Example: Find the electric field at point P

€

rE = k

q

r2ˆ r

ˆ r =r r

r=

(1.2m)ˆ i + (−1.6m) ˆ j

2.0m= 0.6ˆ i − 0.8 ˆ j

r E = (9.0 ×109 Nm2 /C2)

−8.0 ×10−9C

(2.0m)2 (0.6ˆ i − 0.8 ˆ j )

r E = (−11N /C)ˆ i + (14N /C) ˆ j

Rework of this example (21.6 in book)

Forces and fields obey the superposition principle:

Field from a group of particles is a vector sum of fields from each particle

iizzzz

iiyyyy

iixxxx

ii

EEEE

EEEE

EEEE

...

...

...

...

21

21

21

21 EEEE

Electric Field Properties

• A small positive test charge is used to determine the electric field at a given point

• The electric field is a vector field that can be symbolized by lines in space called electric field lines

• The electric field is continuous, existing at every point, it just changes in magnitude with distance from the source

Electric Field Equation

• Electric Field

• For a continuous charge distribution

oqF

E

r

r

qkr

r

qE source

esource

o

ˆˆ4

122

rr

dqkEr

r

dqkEd ee ˆˆ 22

For spatially distributed charges – we can sub-divide the object into the small, “point-like” charges and integrate (sum up) the individual fields.Often, we assign charge density for such spacious charged objects

dzdydx(x,y,z)zzyyxx

xxkzyxE

dVk

ddVdq

ddAdq

ddldq

dq

ex

e

i

2/3222

3

))()()((),,(

)(|-|

)()( :d-3in E.g.

densities charge ingcorrespond,,

)3(volumesfor

)2(surfacesfor

)1(linesfor

......

rrrr

rrE

Examples of field calculations: fields of continuous charge distributions

Field of a ring of charge on the symmetry axis

x xE dE

2 2 2 2 2cosx e e

dQ dQ xE k k

r x a x a

2 2 3/ 2( )x

kQxE E

x a

€

A. the +x-direction.

B. the –x-direction.

C. the +y-direction.

D. the –y-direction.

E. none of the above

Positive charge is uniformly distributed around a semicircle. The electric field that this charge produces at the center of curvature P is in

Field of a disk, uniformly charged on the symmetry axisSurface charge density is , radius R

Area dA of a ring of radius r

2 ; ; 2dQ

dA rdr dQ r drdA

2 2 3/ 20

2( )

R

x erdr

E k xx r

2 2

2

x r z

dz rdr

Changing variable of integration

2 20

12x

xE

R x

For the limit of x<<R, we have an electric field of the infinite plane sheet of charge and it is independent of the distance from the plane (assuming that distance x<<linear dimensions of the sheet) 02planeE

€

Ex = πkeσxdz

z3 / 2x 2

x 2 +R 2

∫ = 2πkeσx−1

z

⎛

⎝ ⎜

⎞

⎠ ⎟ x 2x 2 +R 2

Electric Field Line Properties

• Relation between field lines and electric field vectors:a. The direction of the tangent to a field line is the direction of

the electric field E at that pointb. The number of field lines per unit area is proportional to the

magnitude of E: the more field lines the stronger E

• Electric field lines point in direction of force on a positive test charge therefore away from a positive charge and toward a negative charge

• Electric field lines begin on positive charges and end on negative charges or infinity

• No two electric field lines can cross

Electric field lines

E is tangent to the electric field line – no 2 lines can cross (E is unique at each point)

Magnitude of E is proportional to the density of the lines

Remember, electric field lines are NOT trajectories!

When a particle moves on a curved path, the direction of acceleration (and hence of the force) is not collinear with the tangent to the curve

Electric dipole

Many physical systems are described as electric dipoles – hugely important concept

Water is a good solvent for ionic and polar substances specifically because of its dipole properties. The solvent properties of water are vital in biology, because many biochemical reactions take place only within aqueous solutions

Torque on the electric dipole

Electric field is uniform in space Net Force is zero Net Torque is not zero

( )( sin )qE d

p E

(torque is a vector)

Stable and unstable equilibrium

p E

p E

(electric dipole moment from “-” to “+”)

€

rp = q

r d

Charge #1

Charge #2

Charge #3

+q

+q

–q

x

y

A. clockwise.

B. counterclockwise.

C. zero.

D. not enough information given to decide

Three point charges lie at the vertices of an equilateral triangle as shown. Charges #2 and #3 make up an electric dipole.

The net electric torque that Charge #1 exerts on the dipole is

Electric field of a dipole

+-d

A E

1r

2r

E-field on the line connecting two charges

3

2 ep kE

r

when r>>d

E-field on the line perpendicular to the dipole’s axis

2E

+-d

AE

1E

General case – combination of the above two5 3

3( )p r pE r

r r

€

E = keq1

r22

−1

r12

⎛

⎝ ⎜

⎞

⎠ ⎟

€

E = 2E2 sinα

2= E2

d

r

E2 = ke

q

r2

E = ke

qd

r3

r E = −ke

r p

r3

Dipole’s Potential EnergyE-field does work on the dipole – changes its potential energyWork done by the field (remember your mechanics class?)

sindW d pE d

U p E

Dipole aligns itself to minimize its potential energy in the external E-field.Net force is not necessarily zero in the non-uniform electric field – induced polarization and electrostatic forces on the uncharged bodies

Reading assignment: 22.1 – 22.5