University Physics: Waves and Electricity

University Physics: Waves and Electricity

Ch23. Finding the Electric Field – IILecture 8

Dr.-Ing. Erwin Sitompulhttp://zitompul.wordpress.com

2014

8/2Erwin Sitompul University Physics: Wave and Electricity

Three particles are fixed in place and have charges q1 = q2 = +p and q3 = +2p. Distance a = 6 μm.What are the magnitude and direction of the net electric field at point P due to the particles?

Homework 6: Three Particles

191.602 10 Cp 191.602 10 Ce


Solution of Homework 6: Three Particles

1E

2E

3E

1 2 0E E

• Both fields cancel one another

,net 1 2 3PE E E E

,net 3PE E

332

3

r PP

qkr

1 13 2 2

ˆ î jPr a a

13 2

2Pr a

33

3

ˆ PP

P

rr

r

1 1

2 2ˆ ˆ2i 2 j

199

,net 1 6 22

(2 1.602 10 )8.99 10

( 2(6 10 ))PE

• Magnitude

• Direction

160 N C

,net 45PE

ˆ ˆcos i sin j 45


The Electric Field

The calculation of the electric field E can be simplified by using symmetry to discard the perpendicular components of the dE vectors.

→

→

Instead of considering dE in a given charge distribution, Gauss’ law considers a hypothetical (imaginary) closed surface enclosing the charge distribution.

Gauss’ law relates the electric fields at points on a closed Gaussian surface to the net charge enclosed by that surface.

→

For certain charge distributions involving symmetry, we can simplify even more by using a law called Gauss’ law, developed by German mathematician and physicist Carl Friedrich Gauss (1777–1855).


Flux

Suppose that a wide airstream flows with uniform velocity v flows through a small square loop of area A.

Let Φ represent the volume flow rate (volume per unit time) at which air flows through the loop.

Φ depends on the angle θ between v and the plane of the loop.

→

→

ˆNA A r • Unit vector pointing to the normal

direction of the plane


Flux

If v is perpendicular to the plane (or parallel to the plane’s direction), the rate Φ is equal to vA.

If v is parallel to the plane (or perpendicular to the plane’s direction), no air moves through the loop, so Φ is zero.

For an intermediate angle θ, the rate of volume flow through the loop is:

→

→

( cos )v A

This rate of flow through an area is an example of a flux.

The flux can be interpreted as the flow of the velocity field through the loop.

v A


Flashback: Multiplying Vectors

The Scalar Product The scalar product of the vector a and b is written as a·b

and defined to be:

Because of the notation, a·b is also known as the dot product and is spoken as “a dot b.”

If a is perpendicular to b, means Φ = 90°, then the dot product is equal to zero.

If a is parallel to b, means Φ = 0, then the dot product is equal to ab.

cosa b ab

→ → → →

→ →

→ →

→ →


The dot product can be regarded as the product of the magnitude of the first vector and the projection magnitude of the second vector on the first vector


( cos )( )a b

( )( cos )a b

cosa b ab


Flashback: Multiplying Vectors When two vectors are in unit vector notation, their dot

product can be written as

x x y y z za b a b a b

ˆ ˆ ˆ ˆ ˆ ˆ( i j k) ( i j k)x y z x y za b a a a b b b

ˆ ˆ î j k

i 1 0 0

j 0 1 0

k 0 0 1


ˆ ˆ î j k

i 1 0 0

j 0 1 0

k 0 0 1

Solution:

What is the angle Φ between a = 3i – 4j and b = –2i + 3k ?

x

z

y

3

–4

–2

3

a

b

2 23 ( 4) 5a 2 2( 2) 3 3.606b

ˆ ˆ(3i)( 2i) 6

6 (5)(3.606)cos 1 6

cos(5)(3.606)

109.438


cosa b ab

ˆ ˆ ˆ ˆ(3i 4 j) ( 2i 3k)a b

^ ^ ^ ^ → →


Flux of an Electric Field

The next figure shows an arbitrary Gaussian surface immersed in a nonuniform electric field.

The surface is divided into small squares of area ΔA, each being very small to permit us to consider the individual square to be flat.

The electric field E may now be taken as constant over any given square.

The flux of the electric field for the given Gaussian surface is:

→

E A

• Φ can be positive, negative, or zero, depending on the angle θ between E and ΔA

→ →


Flux of an Electric Field

S

E dA

The exact solution of the flux of electric field through a closed surface is:

The flux is a scalar, and its Si unit is Nm2/C.

• The electric flux through a Gaussian surface is proportional to the net number of field lines passing through that surface

• Without any source of electric field inside the surface as in this case, the total flux through this surface is in fact equal to zero


Checkpoint

The figure below shows a Gaussian cube of face area A immersed in a uniform electric field E that has the positive direction of the z axis. In terms of E and A, determine the flux flowing through:

(a) the front face (xy plane)(b) the rear face(c) the top face(d) the whole cube

Φ = +EAΦ = –EA

Φ = 0Φ = 0


Gaussian Surface: Cylinder

The next figure shows a Gaussian surface in the form of a cylinder of radius R immersed in a uniform electric field E, with the cylinder axis parallel to the field. What is the flux Φ of the electric field through this closed surface?

→

S

E dA

a b c

E dA E dA E dA

( )( )(cos180 )a a

E dA E dA

a

E dA EA

• left cap

• right cap

• cylindrical surface

( )( )(cos 0)c c

E dA E dA

c

E dA EA

( )( )(cos90 )b b

E dA E dA

00 0EA EA • zero net flux


2 2

0 0

(3)(1)z y

dydz

2 2

0 0

(3)(3)z y

dydz

Gaussian Surface: Cube

A nonuniform electric field given by E = 3xi + 4j N/C pierces the Gaussian cube shown here (x in meters). What is the electric flux through the right face, the left face, and the top face?

^ ^

r r

r

E dA

ˆ ˆ ˆ(3 i 4 j) ( i)rr

x dA 2 2

0 0 3

ˆ ˆ ˆ(3 i 4 j) ( i)z y x

x dydz

236 N m C

l l

l

E dA

ˆ ˆ ˆ(3 i 4 j) ( i)ll

x dA 2 2

0 0 1

ˆ ˆ ˆ(3 i 4 j) ( i)z y x

x dydz

212 N m C

2 2

0 0 3

3z y x

x dydz

2 2

0 0 1

3z y x

x dydz

→

• right face • left face


Gaussian Surface: Cube

A nonuniform electric field given by E = 3xi + 4j N/C pierces the Gaussian cube shown here (x in meters). What is the electric flux through the right face, the left face, and the top face?

^ ^

t t

t

E dA

ˆ ˆ ˆ(3 i 4 j) ( j)tt

x dA 2 3

0 1 2

ˆ ˆ ˆ(3 i 4 j) ( j)z x y

x dxdz

216 N m C

2 3

0 1 2

4z x y

dxdz

→

• top face


Example: Flux of an Electric Field

In a three-dimensional space, a homogenous electric field of 10 N/C is directed down to the negative z direction. Calculate the flux flowing through:(a) the square ABCD (xy plane)(b) the rectangular AEFG (xz plane)

x

y

z

0 1 2 3

1

2

3

1

2

3

E

ˆ10k N CE

(a)2

ABCDˆ4k mA

ABCD ABCDE A

ˆ ˆ( 10k) (4k) 240 N m C

(b)2

AEFGˆ6 j mA

AEFG AEFGE A

ˆ ˆ( 10k) (6 j) 0

A B

D C

E

G

F


Gauss’ Law

0 encq

Gauss’ law relates the net flux Φ of an electric field through a closed surface (a Gaussian surface) to the net charge qenc that is enclosed by that surface.

0 enc

S

E dA q

• in vacuum

If you know how much electric field is intercepted by the closed surface, you can calculate how much net charge is inside in the volume of that enclosed surface.


Gauss’ Law

• S1: Electric field is outward for all points flux is positive enclosed charge is positive

• S2: Electric field is inward for all points flux is negative enclosed charge is negative

• S3: No charge enclosed

• S4: Net charge enclosed is equal to zero, the field lines leaving the surface are as many as the field lines entering it


Gauss’ Law: Net Enclosed Charge

What is the net charge enclosed by the Gaussian cube of the previous example?

b b

b

E dA

ˆ ˆ ˆ(3 i 4 j) ( j)b

b

x dA 2 3

0 1 0

ˆ ˆ ˆ(3 i 4 j) ( j)z x y

x dxdz

216 N m C 2 3

0 1 0

4z x y

dxdz

• bottom face

f f

f

E dA

ˆ ˆ ˆ(3 i 4 j) ( k)f

f

x dA

• front face

0

back back

back

E dA

ˆ ˆ ˆ(3 i 4 j) ( k)back

back

x dA

• back face

0


Gauss’ Law: Net Enclosed Charge

r l t b f back

What is the net charge enclosed by the Gaussian cube of the previous example?

236 12 16 16 0 0 24 N m C

enc 0

S

q E dA

0

12(8.854 10 )(24) 102.125 10 C


Applying Gauss’ Law: Cylindrical Symmetry

The next figure shows a section of an infinitely long cylindrical plastic rod with a uniform positive linear charge density λ, in C/m.

We need an expression for the magnitude of the electric field E at a distance r from the axis of the rod.

Due to symmetry, the component of the resultant electric field will be only directed radially outward.

→

cosEA (2 )cos 0E rh (2 )E rh• What about top cap

and bottom cap

0 encq But,

0 (2 )E rh h 02E

r


Applying Gauss’ Law: Planar Symmetry

Imagine a thin, infinite sheet with a uniform (positive) surface charge density σ, in C/m2. Now, let us find the electric field E a distance r in front of the sheet.

A useful Gaussian surface in this case is a closed cylinder with end caps of area A, arranged to pierce the sheet perpendicularly, as shown.

→

0 enc

S

E dA q

0 ( )EA EA A

02E

• What is the direction of the electric field?


Example: Single Thin Plate

Charge of 10–6 C is given to a 2-m2 thin plate. Afterwards, an electron with the mass 9.109×10–31

kg with the charge 1.602×10–19

C is held in a distance 10 cm from the plate.(a) Determine the force acting on the electron.(b) If the electron is released, determine the speed of the

electron when it hits the plate.

q

A (a)

67 210

5 10 C m2

02E

7

412

5 102.824 10 N C

2(8.854 10 )

away from the plate

F qE 19 4(1.602 10 )(2.824 10 ) 154.524 10 N

toward the plate


Example: Single Thin Plate

Charge of 10–6 C is given to a 2-m2 thin plate. Afterwards, an electron with the mass 9.109×10–31

kg with the charge 1.602×10–19

C is held in a distance 10 cm from the plate.(a) Determine the force acting on the electron.(b) If the electron is released, determine the speed of the

electron when it hits the plate.

Fa

m(b)

1515 2

31

4.524 104.967 10 m s

9.109 10

2 2

0 02 ( )v v a x x

toward the plate

2 15(0) 2(4.967 10 )(0 0.1) 149.934 10

73.152 10 m sv


Example: Double Parallel Plates

( )( )

02E

6

512

6.8 103.840 10 N C

(2)(8.854 10 )

Two large, parallel plates, each with a fixed uniform charge on one side, is shown below. The magnitudes of the surface charge densities are σ(+) = 6.8 μC/m2 and σ(–) = 4.3 μC/m2. Find the electric field E (a) to the left (L) of the plates(b) between (B) the plates(c) to the right (R) of the plates

→

away from the (+) plate

( )( )

02E

6

512

4.3 102.428 10 N C

(2)(8.854 10 )

toward the (–) plate


Example: Double Parallel Plates

(a) ( ) ( )LE E E 5 53.840 10 2.428 10 51.412 10 N C


(b) ( ) ( )BE E E 5 53.840 10 2.428 10 56.268 10 N C


(c) ( ) ( )RE E E 5 53.840 10 2.428 10 51.412 10 N C

away from the (–) plate


Applying Gauss’ Law: Spherical Symmetry

A spherical shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell’s charge were concentrated at the center of the shell.

If a charged particle is located inside a spherical shell of uniform charge, there is no electrostatic force on the particle from the shell.

20

1,

4

qE r R

r

0,E r R


Applying Gauss’ Law: Spherical Symmetry

Any spherically symmetric charge distribution, such as on the figure, can be constructed with a nest of concentric spherical shells.

If the entire charge lies within a Gaussian surface, r > R, the charge produces an electric field on the Gaussian surface as if the charge were a point charge located at the center.

If only a portion of the charge lies within a Gaussian surface, r < R, then the charge enclosed q’ is proportional to q.

30

,4

qE r r R

R


Checkpoint

The figure shows two large, parallel sheets with identical (positive) uniform surface charge densities, and a sphere with a uniform (positive) volume charge density. Rank the four numbered points according to the magnitude of the net electric field there, greatest first.

• The electric field contributed by the two parallel sheets is identical at all numbered points.

• The closer to the sphere, the greater the electric field contributed by it.

3 and 4 tie, then 2, 1.


Email Quiz

A long thin cord has a positive a linear charge density of 3.1 nC/m. The wire is to be enclosed by a coaxial, thin cylindrical shell of radius 1.8 cm. The shell is to have negative charge on its surface with a surface charge density that makes the net external electric field zero. Calculate the surface charge density σ of the cylindrical shell.

0E ��


Homework 7

(a) The rectangle ABCD is defined by its corner points of A(2,0,0), B(0,3,0), C(0,3,2.5), and D(2,0,2.5). Draw a sketch of the rectangular.

(b) Given an electric field of E = –2i + 6j N/C, draw the electric field on the sketch from part (a).

(c) Determine the number of flux crossing the area of the rectangular ABCD.

→ ^ ^


Homework 7A

1. (a) The triangle FGH is defined by its corner points of F(2,0,0), G(0,3,0), and H(0,0,4). Draw a sketch of the triangle.(b) Given an electric field of E = –2i + 6j N/C, draw the electric field on

the sketch from part (a).(c) Determine the number of flux crossing the area of the triangle FGH.

→ ^ ^

2. A rectangle is under the influence of electric field of E = 2xyi + 4zk N/C. The dimension of the rectangle is 1 m × 2 m × 3 m, with x1 = 5 m and y1 = 4 m. What is the electric flux flowing through the front face and top face?

→ ^ ^

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University Physics: Waves and Electricity