lecture 5 r 2r2r yesterday we introduced electric field lines today we will cover some extra topics...

Lecture 5

R

2R

Lecture 5

• Yesterday we introduced electric field lines• Today we will cover some extra topics on

Electric Fields before going on to Electric Flux and Gauss’ Law next week:– Continuous Charge Distributions– Infinite line of charge– Motion of a charge in an electric field

Question 1

• Consider a circular ring with total charge +Q. The charge is spread uniformly around the ring, as shown, so there is λ = Q/2R charge per unit length.

• The electric field at the origin is

R

x

y++++

+ +

+++

++

+ + + + ++++++

+

(a) zeroR

2

4

1

0

(b) (c)

Question 1

a b c

77%

7%15%

1. a

2. b

3. c

Question 1• Consider a circular ring with total charge +Q.

The charge is spread uniformly around the ring, as shown, so there is λ = Q/2R charge per unit length.

• The electric field at the origin is

R

x

y++++

+ +

+++

++

+ + + + ++++++

+

(a) zeroR

2

4

1

0

(b) (c)

• The key thing to remember here is that the total field at the origin is given by the VECTOR SUM of the contributions from all bits of charge.

• If the total field were given by the ALGEBRAIC SUM, then (b) would be correct. (exercise for the student).

• Note that the electric field at the origin produced by one bit of charge is exactly cancelled by that produced by the bit of charge diametrically opposite!!

• Therefore, the VECTOR SUM of all these contributions is ZERO!!

Electric Fieldsfrom

Continuous Charge Distributions

• Principles (Coulomb’s Law + Law of Superposition) remain the same.

• Only change:

Examples: • line of charge• charged plates• electron cloud in atoms, …

++++++++++++++++++++++++++

rE(r) = ?

Charge Densities• How do we represent the charge “Q” on an extended object?

total chargeQ

small piecesof charge

dq

• Line of charge:= charge per

unit lengthdq = dx

• Surface of charge:= charge per

unit area

dq = dA

• Volume of Charge:

= charge per unit volume

dq = dV

How We Calculate (Uniform) Charge Densities:

Take total charge, divide by “size”

Examples:10 coulombs distributed over a 2-meter rod.

10Cλ 5 C/m

2m

14 pC (pico = 10-12) distributed over the surface of a sphere of radius 1 μm. 12

2-6 2

14 10 C 14σ C/m

4π(10 m) 4π

14 pC distributed over the volume of a sphere of radius 1 mm.

123 3

-3 343

14 10 C (3) 14ρ 10 C/m

π(10 m) 4π

Electric field from an infinite line charge

Approach:“Add up the electric field contribution from each bit of

charge, using superposition of the results to get the final field.”

In practice:• Use Coulomb’s Law to find the E-field per segment of charge• Plan to integrate along the line…

– x: from toOR : from to

++++++++++++++++++++++++++r

E(r) = ?

Any symmetries ? This may help for easy cancellations

+++++++++++++++++++++++++++++

x

Infinite Line of Charge

We need to add up the E-field components dE at the point r given by contributions from all segments dx along the line.

and x are not independent, choose to work in terms of Write dE in terms of , r, and Integrate from -/2 to +/2

Charge density =

++++++++++++++++ x

y

dx

r'r

dE


We use Coulomb’s Law to find dE:

What is r’ in terms of r ?

What is dq in terms of dx?

Therefore,

++++++++++++++++ x

y

dx

r'r

dE

What is dx in terms of ?

2tan secx r dx r d


• Components:

• Integrate:

Ex

++++++++++++++++ x

y

dx

r'r

dE Ey


• Now

• The final result:

/ 2

/ 2

sin 0d

/ 2

/ 2

cos 2d

Ex

++++++++++++++++ x

y

dx

r'r

dE Ey


Conclusion:

• The Electric Field produced by an infinite line of charge is:

- everywhere perpendicular to the line (Ex=0)- is proportional to the charge density ()- decreases as- Gauss’ Law makes this trivial!!

1r

++++++++++++++++ x

y

dx

r'r

dE

Question 2

•Examine the electric field lines produced by the charges in this figure.•Which statement is true?

(a) q1 and q2 have the same sign(b) q1 and q2 have the opposite signs and q1 > q2 (c) q1 and q2 have the opposite signs and q1 < q2

q1 q2

Question 2

a b c

1%

86%

13%

1. a

2. b

3. c

Question 2

Field lines start from q2 and terminate on q1.

This means q2 is positive; q1 is negative; so, … not (a)

Now, which one is bigger?

Note that more field lines emerge from q2 than end on q1

This indicates that q2 is greater than q1

•Examine the electric field lines produced by the charges in this figure.•Which statement is true?

(a) q1 and q2 have the same sign(b) q1 and q2 have the opposite signs and q1 > q2 (c) q1 and q2 have the opposite signs and q1 < q2

q1 q2

Electric Dipole: Lines of Force

Consider imaginary spheres centered on :

aa) +q (blue)

b

b) -q (red)c

c) midpoint (yellow)

• All lines leave a)

• All lines enter b)

• Equal amounts of leaving and entering lines for c)

Electric Field LinesElectric Field Patterns

Dipole ~ 1/R3

Point Charge ~ 1/R2

~ 1/RInfinite

Line of Charge

Distance dependence

Motion of a Charge in a Field

•An electron passes between two charged plates (cathode ray tube in your television set)

•While the electron is between the plates, it experiences an acceleration in the y-direction due to the electric field E


The only force is in the y direction and the acceleration is:

The initial velocity is in the x-direction, so the velocity as a function of time (t) is:

The time (T) taken by the charge to traverse the plates is determined only by the initial velocity in the x direction.

ˆ qE

F ma F qE a jm

0ˆ ˆ ˆ ˆ

x y

qEv v i v j v i t j

m

1

0

LT

v


•Therefore, the particle’s deflection in the y-direction is:

•It exits the field making an angle θ with its original direction, where:

22 1

20

1 1

2 2y

LqEy a T

m v

1

0 12

0 0

tan y

x

LqEv m v qEL

v v mv

•Exercise for the student; calculate where it hits the screen after a distance L2

The Story Thus Far

Two types of electric charge: opposite charges attract, like charges repel

Coulomb’s Law: 1 22

0

1ˆ

4

q qF r

r

Electric Fields• Charges respond to electric fields:

• Charges produce electric fields:

1F q E

22

0

1ˆ

4

qE r

r

The Story Thus Far

We want to be able to calculate the electric fields from various charge arrangements. Two ways:

2. Gauss’ Law: The net electric flux through any closed surface is proportional to the charge enclosed by that surface.

In cases of symmetry, this will be MUCH EASIER than the brute force method.

1. Brute Force: Add up / integrate contribution from each charge.

Often this is pretty difficult.

Ack!Ex: electron cloud around nucleus

• Finished Coulomb’s Law • Next lecture: Electric field Flux and Gauss’

Law• Read Chapter 23• Try Chapter 22, Problems 28,45,70

lecture 5 r 2r2r yesterday we introduced electric field lines today we will cover some extra topics...

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