Lecture 3-1

Electric Field

Define electric field, which is independent of the test charge, q, and depends only on position in space:

FE

q��������������

��������������

dipole

• One is > 0, the other < 0

-q q

electric dipole of dipole moment:

p q d����������������������������

d��������������

- +

Lecture 3-2

Dipole in uniform electric fields

• No net force. The electrostatic forces on the constituent point charges are of the same magnitude but along opposite directions. So, there is no net force on the dipole and thus its center of mass should not accelerate.

• Net torque! There is clearly a net torque acting on the dipole with respect to its center of mass, since the forces are not aligned.

http://qbx6.ltu.edu/s_schneider/physlets/main/dipole_torque.shtml

Ep

Lecture 3-3

Electric Field from Coulomb’s Law

20

1ˆ

4i

ii i

qE r

r

++

+ +

+

--

-+-

-

Bunch of Charges

dV

dq dA

dL

(volume charge)

(surface charge)

(line charge)

Continuous Charge Distribution

dq2

0

1ˆ

4

dqE r d E

r

��������������P

r

Summation over discrete charges

Integral over continuous charge distribution

Pir

iq

k

http://www.falstad.com/vector3de/

Lecture 3-4

Reading Quiz 1

Which one of the following statements is incorrect ?

• A) Electric fields leave positive charges and end on negative charges

• B) Electric field lines can intersect at some points in space.

• C) Electric field field lines from a dipole fall off faster than 1/r2.

• D) Electric fields describe a conservative force field.

Lecture 3-5

SUMMARY: FIND THE ELECTRIC FIELD

GIVEN THE CHARGES

1) GEOMETRY FOR qi or dqi

2) DISCRETE CHARGES q i

3) CONTINUOUS CHARGES dqi

line charge density λ (x)

surface charge density σ (x.y)

volume charge density ρ (x,y,z)

Geometry may suggest other coordinate systems, R,θ,Φ or R,θ,Z

Lecture 3-6

Continuous Charge Distribution 1 Charged Line

At a point P on axis:

Q

L

/ 2px LEx = k λ (1/r1 – 1/r2 ) ( Eqn 22-2a)

Ex = k λ (1/( XP – L/2 ) - 1/( XP + L/2 )

Ex = k ( Q/L) L [ XP2 – (L/2)2 ] -1

For XP2 >> (L/2)2 Ex = k Q / XP

2

For XP = 0 Ex = 0

Lecture 3-7

Again: Continuous Charge Distribution 1: Charged Line

At a point P on perpendicular axis:

/ 2

2 2/ 2

/ 2

2 2 3/ 2/ 2

2 2

2 2 2 3/ 2

cos

tan( )

sec

( tan )

cos (2sin )

/ 2 2 /

L

y L

L

L

dxE E k

x y

yk dx x y

x y

yk d

y y

k

L and E k y

kd

y y

1 2

x

2 2

1 1

2

1

2 2 2 3/ 2

2

2 12 3/ 2

cos( )

secsin sin tan

(tan 1)

x x

y x x

dx yE k k dx

r x y

k kd x y

y y

Lecture 3-8Physics 241 –Warm-up quiz 2

The rod is uniformly charged with a positive charge density . What is the direction of the electric field at a point P on a line perpendicular to the rod? Note that the line and the rod are in the same plane.

a) to the right b) to the leftc upd) downe) lower right

p

Lecture 3-9General location of P: Charged Line

2 2

1 1

2

1

2 2 2 3/ 2

2 12 2 1/ 2

2 1

sin( )

1 cos cos

( )

cos cos

x x

x x x

x

x

dx xE k k dx

r x y

k kx y y y

k

y

2 2

1 1

2

1

2 2 2 3/ 2

2

2 12 3/ 2

cos( )

secsin sin tan

(tan 1)

x x

y x x

dx yE k k dx

r x y

k kd x y

y y

At a point P off axis:

Lecture 3-10

Continuous Charge Distribution 2: Charged Ring

Use symmetry!

2

3(

)

( )

(

/ )

/

kQ

kQ x x a

a x x a

2

Q

R

At point P on axis of ring:

ds

2 2

2 2 3/ 2

2 2 3/ 2

cos

( )

( )

x

dsE E k

x ax

k dsx a

xk Q

x a

Ex = k Q x ( x2 + a2 )-3

Lecture 3-11Continuous Charge Distribution 3: Charged Disk

2 2 3/ 2

2 2 3/ 20

2 2 3/ 20

2 2 1 2

2

0

/

0

( )

2( )

2( )

2 ( )

12 1 0

1 ( / )

/ 2

x x

R

R

R

R whole plane and

xE E dE k dq

x a

xk ada

x a

akx da

x a

E

kx x a

k xR x

<= Independent of x

Superposition of rings!

At a point P on axis:Use the ring with radius a EX value

2dq a da

0

2

2 2

22

( )

k x R

Ek R kQ

x Rx x

dEx = k dq x ( x2 + a2 ) -3

Integrate rings from 0 to R

Ex = -2πσkx ( 1/( x2 + R2 )1/2 – 1/x )

E = σ/2εo

Lecture 3-12

Continuous Charge Distribution 4: Charged Sheets

( ) ( ) E=const in each region Superposition!

Capacitor geometry

Lecture 3-13

MULTIPLE CHARGE SHEET EXAMPLE

DOCCAM 2

Lecture 3-14

Gauss’s Law: Qualitative Statement

Form any closed surface around charges

Count the number of electric field lines coming through the surface, those outward as positive and inward as negative.

Then the net number of lines is proportional to the net charges enclosed in the surface.

Lecture 3-15 Electric flux

N E A E An ����������������������������

# of field lines N= density of field lines x “area”

where “area” = A2 x cos

General definition of electric flux: E

S

E n dA ��������������

(must specify sense, i.e., which way)

To state Gauss’s Law in a quantitative form, we first need to define Electric Flux.

E An��������������

Sum over surface

Lecture 3-16

Electric Flux through Closed SurfaceE

• The integral is over a CLOSED surface.E n

• Since is a scalar product, the electric flux is a

SCALAR quantity

E nS S

E n dA E dA ��������������

• The integration element is a vector normal to the surface and points OUTWARD from the surface. Out is +, In is -

n

E proportional to # field lines coming through outward

Lecture 3-17

Why are we interested in electric flux?

E is closely related to the charge(s) which cause it.

2

22

0

4

E

rE ndA kq ndA

rkQ Q

rr

��������������

Consider Point charge Q

If we now turn to our previous discussion and use the analogy to the number of field lines, then the flux should be the same even when the surface is deformed. Thus should only depend on Q enclosed.

Lecture 3-18

Gauss’s Law: Quantitative Statement

The net electric flux through any closed surface equals the net charge enclosed by that surface divided by 0.

How do we use this equation??The above equation is TRUE always but it doesn’t look easy to use.

BUT - It is very useful in finding E when the physical situation exhibits a lot of SYMMETRY.

0

enclosedE

QE ndA

Lecture 3-19Physics 241 – 10:30 Quiz 3

The left half of a rod is uniformly charged with a positive charge density , whereas the right half is uniformly charged with a charge density of . What is the direction of the electric field at a point on the perpendicular bisector and above the rod as shown?

a) to the right b) to the leftc upd) downe) E is zero.

Lecture 3-20Physics 241 – 11:30 Quiz 3

The upper half of a ring is uniformly charged with a positive charge density , whereas the lower half is uniformly charged with a charge density of . What is the direction of the electric field at a point on the perpendicular axis and to the left of the ring as shown?

a) to the right b) to the leftc upd) downe) E is zero.

Lecture 3-21

1 22 2

1 2

cosA A

r r

Lecture 3-22

Gauss’ Law - Examples

2

QE k

r

Shell Theorem

• Outside shell: E is as if Q at center

• Inside: E is zero

Lecture 3-23

Proof of the Shell Theorem

• By symmetry, the electric fieldmust only depend on r and isalong a radial line everywhere.

• Apply Gauss’s law to the blue surface , we get

2

0

20

(4 )

1

4

QE r

QE

r

Electric Field Outside

Lecture 3-24

Uniformly charged thin shell: Inside

• By symmetry, the electric fieldmust only depend on r and isalong a radial line everywhere.

• Apply Gauss’s law to the blue surface , we get E = 0.

E = 0 inside

Discontinuity in E

• Equal and opposite contributions from charges on diagonally opposite surface elements.

Lecture 3-25

Electric Field of a Uniformly Charged Sphere

Apply Gauss’s Law directly or use superposition of the shell results

Lecture 3-26Physics 241 –Quiz 2a

Two identical point charges are each placed inside a large cube. One is at the center while the other is close to the surface. Which statement about the net electric flux through the surface of the cube is true?

a) The flux is larger when the charge is at the center.b) The flux is the same (and not zero).c) The flux is larger when the charge is near the

surface.d) Not enough information to tell.e) The flux is zero in both cases.

+Q+Q

Lecture 3-27Physics 241 –Quiz 2b

Two identical point charges are each placed inside a large sphere. One is at the center while the other is close to the surface. Which statement about the net electric flux through the surface of the sphere is true?

a) The flux is larger when the charge is at the center.b) The flux is larger when the charge is near the

surface.c) The magnitude of the flux is the same (and not zero).d) Not enough information to tell.e) The flux is zero in both cases.

+Q+Q

Lecture 3-28Physics 241 –Quiz 2c

Two identical point charges are placed, at the center of a large sphere in one case, and outside an identical sphere in the other case. Which statement about the net electric flux through the surface of the sphere is true?

a) The flux is larger when the charge is inside.b) The flux is larger when the charge is outside.c) The flux is the same (and not zero).d) Not enough information to tell.e) The flux is zero in both cases.

Q>0

Q>0