electromagnetics lecture 5v2€¦ · dr. essam sourour. 2 example: line charge on a short wire •...
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EE2030, Electromagnetics ILecture 5
Electrostatic Fields 2Dr. Essam Sourour
2
Example: Line charge on a short wire
• Consider a line charge with uniform charge density L
• The charge extends from a to b on the z axis• We want to calculate the total charge Q• We want to calculate at point P (0, c, h)
• The differential charge size is dz=dl• The vector from dz to point P is
L Lb
L LL a
dQ dl dz
Q dQ dz L
22
ˆ ˆ0, , 0,0, y zR c h z ca h z a
R c h z
dE R
R
22
ˆ ˆˆ y z
R
ca h z aRaR c h z
3
Line charge on a short wire, cont.
• Now we use the equation for
• We have to solve the above integration
2 3
3 222
3 2 3 22 22 2
ˆ4 4
ˆ ˆ4
ˆ ˆ4
L LR
o oL L
by zL
o a
b by zL
o a a
dl RE a dlR R
ca h z adz
c h z
ca h z adz dz
c h z c h z
dE R
R
4
Line charge on a short wire, cont.
• By changing the variables: put (h‐z)=v
• Now we use the famous integrations:
3 2 3 22 2 2 2
3 2 3 22 2 2 2
We replace:
1ˆ ˆ4
ˆ ˆ4 4
h b h bL
y zo h a h a
h a h aL L
y zo oh b h b
v h zdv dz
vE ca dv a dvc v c v
c dv v dva ac v c v
3 2 3 22 2 2 2 22 2 2 2
1anddx x x dxa a x a xa x a x
5
Line charge on a short wire, cont.
• Going back to the integration for :
1 2 1 2ˆ ˆsin sin cos cos4
Ly z
o
E a ac
3 2 3 22 2 2 2
2 2 2 2 2
2 2 2 22 2 2 2
ˆ ˆ4 4
1ˆ ˆ4 4
ˆ ˆ4 4
4
h a h aL L
y zo oh b h b
h a h a
L Ly z
o oh b h b
L Ly z
o o
L
o
c dv v dvE a ac v c v
c va ac c v c v
h a h b c ca ac cc h a c h b c h a c h b
c
1 2 1 2ˆ ˆsin sin cos cos4
Ly z
o
a ac
dE R
R
6
Line charge on a short wire, cont.
• A special case is infinite line where point a at (0, 0, ‐) and point b at (0, 0, ).
• This makes 1 = /2 and 2 = ‐/2
ˆ fo infinite linr e2
Ly
o
E ac
dE R
R
1 2 1 2ˆ ˆsin sin cos cos4
ˆ2
Ly z
o
Ly
o
E a ac
ac
sin 2 1 , sin 2 1, cos 2 cos 2 0
7
Example: Surface charge on infinite surface
• Consider an infinite sheet in the x‐y plane with uniform charge density function S C/m2
• The differential charge dQ=S dS
• We need to find R and
• Due to symmetry the component in direction cancels
4
2 2
ˆ ˆ ˆ
ˆ ˆz x y
z
R ha xa ya
a ha
R h
3 22 2
ˆ ˆ
4zS
o S
a ha d dE
h
8
Surface charge on infinite surface, cont.
• Performing the integration:
3 22 2
3 22 2
2
3 22 20 0
3 22 20
2 20
ˆ ˆ
4
ˆNow weignore thedirection ofˆ
4
ˆ4
ˆ2
ˆ2
ˆ2
zS
o S
S z
o S
S z
o
S z
o
S z
o
Sz
o
a ha d dE
h
ah a d dE
h
h a d d
h
h a d
h
h ah
a
Sn
o
ˆ2
In general for anyinfinitesurface:ρ ˆE= a2ε
Sz
o
E a
3 2 2 22 2
1x dxa xa x
9
Surface charge on infinite surface, cont.
• Note that, for infinite sheet of charge, does not depend on the height h
• If we have a capacitor with two plates with equal and opposite charges (capacitor)
S Sn n
o o
Sn
o
ρ ρˆ ˆE= a a2ε 2ερ aε
10
Volume charge on a sphere
• Consider a sphere with radius a centered at the origin• Sphere is charged uniformly with volume charge density v
• The total charge is
• This problem is much easier to solve with Gauss’s law. So we will solve it with Gauss law later
11
Electric flux density
• From the electric field intensity we define the electric flux density
• (simply multiply by the permittivity )
• The electric flux density gives the density of the electric flux (how many lines of electric flux per m2)
• The total flux passing from a certain surface S is calculated from the integration:
S
D dS
E and D S
12
Example
Determine at point (4, 0, 3) if there is a point charge of ‐5mC located at (4, 0, 0) and an infinite line charge of line charge density 3mC/m along the y axis.Solution:We have two sources of electric field intensity and electric field density
due to point charge and due to line charge
2 3
3
2
ˆ4 4
ˆ4,0,3 4,0,0 3
3, 275 3 ˆ ˆ0.139
4 27
Q R
z
Q z z
Q QRD aR R
R a
R R
D a a mC m
Use theinifinite lineexamplebut with cylinderical coordinates on the y axis:
ˆ2
5 , 36.8
We can divide into two components, in x and z directions:
ˆ ˆcos sin2 2
0.
LL
o
L
L LL x z
L
D a
D
D a a
D
2
ˆ ˆ24 0.18
ˆ ˆ, 0.24 0.041x z
Q L x z
a a
Hence D D D a a mC m