Chapter 26 - Electric Chapter 26 - Electric FieldField

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity© 2007

Objectives: After finishing Objectives: After finishing this unit you should be able this unit you should be able to:to:

• Define the electric field and explain what determines its magnitude and direction.

• Discuss electric field lines and the meaning of permittivity of space.

• Write and apply formulas for the electric field intensity at known distances from point charges.

• Write and apply Gauss's law for fields around surfaces of known charge densities.

The Concept of a FieldThe Concept of a Field

A A fieldfield is defined as a is defined as a property of spaceproperty of space in which a in which a material object experiences a material object experiences a forceforce..

.PAbove earth, we say there is Above earth, we say there is a a gravitational fieldgravitational field at at PP..

Because Because a mass a mass mm experiences a experiences a downward downward forceforce at that point. at that point.

No force, no field; No field, no force!

m

F

The The directiondirection of the field is determined by the of the field is determined by the forceforce..

The Gravitational FieldThe Gravitational Field

Note that the force Note that the force F F is is realreal, but , but the field is just a convenient way the field is just a convenient way of of describing spacedescribing space..

The field at points A or B might The field at points A or B might be found from:be found from:

Fg

m

Fg

m

IfIf g g is known at is known at every point above every point above the earth then the the earth then the force force F F on a given on a given mass can be found.mass can be found.

The The magnitudemagnitude and and directiondirection of of the field the field gg is depends on the is depends on the weight, which is the force weight, which is the force F.F.

A

B Consider points Consider points AA and and BB above above the surface of the earth—just the surface of the earth—just points in points in spacespace..

FF

FF

The Electric FieldThe Electric Field1. Now, consider point 1. Now, consider point PP a a

distance distance rr from from +Q+Q..

2. An electric field 2. An electric field EE exists exists at at PP if a if a testtest charge charge +q +q has a force has a force FF at that point. at that point.

3. The 3. The directiondirection of the of the E E is is the same as the direction of the same as the direction of a a forceforce on on + (pos)+ (pos) charge. charge.

E

4. The 4. The magnitudemagnitude of of EE is is given by the formula:given by the formula:

N; Units

C

FE

q

N; Units

C

FE

q

Electric Field

++++ +

+++Q

.P

r

+qF

+

Field is Property of SpaceField is Property of Space

E

Electric Field

++++ +

+++Q

.

r

The field The field EE at a point exists whether there is a at a point exists whether there is a charge at that point or not. The charge at that point or not. The directiondirection of the of the field is field is awayaway from the from the +Q+Q charge. charge.

E

Electric Field

++++ +

+++Q

.

r++q --q

F

F

Force on Force on +q+q is with is with field direction.field direction.

Force on Force on -q-q is is against field against field

direction.direction.

Field Near a Negative Field Near a Negative ChargeCharge

Note that the field Note that the field EE in the vicinity of a in the vicinity of a negative negative chargecharge –Q –Q is is toward toward the charge—the direction that a the charge—the direction that a +q+q test charge would move. test charge would move.

Force on Force on +q+q is with is with field direction.field direction.

Force on Force on -q-q is is against field against field

direction.direction.

E

Electric Field

.

r++q

F

---- -

----Q

E

Electric Field

.

r

--qF

---- -

----Q

The Magnitude of E-FieldThe Magnitude of E-Field

The The magnitudemagnitude of the electric field intensity at a of the electric field intensity at a point in space is defined as the point in space is defined as the force per unit force per unit chargecharge (N/C)(N/C) that would be experienced by any that would be experienced by any test charge placed at that point.test charge placed at that point.

Electric Field Intensity E

Electric Field Intensity E

N; Units

C

FE

q

N; Units

C

FE

q

The The directiondirection of of EE at a point is the same as the at a point is the same as the direction that a direction that a positivepositive charge would move charge would move IFIF placed at that point.placed at that point.

Example 1.Example 1. A A +2 nC+2 nC charge is placed at a charge is placed at a distance distance rr from a from a –8 –8 CC charge. If the charge charge. If the charge experiences a force of experiences a force of 4000 N4000 N, what is the , what is the electric field intensity E at electric field intensity E at point P?point P?

Electric Field

.

---- -

----Q

P

First, we note that the direction of First, we note that the direction of E is toward –Q (down).E is toward –Q (down).

–8 C

E

++q

E4000 N

-9

4000 N

2 x 10 C

FE

q

+2 nC

r

E = 2 x 1012

N/C Downward

Note: The field Note: The field E E would be the would be the samesame for for any any charge charge placed at point placed at point PP. It is a property of that . It is a property of that spacespace..

Example 2.Example 2. A constant A constant E E field of field of 40,000 N/C40,000 N/C is maintained between the is maintained between the two parallel plates. What are the two parallel plates. What are the magnitude and direction of the force magnitude and direction of the force on an electron that passes horizontally on an electron that passes horizontally between the plates.cbetween the plates.c

E.FThe E-field is The E-field is

downward, and the downward, and the force on eforce on e-- is up. is up.

; F

E F qEq

-19 4(1.6 x 10 C)(4 x 10 )NCF qE

F = 6.40 x 10-15 N, Upward

F = 6.40 x 10-15 N, Upward

+ + + + + + + + +

- - - - - - - - -

-ee-- -ee-- -ee--

The E-Field at a distance r The E-Field at a distance r from a single charge Qfrom a single charge Q

++++ +

+++Q

.r

P

Consider a test charge Consider a test charge +q+q placed placed at at PP a distance a distance rr from from QQ..

The outward force on +q is:The outward force on +q is:

The electric field The electric field EE is therefore: is therefore:

2F kQq rE

q q 2

kQE

r 2

kQE

r

++qF

2

kQqF

r

++++

+

+++Q

.r

P

E

2

kQE

r

Example 3.Example 3. What is the electric field What is the electric field intensity intensity EE at point at point PP, a distance of , a distance of 3 3 mm from a negative charge of from a negative charge of –8 nC–8 nC??

.r

P

-Q3 m

-8 nC

E = ? First, find the First, find the magnitude:magnitude: 2

2

9 -9NmC

2 2

(9 x 10 )(8 x 10 C)

(3 m)

kQE

r

E = 8.00 N/CE = 8.00 N/C

The direction is the same as the force on a The direction is the same as the force on a positive charge positive charge ifif it were placed at the point it were placed at the point P: P: toward –Qtoward –Q..

EE = 8.00 N, toward - = 8.00 N, toward -QQ

The Resultant Electric The Resultant Electric Field.Field.The resultant field The resultant field EE in the vicinity of a number in the vicinity of a number of point charges is equal to the of point charges is equal to the vector sumvector sum of the of the fields due to each charge taken individually.fields due to each charge taken individually.

Consider E for each charge.Consider E for each charge.

+

- q1

q2q3

-

AE1

E3

E2

ER Vector Sum:

E = E1 + E2 + E3

Vector Sum:

E = E1 + E2 + E3

Directions are based on positive test charge.

Magnitudes are from:

2

kQE

r

Example 4.Example 4. Find the resultant field at Find the resultant field at pointpoint A A due to the due to the –3 nC–3 nC charge and the charge and the +6 nC+6 nC charge arranged as shown. charge arranged as shown.

+

-

q1

q24 cm

3 cm 5 cm

-3 nC

+6 nC

E for each q is shown E for each q is shown with direction given.with direction given.

E2

E1

1 21 22 2

1 2

; kq kq

E Er r

1 21 22 2

1 2

; kq kq

E Er r

A

2

2

9 -9NmC

1 2

(9 x 10 )(3 x 10 C)

(3 m)E

2

2

9 -9NmC

2 2

(9 x 10 )(6 x 10 C)

(4 m)E

Signs of the charges are used only to find direction of ESigns of the charges are used only to find direction of E

Example 4.Example 4. (Cont.) (Cont.)Find the resultant Find the resultant field at pointfield at point A A. The magnitudes are:. The magnitudes are:

+

-

q1

q24 cm

3 cm 5 cm

-3 nC

+6 nC

E2

E1

A

2

2

9 -9NmC

1 2

(9 x 10 )(3 x 10 C)

(3 m)E

2

2

9 -9NmC

2 2

(9 x 10 )(6 x 10 C)

(4 m)E

EE11 = = 3.00 N, West3.00 N, West EE22 = = 3.38 N, North3.38 N, North

E2

E1Next, we find vector resultant ENext, we find vector resultant ERR

ER

2 2 12 1

2

; tanR

EE E R

E

2 2 12 1

2

; tanR

EE E R

E

Example 4.Example 4. (Cont.) (Cont.)Find the resultant Find the resultant field at pointfield at point A A using vector using vector mathematics.mathematics.

E1 = 3.00 N, West

E2 = 3.38 N, North

Find vector resultant EFind vector resultant ERRE2

E1

ER

2 2(3.00 N) (3.38 N) 4.52 N;E 3.38 N

tan3.00 N

= 48.4= 48.400 N of W; or N of W; or = = 131.6131.600

Resultant Field: ER = 4.52 N; 131.60

Resultant Field: ER = 4.52 N; 131.60

Electric Field LinesElectric Field Lines

++++ +

+++Q --

-- -

----Q

Electric Field LinesElectric Field Lines are imaginary lines drawn in are imaginary lines drawn in such a way that their direction at any point is the such a way that their direction at any point is the same as the direction of the field at that point.same as the direction of the field at that point.

Field lines go Field lines go awayaway from from positivepositive charges and charges and towardtoward negativenegative charges. charges.

1. The direction of the field line at any point is 1. The direction of the field line at any point is the same as motion of +q at that point.the same as motion of +q at that point.

2. The spacing of the lines must be such that they 2. The spacing of the lines must be such that they are close together where the field is strong and are close together where the field is strong and far apart where the field is weak.far apart where the field is weak.

+ -qq11 qq22

EE11

EE22

EERR

Rules for Drawing Field LinesRules for Drawing Field LinesRules for Drawing Field LinesRules for Drawing Field Lines

Examples of E-Field LinesExamples of E-Field LinesTwo equal but Two equal but oppositeopposite charges. charges.

Two Two identical identical charges (both +).charges (both +).

Notice that lines Notice that lines leave +leave + charges and charges and enterenter - - charges.charges.

Also, Also, EE is is strongeststrongest where field lines are where field lines are most densemost dense..

The Density of Field LinesThe Density of Field Lines

NGaussian Surface

N

A

Line density

Gauss’s Law: The field E at any point in space is proportional to the line density at that point.

Gauss’s Law: The field E at any point in space is proportional to the line density at that point.

A

Radius r

rr

Line Density and Spacing Line Density and Spacing Constant Constant

Consider the field near a positive point charge q:Consider the field near a positive point charge q:

Gaussian Surface

Radius r

rr

Then, imagine a surface (radius r) surrounding q.Then, imagine a surface (radius r) surrounding q.

EE is proportional to is proportional to N/N/AA and and is equal to is equal to kq/rkq/r22 at any point. at any point.

2;

N kqE E

A r

Define Define as spacing constant. as spacing constant.

Then:Then:

0 0 Where is:N

EA

0

1

4 k

0

1

4 k

Permittivity of Free SpacePermittivity of Free SpaceThe proportionality constant for line density is The proportionality constant for line density is known as the known as the permittivity permittivity and it is defined by:and it is defined by:

2-12

0 2

1 C8.85 x 10

4 N mk

2-12

0 2

1 C8.85 x 10

4 N mk

Recalling the relationship with line density, we have:Recalling the relationship with line density, we have:

0 0 N

E or N E AA

Summing over entire area Summing over entire area A gives the total lines as:A gives the total lines as: N = oEAN = oEA

Example 5. Example 5. Write an equation for Write an equation for finding the total number of lines finding the total number of lines N N leaving a single positive charge leaving a single positive charge qq..

Gaussian Surface

Radius r

rr

Draw spherical Gaussian surface:Draw spherical Gaussian surface:

0 0 and EA N E A N

22 2

; A = 4 r4

kq qE

r r

Substitute for E and A from:Substitute for E and A from:

20 0 2

(4 )4

qN EA r

r

N = oqA = q N = oqA = q

Total number of lines is equal to the enclosed charge q.Total number of lines is equal to the enclosed charge q.

Gauss’s LawGauss’s Law

Gauss’s Law:Gauss’s Law: The net number of electric field The net number of electric field lines crossing any closed surface in an outward lines crossing any closed surface in an outward direction is numerically equal to the net total direction is numerically equal to the net total charge within that surface.charge within that surface.

Gauss’s Law:Gauss’s Law: The net number of electric field The net number of electric field lines crossing any closed surface in an outward lines crossing any closed surface in an outward direction is numerically equal to the net total direction is numerically equal to the net total charge within that surface.charge within that surface.

0N EA q 0N EA q

If we represent If we represent q q as as net enclosed net enclosed positive chargepositive charge, we can write , we can write rewrite Gauss’s law as:rewrite Gauss’s law as: 0

qEA

0

qEA

Example 6.Example 6. How many electric How many electric field lines pass through the field lines pass through the Gaussian surface drawn below.Gaussian surface drawn below.

+

-q1

q4

q3-

+q2

-4 C

+5 C

+8 C

-1 C

Gaussian surfaceFirst we find the NET First we find the NET charge charge qq enclosed enclosed by the surfaceby the surface::

q = (+8 –4 – 1) = +3 q = (+8 –4 – 1) = +3 CC

0N EA q

N = +3 C = +3 x 10-6 linesN = +3 C = +3 x 10-6 lines

Example 6.Example 6. A solid sphere (R = 6 cm) A solid sphere (R = 6 cm) having net charge +8 having net charge +8 C is inside a C is inside a hollow shell (R = 8 cm) having a net hollow shell (R = 8 cm) having a net charge of –6 charge of –6 C. What is the electric C. What is the electric field at a distance of 12 cm from the field at a distance of 12 cm from the center of the solid sphere?center of the solid sphere?

q = (+8 – 6) = +2 q = (+8 – 6) = +2 CC

0N EA q -6 C

+8 C--

--

-

-- -

Draw Gaussian sphere at Draw Gaussian sphere at radius of 12 cm to find E.radius of 12 cm to find E.

8cm

6 cm

12 cm

Gaussian surface

00

; net

qAE q E

A

2

2

-6

2 -12 2Nm0 C

2 x 10 C

(4 ) (8.85 x 10 )(4 )(0.12 m)

qE

r

Example 6 (Cont.)Example 6 (Cont.) What is the electric What is the electric field at a distance of 12 cm from the field at a distance of 12 cm from the center of the solid sphere?center of the solid sphere?Draw Gaussian sphere at Draw Gaussian sphere at radius of 12 cm to find E.radius of 12 cm to find E.

q = (+8 – 6) = +2 q = (+8 – 6) = +2 CC

0N EA q

00

; net

qAE q E

A

6 NC2

0

2 C1.25 x 10

(4 )E

r

-6 C

+8 C--

--

-

-- -

8cm

6 cm

12 cm

Gaussian surface

E = 1.25 MN/CE = 1.25 MN/C

Charge on Surface of Charge on Surface of ConductorConductor

Charged Conductor

Gaussian Surface just inside conductor

Since like charges Since like charges repel, you would repel, you would expect that all charge expect that all charge would move until they would move until they come to rest. Then come to rest. Then from Gauss’s Law . . .from Gauss’s Law . . .

Since charges are at rest, E = 0 inside conductor, thus:Since charges are at rest, E = 0 inside conductor, thus:

0 or 0 = N EA q q

All charge is on surface; None inside Conductor

All charge is on surface; None inside Conductor

Example 7.Example 7. Use Gauss’s law to find the Use Gauss’s law to find the E-field just outside the surface of a E-field just outside the surface of a conductor. The surface charge density conductor. The surface charge density = = q/Aq/A..Consider Consider q insideq inside thethe pillboxpillbox. . EE-lines -lines through all areas through all areas outward.outward.

Surface Charge Density

++

+ ++

+ ++

+

+ +++A

E2

E1

0 AE q E-lines through E-lines through sidessides cancel by cancel by symmetry.symmetry.

E3

E3 E3

E3

ooEE11A + A + ooEE22AA = = qq

The field is zero inside the conductor, so The field is zero inside the conductor, so EE22 = 0 = 0 00

0 0

qE

A

0 0

qE

A

Example 7 (Cont.)Example 7 (Cont.) Find the field just Find the field just outside the surface if outside the surface if = = q/A = q/A = +2 +2 C/mC/m22..

Surface Charge Density

++

+ ++

+ ++

+

+ +++A

E2

E1 E3

E3 E3

E3

10 0

qE

A

10 0

qE

A

Recall that side fields Recall that side fields cancel and inside cancel and inside field is zero, so thatfield is zero, so that

2

2

-6 2

-12 NmC

2 x 10 C/m

8.85 x 10E

E = 226,000

N/C E = 226,000 N/C

Field Between Parallel Field Between Parallel PlatesPlates

Equal and opposite charges.Equal and opposite charges.

Draw Gaussian pillboxes Draw Gaussian pillboxes on each inside surface.on each inside surface.

+++++

Q1 Q2

-----

Field EField E11 and E and E22 to right. to right.

E1

E2

E1

E2

Gauss’s Law for either box Gauss’s Law for either box gives same field (Egives same field (E11 = E = E22).).

0 AE q 0 0

qE

A

0 0

qE

A

Line of ChargeLine of Charge

r

E

2r

L

q

L

A1

A

A2

0

q; =

2 L

qE

rL

02E

r

02

Er

Field due to AField due to A11 and and AA2 2 Cancel out due Cancel out due to symmetry.to symmetry.

0

; (2 )q

EA A r L

0 AE q

Example 8:Example 8: The Electric field at a The Electric field at a distance of 1.5 m from a line of charge distance of 1.5 m from a line of charge is 5 x 10is 5 x 1044 N/C. What is the linear N/C. What is the linear density of the line?density of the line?

r

EL

q

L

02E

r

02

Er

02 rE

2

2

-12 4CNm

2 (8.85 x 10 )(1.5 m)(5 x 10 N/C)

E E = 5 x 10= 5 x 1044 N/CN/C r = 1.5 r = 1.5 mm

4.17 C/m

Concentric CylindersConcentric Cylinders

+ + ++ + +

++ +

+ + + + ++ + + +

+ +

+ +

a

b

a

b

r1r2

-6 Cra

rb

12 cm

Gaussian surface

a

b

Outside is like Outside is like charged long wire:charged long wire:

For r >

rb02

a bEr

For rb > r >

ra

02aE

r

Example 9.Example 9. Two concentric cylinders of Two concentric cylinders of radii radii 3 and 6 cm3 and 6 cm. Inner linear charge . Inner linear charge density is density is +3 +3 C/mC/m and outer is and outer is -5 -5 C/mC/m. . Find E at distance of Find E at distance of 4 cm4 cm from center. from center.

+ + +

+ + + ++ +

+ + + + +

+ + + +

+ +

+ + a = 3

cm

b=6 cm

-7 C/m

+5 C/m

E = 1.38 x 106 N/C, Radially outE = 1.38 x 106 N/C, Radially out

rr

Draw Gaussian Draw Gaussian surface between surface between

cylinders.cylinders.

02bE

r

0

3 C/m

2 (0.04 m)E

E = 5.00 x 105 N/C, Radially inward

E = 5.00 x 105 N/C, Radially inward

+ + +

+ + + ++ +

+ + + + +

+ + + +

+ +

+ + a = 3 cm

b=6 cm

-7 C/m

+5 C/m

rr

Gaussian outside of Gaussian outside of both cylinders.both cylinders.

02a bE

r

0

( 3 5) C/m

2 (0.075 m)E

Example 8 (Cont.)Example 8 (Cont.) Next, find E at a Next, find E at a distance of 7.5 cm from center (outside distance of 7.5 cm from center (outside both cylinders.)both cylinders.)

Summary of FormulasSummary of Formulas

The Electric Field Intensity E:

The Electric Field Intensity E: 2

N Units are

C

F kQE

q r 2

N Units are

C

F kQE

q r

The Electric Field Near several charges:

The Electric Field Near several charges: 2

Vector SumkQ

Er

2 Vector Sum

kQE

r

Gauss’s Law for Charge distributions.

Gauss’s Law for Charge distributions. 0 ;

qEA q

A 0 ;

qEA q

A

CONCLUSION: Chapter 24CONCLUSION: Chapter 24The Electric FieldThe Electric Field