Lecture 18 - UMD Department of Physics 18 • Electric field due to point charge • Chapter 27 (The Electric Field) due to configuration of (source) charges (today) parallel plate

Lecture 18

• Electric field due to point charge

• Chapter 27 (The Electric Field)

due to configuration of (source) charges (today)

parallel plate capacitor (uniform )

motion of (other) charges in

E

E

E

Electric Field of Point Charge

• Use Coulomb’s law:

• field diagram (sample vectors): at tail of vector; does not stretch...

E = Fon q!

q! =!

14!"0

qr2 , away from q

"

Unit Vector Notation

• mathematical notation for “away from q’’

• : magnitude 1 (no units), purely directional information

• : unit vector pointing from origin to point (“straight outward from point” like )

• applies to q < 0 ( points towards charge)

i, j, k

r

!r

E

• good approx. for small charged object, real wire

• Coulomb’s law

• due to point charge

• limiting cases (check + simpler formula): like point charge very far ( size)

Electric Field Models

E

E

!K = 1

4!"0

"

!

Electric Field of Multiple Charges• Principle of superposition

• Electric field of a dipole (two opposite charges separated by small distance s): no net charge, but does have a net

+Fon q = F1 on q + F2 on q ! Enet =!

i Ei

E

Strategy: picture..., identify P, find Ei,!

i Ei (use symmetry)

(Edipole)y

= (E+)y + (E!)y

=1

4!"0

q

(y ! 1/2s)2+

14!"0

!q

(y + 1/2s)2

=q

4!"0

2ys

(y ! 1/2s)2 (y + 1/2s)2

• For y >> s:

• dipole moment:

• draw vectors or lines

• tangent: direction of

• closer together for larger

• starts on +ve, end -ve

• cannot cross (unique direction at point)

Picturing

p = (qs, from negative to positive)

E

E

(Edipole)y !1

4!"02qsy3

Electric Field of a Dipole

E

Electric Field of a Continuous Charge Distribution• even if charge is discrete, consider it

continuous, describe how it’s distributed (like density, even if atoms

• Strategy (based on of point charge and principle of superposition)

divide Q into point-like charges

find due to

convert sum to integral:

E

!Q

!Q

!Q ! density "dx(x describes shape of !Q)

Electric Field of a Line of Charge• Problem

• Strategy

• Solution...:

Infinite line of charge:Eline = limL!"1

4!"0

|Q|r!

r2+(L/2)2= 1

4!"0

|Q|rL/2 = 1

4!"0

2|#|r

Erod = 14!"0

|Q|r!

r2+(L/2)2

Review of Line of charge

• due to segment i:

• Sum over segments:

• Relate to coordinate:

• Sum to integral:

Ex = Q/L4!"0

! L/2!L/2

rdy(r2+y2)3/2

!Q

N !"; !y ! dy;!N

i !" L/2!L/2

Ex =!N

i (Ei)x = 14!"0

!Ni

r!Q

(r2+y2i )

3/2

Ex = Q/L4!"0

!Ni

r!y

(r2+y2i )

3/2

!Q = !!y = (Q/L) !y

!Ei

"x

=1

4!0

!Q

r2i

cos "i

=1

4!0

r!Q

(r2 + y2i )3/2

Ring and Disk of charge

• On axis of ring:

• Divide disk of surface charge density into rings; sum:

(Ering)z = 14!"0

zQ(z2+R2)3/2

! = QA = Q

!R2

!Ering i

"

z= 1

4!"0z!Qi

(z2+r2i )

3/2(Edisk)z =

!Ni

"Ering i

#

z= z

4!"0

!Ni

!Qi

(z2+r2i )

3/2

Disk of Charge• Relate to coordinate:

• Sum to integral:

• For z >> R, (point charge), as expected

!Q !Qi = !!Ai = !2"ri!r

(Edisk)z = 2!"z4!#0

!Ni

ri!r

(z2+r2i )

3/2

(Edisk)z = !z2"0

! R0

rdr(z2+r2)3/2

N !"; !r ! dr;!N

i !" R0

(Edisk)z = Q4!"0z2

Plane and Sphere of Charge

• Electrodes (charged surfaces used to steer electrons): model as infinite plane if distance to plane, z << distance to edges

• Outside sphere (or shell) of uniform charge