Lecture 18 - UMD Department of Physics 18 • Electric field due to point charge • Chapter 27...
Transcript of Lecture 18 - UMD Department of Physics 18 • Electric field due to point charge • Chapter 27...
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