University Physics: Waves and Electricity

Ch23. Finding the Electric Field – IILecture 8

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

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

8/3Erwin Sitompul University Physics: Wave and Electricity

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 ˆi jPr a a

13 2 2Pr a

33

3

ˆ PP

P

rr

r

1 12 2

ˆ 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

8/4Erwin Sitompul University Physics: Wave and Electricity

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.

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).

→

→

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.

→

8/5Erwin Sitompul University Physics: Wave and Electricity

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

8/6Erwin Sitompul University Physics: Wave and Electricity

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

8/7Erwin Sitompul University Physics: Wave and Electricity

Flashback: Multiplying VectorsThe 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

→ → → →

→ →

→ →

→ →

8/8Erwin Sitompul University Physics: Wave and Electricity

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

Flashback: Multiplying Vectors

( cos )( )a b

( )( cos )a b

cosa b ab

8/9Erwin Sitompul University Physics: Wave and Electricity

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 ˆi j ki 1 0 0j 0 1 0k 0 0 1

8/10Erwin Sitompul University Physics: Wave and Electricity

ˆ ˆ ˆi j ki 1 0 0j 0 1 0k 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 6cos

(5)(3.606)

109.438

Flashback: Multiplying Vectors

cosa b ab

ˆ ˆ ˆ ˆ(3i 4 j) ( 2i 3k)a b

^ ^ ^ ^ → →

8/11Erwin Sitompul University Physics: Wave and Electricity

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

→ →

8/12Erwin Sitompul University Physics: Wave and Electricity

Flux of an Electric Field

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

8/13Erwin Sitompul University Physics: Wave and Electricity

CheckpointThe 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

8/14Erwin Sitompul University Physics: Wave and Electricity

Example: Flux of an Electric FieldIn 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 V mE

(a)2

ABCDˆ4k mA

ABCD ABCDE A

ˆ ˆ( 10k) (4k) 240 N m C

(b) 2AEFG

ˆ6 j mA

AEFG AEFGE A

ˆ ˆ( 10k) (6 j) 0

A B

D CE

G

F

8/15Erwin Sitompul University Physics: Wave and Electricity

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.

→ ^ ^

8/16Erwin Sitompul University Physics: Wave and Electricity

Homework 7

(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.

→ ^ ^

New