylindrically symmetric situation depends only on the distance from the axis. The direction of the electric field is pointed away from the axis for positively charged and towards the axis for negative charges. The flux through a cylindrical surface, having the same axis as the axis of the cylinder of charge distribution, is easy to compute if we divide our task into two parts – flux through the flat ends and flux through the curved surface (Figure 2-11). Figure 2-11: Gaussian surface in the case of a cylindrical symmetry. Electric field at a patch is either parallel or perpendicular to the normal.

Axis

A

r!

E

r

A

r!

E

r

L

Axis

s s

19

The flux through the flat ends must be zero since electric field is perpendicular to the area vectors of the patches on it.

!

"E

ends=

r E # \$

r A = 0

patches

% Equation 2-17

The flux through the curved side is simply the product of electric field at any point of this surface, since all the points on it are same distance from the axis), and the area of the surface, 2πsL.

! "=#\$=%patches

sideE sLsEAE &2)(

rr Equation 2-18

Therefore, the flux through the closed cylindrical surface is given by the following.

E

ends

EE!2)( "=#+#=\$=# %%

rr Equation 2-19

According to the Gauss’s law, this must equal the amount of charge residing in the enclosed volume divided by the permittivity of free space. Hence, Gauss’s law for any cylindrically symmetric charge distribution yields the following result for the electric field a distance s away from the axis.

ssE

enc1

2)(

0!"

#= Equation 2-20

where λenc will depend on whether the field point is inside the cylinder or outside. If s>R, the radius of the charge distribution, the enclosed charge will be all the charge in the cylinder of radius R and of length L. On the other hand, if r<R, then only the charged within a cylinder of radius s and length L are enclosed by the Gaussian surface.

!"#

<

>=

Rsif

Rsifqenc

where R is the radius of the charged cylinder. When you do the calculation for a cylinder of length L, you will find that qenc is directly proportional to L. Let us write it as charge per unit length ( λenc) times L.

!

qenc = "encL Equation 2-22

You should notice that electric field for a cylindrical charge distribution does not drop as 1/distance squared, but instead as 1/distance. This has to do with the charges on the cylinder extending to infinity in one direction. In the case of planar charge distribution, below, where charges extend to infinity in an entire plane, we shall find that electric field is constant and does not depend on the distance at all.

20

EXAMPLE 2-7 Uniformly Charged Wire

Consider a uniformly charged thin straight wire with charge density λ0 (coulomb/meter). Find the electric field at a distance d (meters) from the wire. Solution: The charge distribution has a cylindrical symmetry of the simplest kind. Figure Uniformly charged infinite wire To apply Gauss’s law, we imagine a cylindrical surface passing through P. Then the enclosed charge is simply λ0 L, which means that enclosed charge per unit length is:

0

0 !!

! ==L

L

enc (1)

It is kind of silly to do this calculation of finding charge per unit length, since we already knew it was λ0 all the time. We did this so that we will use the same procedure in the next example. Now, that we know λenc, we can write electric field at P right away.

sd

E ˆ1

20

0

!"

#=

r (2)

where s is a unit vector, perpendicular to the axis and pointing away from it as shown in Figure (b).

L P d

(a)

d s

(b)

21

EXAMPLE 2-8 Uniformly Charged Cylindrical Shell

Consider a uniformly charged non-conducting cylindrical shell of radius R (meter) and charge density σ0 (coulomb/square meter). (a) Find the electric field at a point outside the shell. (b) Find the electric field at a point inside the shell. Solution: For a point on the outside of the cylindrical shell, the Gaussian surface will be the surface of a cylinder of radius s>R and length L (Figure a) Figure (a) Uniformly charged cylindrical shell The charge enclosed per unit length of the Gaussian cylinder is equal to the charge on the cylindrical shell of length L by its length L.

!

"enc

=(#

0)(2\$RL)

L= 2\$R#

0 (1)

Hence the electric field at a point outside the shell at a distance s away from the axis is:

!

r E =

2"R#0

2"\$0

1

s

ˆ s =R#

0

\$0

1

s

ˆ s (s > R) (2)

where s is a unit vector, perpendicular to the axis and pointing away from it. For a point inside the cylindrical shell, the Gaussian surface will be a cylinder whose radius s is less than R (Figure b)

R

s

Gaussian surface

Charged cylindrical shell

L

22

Figure (b) Uniformly charged cylindrical shell – Gaussian surface inside (s<R). Now, there are no charges included inside the Gaussian surface, since all the charges are on the cylindrical shell at R. 0=

enc! (3)

Therefore, the electric field inside a uniformly charged cylinder is zero.

)(0 RsE <=r

(4)

EXAMPLE 2-9 Non-Uniformly Charged Cylinder

Consider a non-conducting infinite cylinder of radius R (meter) which has a nonuniform charge density ρ (coulomb/cubic meter). The charge density changes in the cylinder but only depends upon the distance from the axis and not along the cylinder or on the direction. The distance dependence of the charge density is known to be given as follows.

R

ss != 0)( "" (1)

Find the electric field (a) at a point outside the cylinder, and (b) at a point inside the cylinder. Solution: This problem will go similar to the problem on the cylindrical shell done above. Therefore referring to the first figure in that example, the charge enclosed by the Gaussian surface is the charge in the cylinder of length L (Figure).

Gaussian Surface in space inside the shell

s

Charged cylindrical shell

R

23

Figure Cylinder, integration variable and Gaussian surface shown We can find net charges in this cylinder by adding up the charges in infinitesimally thin cylindrical shells. One such thin shell between the s’ and s’+ds’ is shown in the Figure . The charge density in the shell given in the problem statement by

!

"0#s'

R (coulomb/cubic meter)

The volume of the thin shell is:

!

2"s'( ) L( )(d # s ) Therefore charges enclosed in the spherical shell element is:

!

"0#2\$L

R# % s

2d % s

The integration variable is s’ and to find the total charge enclosed by the Gaussian surface we need to integrate from s’=0 to R. We obtain the following result for the net enclosed charge for a point outside the charged cylinder.

L

s’ ds’

s

R

24

3

23

0

R

R

Lqenc !!=

"# (2)

Therefore charge enclosed per unit length of the Gaussian cylinder is:

2

03

2R

enc!"# \$= (3)

Hence electric field

outE

rat a point outside the cylinder located at a distance s from the axis is:

ss

Eenc

outˆ1

3ˆ1

20

2

0

0!

"

#!

\$==

r (4)

Field inside the cylinder is obtained by choosing the Gaussian surface whose radius s is smaller than R. The charge enclosed will depend on the distance s of the inside point.

!

qenc = "0#2\$L

R#s3

3 (5)

The corresponding electric field

inE

r is found to depend on s as follows.

ssR

Ein

ˆ3

2

0

0

!

"=

r (6)

The electric field increases from the center to a maximum at the surface.

2.3.3 The Charge Distribution With A Planar Symmetry A planar symmetry of charge density is obtained when charges are uniformly spread over a flat surface. In a planar symmetry, all points in a plane parallel to the plane of charge are in identical situation with respect to the charges. Hence, the electric field will be expected to at most depend on the distance from the charged plane. Let xy-plane be the plane of charge, then we expect electric field to be:

!

r E = ˆ k E(z) (independent of x and y) Equation 2-23

The Gausian surface will need to be a box, two of whose sides should be parallel to the charged plane. We illustrate the planar symmetry with the following examples.

25

EXAMPLE 2-10 A Thin Infinite Plane

A charge distribution has a planar symmetry if the charges are uniformly spread on an infinite plane. We have already worked out the electric field of an infinite plane with uniform charge density σ0 (coulomb per square meter) (see last chapter). The electric field was found to be constant independent of position of the space point from the plane as shown in Figure (a). Figure (a) Electric field of a positively charges infinite plane with uniform charge density. In this section, we want to re-do this problem using Gauss’s law and the associated symmetry argument. We shall take the plane of the charge distribution to be the xy-plane and we wish to find electric field at a space point P with coordinates (x,y,z). Since charge density is same at all (x,y) coordinates in z= 0 plane, electric field at P cannot depend on its x and y -coordinates. Furthermore, the symmetry in the charge distribution tells us that electric field at point P will not have any components along x or y-axis, because the contribution from charge at one place of the plane will cancel the contribution from another place situated exactly opposite to it. Therefore the electric field at P can only depend on the distance from the plane and point either towards the plane or away from it.

!

Planar symmetry :r E = ˆ k E(z) (1)

where z is the distance from the plane and

!

ˆ k is unit vector normal to the plane. To make use of Gauss’s law, we need to imagine a closed surface that will make use of this property of electric field to simplify the flux integral. A rectangular parallelepiped whose two sides are parallel to the plane and which spans the plane turns out to be a good choice here as we shall see below. See Figure (b).

0

0

2!

"=E

0

0

2!

"=E

26

Figure (b) A thin charged sheet and the Gaussian box. The sides I and II of the Gaussian surface (box here) that are parallel to the infinite plane has been shaded. They are the only surfaces that have a non-zero flux through them. Let the area of the shaded surface on each side of the plane be A. Since sides I and II are same distance from the plane, the electric field has the same value at their location. If the charge on the plane is positive then, the direction of electric field and the area vector are the same. Therefore, we get the following for the flux of electric field through the box.

AEAEAEE

20000 =+++++=! (2) where the zeros are for the flux through the other sides of the box. According to the Gauss’s law, this must equal qenc/ε0. The charge enclosed is on the area A of the infinite plane. Hence,

Aqenc 0!= (3)

Using (2) and (3) in Gauss’s law, we can immediately determine the electric field of a uniformly charged infinite plane.

0

0

2!

"=E (4)

The direction of the field depends on the sign of the charge on the plane.

side I side II (behind the plane)

27

EXAMPLE 2-11 A Thick Infinite Plane

The planar symmetry applies also in the case of an infinite thick slab with volume charges such that they are constant in each plane perpendicular to the slab. Consider an infinite planar thick slab of thickness d (meter) with uniform charge density ρ0 (coulomb/cubic meter). Let us find the electric field at a point inside the slab and at a point outside the slab. Electric field at a point outside the slab We first find the electric field at a point outside the slab. The charges enclosed are in a volume A d. Therefore the total charge enclosed by the Gaussian surfaces shown in Figure is:

dAqenc 0!= (1)

The flux through the Gaussian surface is again:

AEAEAEE

20000 =+++++=! (2) Hence the electric field is given by the following.

0

0

2!

" dE = (3)

Figure (a) A thick charged plate and the Gaussian box.

side I side II (behind)

d

28

Electric field at a point inside the slab We construct a Gaussian box symmetrically inside the thick slab so that the two sides with non-zero flux are the same distance from the center of the slab (Figure). Figure (b) A thick charged plate and the Gaussian box inside. The flux still is given by

AEE

2=! (4) where A is the area of the shaded side where the flux is non-zero as before. The charge enclosed however is given by a different formula. The charge enclosed is equal to the charges in the box of volume A (2 z), where z is the distance of the parallel side of the box from the center of the slab.

zAqenc 20!= (5)

Hence electric field at a point inside the slab is give as follows.

0

0

!

" zE = (6)

The electric field inside depends on the distance of the point from the center of the thick slab.

Side I (Front inside)

Side II (Back inside)

d

29

CHAPTER SUMMARY

1. Gauss’s law Flux of a vector field

!

r E through a surface S is a two-dimensional integral of normal

component of the field integrated over the surface.

!

" =r E #d

r A

Surface S

\$\$ = E%

Surface S

\$\$ dA

Gauss’s law states that Electric flux through any closed surface is proportional to the amount of charge enclosed by the surface. There is no Gauss’s law for open surfaces, i.e., a surface that does not separate the space into inside and outside regions.

!! ="

SurfaceClosed

rr

2. Electric field by Gauss’s law

Gauss’s law can be used to find electric field in three symmetry situations.

i. Spherical symmetry

Electric field is radial as in spherical coordinates, and has the following values.

2

04)(

r

qrE enc

!"= (where r is distance from the center)

ii. Cylindrical symmetry

Electric field is radial as in polar coordinates, and has a magnitude,

ssE

enc1

2)(

0!"

#= (where s is the distance from the axis, and λenc is charge per unit length)

iii. Planar symmetry –

a. Electric field from a single plane of charge of charge density σ Coulomb per square meter is constant given by

!

E ="

2#0

(single sheet)

b. Electric field between two large sheets oppositely charged is

!

E ="

#0

(between two oppositely charged sheets)

30

EXERCISES AND PROBLEMS

Exercises 2-1 Flux calculations

1. Find the electric flux through a rectangular area 3 cm x 2 cm between two parallel plates

where there is a constant electric field of 30 N/C for the following orientations of the area. a. parallel to the plates b. perpendicular to the plates c. the normal to the area makes a 30-degrees angle with the direction of the electric

field. Ans: (c)

!

0.016 Nm2/C .

2. The electric flux through a square shaped area of side 5 cm near a large charged sheet is found to be 3 x 10-5 N.m2/C when the area is parallel to the plate. Find the charge density on the sheet. Ans:

!

4.43"10#14C /m

2 3. Two large rectangular aluminum plates of area 150 cm2 face each other with a separation

of 3 mm between them. The plates are charged with equal amount of opposite charges, ± 20 µC. The charges on the plates face each other. Find the flux through a circle of radius 3 cm between the plates when the normal to the circle makes an angle of 5-degrees with the normal to the plates.

Ans:

!

4.24 "105Nm

2/C .

4. A square surface of area 2 cm2 is in a space of uniform electric field of magnitude 103 N/C. Amount of flux through it depends on how the square is oriented relative to the direction of the electric field. Find the electric flux through the square, when normal to it makes the following angles with electric field. a. 300, b. 900 and c. 00

Ans: (a)

!

3

10Nm

2/C

5. Velocity in a fluid flowing in the y-direction is a function of x given by:

L

xvvy 0

= , where v0 and L are constants.

a. Find the flux through a rectangle in the yz-plane between a<y<b and c<z<d. b. Find the flux through a rectangle in the xz-plane between a<x<b and c<z<d.

Ans: (b)

!

±v0

L(d " c)

b2 " a2

2

#

\$ %

&

' (

6. A vector field is pointed along the z-axis: kyx

VkV zˆˆ

22+

==!r

.

a. Find the flux through a rectangle in the xy-plane between a<x<b and c<y<d. b. Find the flux through a rectangle in the yz-plane between a<z<b and c<y<d.

31

Exercises 2-2 Spherical symmetry problems

1. A non-conducting sphere of radius 3 cm has a uniform charge density of 3

nanoCoulombs per unit meter cubed. a. Find the total charge contained in the sphere. b. Find charge contained within 2 cm of its center. c. Find the flux through a spherical surface of radius 4 cm centered at the center of the

charged sphere. d. Find the flux through a spherical surface of radius 2 cm centered at the center of the

charged sphere. e. Find the flux through a spherical surface of radius 2 cm centered about a point 6 cm

from the center of the charged sphere. Ans: (a)

!

2.26 "10#13C , (b)

!

6.70 " 0#14C , (c)

!

2.56 "10#2Nm

2/C , (d)

!

7.57 "10#3Nm

2/C , (e) 0.

2. A copper spherical ball of radius 2 cm has some charge put on it. The electric flux

through a 30-cm radius spherical surface concentric with the spherical ball is

!

"3#104N.m

2/C .

a. Find the electric flux through a 5-cm radius spherical surface concentric with the copper ball.

b. Find the total charge on the copper ball. c. Determine the number of electrons and protons in the charged copper ball. Density

of copper = 8.5 g/cc. Atomic weight of copper = 63.5. Atomic number of copper = 29.

Ans: (a)

!

"3#104N.m

2/C , (b)

!

"2.67 #10"7C , (c)

!

Ne

= (7.83"1025

+1.67 "1012)electrons

3. When a charge is placed on a metal sphere, it ends up at the outer surface. Use this

information to determine the electric field of +3µC charge put on a 5 cm aluminum spherical ball at the following two points in space: a. a point 1 cm from the center of the ball (an inside point), and b. a point 10 cm from the center of the ball (an outside point).

Ans: (a) 0, (b)

!

2.7 "106N /C .

4. A gold spherical shell of radius 3 cm has a cavity of radius 2 cm. There lies a small

copper ball with a charge of +1.5 nC at the center of the cavity. As a result, the inner surface of the gold shell has a charge -1.5 nC and the outer surface has a charge of +1.5 nC. a. Determine the electric flux through the spherical surfaces of radii (i) 2.5 cm, and (ii)

4 cm concentric with the gold shell. b. Find the value of electric field at a point on these spherical surfaces.

Ans: (a) (ii)

!

1.69 "102Nm

2/C .

32

Exercises 2-3 Cylindrical Symmetry

1) Determine if approximate cylindrical symmetry holds for the following situations. State

why or why not. a) A 300-cm long copper rod of radius 1 cm is charged with +500 nC of charge and we

seek electric field at a point 5 cm from the center of the rod. b) A 10-cm long copper rod of radius 1 cm is charged with +500 nC of charge and we

seek electric field at a point 5 cm from the center of the rod. c) A 150-cm wooden rod is glued to a 150-cm plastic rod to make a 300-cm long rod,

which is then painted with a charged paint so that one obtains a uniform charge density. The radius of each rod is 1 cm, and we seek an electric field at a point that is 4 cm from the center of the rod.

d) Same rod as (c), but we seek electric field at a point that is 500-cm from the center of the rod.

2) A long silver rod of radius 3 cm has a charge of

!

"5µC /cm on its surface. a) Find the electric field at a point 5 cm from the center of the rod (an outside point). b) Find electric field at a 2-cm from the center of the rod (an inside point). Ans: (a)

!

1.8 "108N /C .

3) Electric field at 2-cm from the center of long copper rod of radius 1 cm is found to be 3

N/C. How much charge per unit length resides on the copper rod? Ans:

!

3.33"10#12C /m .

4) A large aluminum rod of diameter ½ cm has an electric field of -20 N/C at a point 3 cm

from the center of the rod. a) Find the electric flux through a cylindrical surface of radius 2 cm radius and height 5

cm. b) Find the electric flux through a cylindrical surface of radius 4 cm radius and height 5

cm. c) Find the electric flux through a cylindrical surface of radius 2 cm radius and height

2.5 cm. 5) A copper cylindrical shell of inner radius 2 cm and outer radius 3 cm surrounds

concentrically a charged aluminum rod of radius 1 cm with a charge density of 4 pC/m. All charges on the aluminum rod reside at its surface. The inner surface of the copper shell has exactly opposite charge to that of the aluminum rod while the outer surface of the copper shell has the same charge as the aluminum rod. Find the electric field at the following distances from the center of the aluminum rod. a) ½ cm b) 1.5 cm c) 2.5 cm d) 3.5 cm e) 7 cm

33

Exercises 2-4 Planar Symmetry

1. A 10 cm x 10 cm aluminum foil 0.1 mm thickness has a charge of 20 µC.

a. Find the charge density. b. Find electric field 1 cm from the center assuming approximate planar symmetry. Ans: (a)

!

1"10#3C /m

2, (b)

!

1.13"108N /C .

2. Two 10 cm x 10 cm aluminum foil pieces of thickness 0.1 mm face each other with a

separation of 5 mm. One of the foils has a charge of +30 µC and the other has −30 µC. a. Find the charge density at all surfaces, i.e., on those facing each other and those

facing away. b. Find electric field between the plates near the center assuming planar symmetry.

3. A thin copper plate of radius 15 cm is charged with an unknown charge. An electric

field of 10 N/C is found at all points near the center of the plate but outside of it. Determine the amount of charge on the plate.

4. Two large copper plates facing each other have charge densities

!

± 4C /m2 on the surface facing the other plate, and zero elsewhere.

Find the electric flux through a 3 cm x 4 cm rectangular are between the plates for the following orientations of the area. a. If the area is parallel to the plates, and b. If the area is tilted 30-degrees from the parallel direction. Ans: (a)

!

5.42 "108Nm

2/C , (b)

!

4.69 "108Nm

2/C .

PROBLEMS 1. A point charge of 10 µC is at an unspecified location inside a cube of side 2 cm. Find

the electric flux though the cube’s surface. Ans: 1.13 x 106 N m2/C.

2. A point charge –q is at the center of a cube of side a. Find the electric flux through one

of its sides. 3. It is found that there is a net flux of 104 N.m2/C inward through the surface of a sphere

of radius 5 cm. a. How much charge is inside the sphere? b. Where is the charge located? Ans: (a) 8.85 x 10-8 C, (b) Trick question.

300

34

4. A charge +q is placed at one of the corners of a cube of side a. Find the electric flux

through the shaded face shown in the figure.

5. A non-conducting spherical shell of inner radius a1 and outer radius b1 is uniformly

charged with charged density ρ1 inside another non-conducting spherical shell of inner radius a2 and outer radius b2 which is also uniformly charged with charge density ρ2. Find electric field at space point P at a distance r from the common center such that (a) r>b2, (b) a2<r<b2, (c) b1<r<a2, (d) a1<r<b1, and (e) r<a1.

6. A non-conducting sphere of radius R has non-uniform charge density given by the

following function of the distance r from the center of the sphere.

R

r

0!! =

where ρ0 is constant with units of charge density. Find the electric field (a) at an arbitrary point outside the sphere, and (b) at an arbitrary point inside the sphere.

Ans: (a)

!

"0

4#0

R3

r2

, (b)

!

"0

4#0Rr2.

7. Two non-conducting spheres of radii R1 and R2 are uniformly charged with charge

densities ρ1 and ρ2 respectively. They are separated at center-to-center distance a. Find the electric field at point P located at a distance r from the center of sphere 1 and is in the direction θ from the line joining the two spheres assuming their charge densities are not affected by the presence of the other cylinder.

a

R1

r

θ

P R2

35

8. Two non-conducting uniformly charged thin straight wires of charge densities λ1 and λ2 (coulomb per unit meter) respectively are placed a distance d apart. Find the electric field at a distance s from wire #1 and at an angle θ from the line joining 1 and 2.

9. A spherical cavity of radius a is cut in a non-conducting sphere R of uniform charge

density ρ. The center of the cavity is at a distance d from the center of the sphere. Find the electric field at points P1, P2 and P3 shown in the figure. Point P1 is at a distance d-a from the center, point P2 is directly above the center of the cavity perpendicular to the line joining the centers of the sphere and the cavity, and point P3 is a distance d above the center of the sphere above the center perpendicular to the line joining the centers of the sphere and cavity.

10. A disk of radius R is cut in a non-conducting large plate that is uniformly charged with

charge density σ coulomb per square meter. Find the electric field at a height h above

the center of the disk. (h>>R, h<<l or w). Ans:

!

"h

R2

+ h2

.

P1

P2

P3

h

P

l

w

36

37

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