Lecture 3-1

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 3-2 Torque on the Dipole

The magnitude of the torque:

sin

sinsin

sin)(sin

pE

qEdFd

xdFFx

The torque points into the screen.

Ep

x

F

F

CM

Fr

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

Continuous Charge Distribution 1: Charged Line

/ 2

2/ 2

/ 2

2/ 2

/ 2

2

2

/

2 2

( )

1

( /

/ /

2)

p

p

p

p

L

x Lp

x L

px L

x L

x L

p

p p

dxE E k

x x

duk u x x

u

k

x

u

L k

k L

x L

Q x

At a point P on axis:

Q

L

/ 2px L

Lecture 3-5

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

Lecture 3-6Physics 241 –Warm-up quiz

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 3-7General 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 3-8

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

Lecture 3-9Continuous 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 result

2 2 3/ 2( )x

xdE

aqk

xd

2dq a da

0

2

2 2

22

( )

k x R

Ek R kQ

x Rx x

Lecture 3-10

Continuous Charge Distribution 4: Charged Sheets

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

Capacitor geometry

Lecture 3-11Physics 241 –Quiz 1A

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 zeo.

Lecture 3-12Physics 241 –Quiz 1b

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 through the center of the ring and to the left of the ring as shown?

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

Lecture 3-13Physics 241 –Quiz 1C

The upper half of a rod 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 bisector and to the right of the rod as shown?

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