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Vector analysis
Submitted by: I.D. 061110185
The problem:
~ = (2xy + z 3 )x̂ + x2 ŷ + 3xz 2 ẑ find:
For the electric field E
1 ~
~
1. ρ = 4π
∇·E
R
~ · d~r, where the path C is
2. Φ = C E
• a straight line from (0, 0, 0) to (1, 1, 1)
• a curve y = x2 , z = 21 (y + x) between these two points
The solution:
∂Ex ∂Ey
∂Ez
1
+
+
= 2y + 6xz ⇒ ρ =
(2y + 6xz)
∂x
∂y
∂z
4π
x̂
ŷ
ẑ ∂
∂
∂
~
~
∇×E = ∂x
∂y
∂z = 0
2xy + z 3 x2 3xz 2 ~ ·E
~ =
∇
(1)
(2)
~ ×E
~ = 0 the field is conservative.
Since ∇
First path
Parametrization:
x=y=z=t
(3)
Line element:
d~l =
3
X
∂xi
i=1
∂t
x̂i dt =
∂x
∂y
∂z
x̂ +
ŷ +
ẑ dt = (x̂ + ŷ + ẑ)dt
∂t
∂t
∂t
~ · d~l = (Ex + Ey + Ez )dt = (2xy + z 3 + x2 + 3xz 2 )dt = (3t2 + 4t3 )dt
E
Z
Z1
~
~
E · dl =
(3t2 + 4t3 )dt = 2
(4)
(5)
(6)
0
Second path
y = x2
(7)
z = 0.5(y + x)
(8)
Parametrization:
1
z = (t + t2 )
2
1
dx = dt, dy = 2tdt, dz = (1 + 2t)dt
2
1
d~l = (x̂ + 2tŷ + (1 + 2t))dt
2
#
3
2
Z
Z 1"
1
1
1
2
2
2
2
~
~
E · dl =
(2tt +
(t + t ) + 2tt + 3t
(t + t )
(1 + 2t) dt = 2
2
2
2
0
x = t,
y = t2 ,
(9)
(10)
(11)
(12)
We can see that the two integrals for two different paths give the same answer, and this is because
the field is conservative.
1
Electric force
Submitted by: I.D. 061110185
The problem:
Given 4 charges arranged at the corners of a square with sides of a length 2a (as on the picture)
find the force acting on the charge Q at some point on the x-axis.
2
sign(x)
ŷ.
Show that for |x| a F~ = 3Qqa
πε0 ·
x4
The solution:
Coulomb law
F~
= kq1 q2
k =
~r
|r|3
(1)
c2
1
ε0 = 8.85 · 10−12
4πε0
N m2
(2)
We use the superposition principle and find the forces acting from each one of the charges. We
define ~ri as a vector pointing from the charge qi to the charge Q. Then
~r1 = (x − a)~i − aĵ
~r2 = (x + a)~i − aĵ
(3)
~r3 = (x + a)~i + aĵ
~r4 = (x − a)~i + aĵ
(5)
(4)
(6)
where the the charges qi are counted from the top right one in the counterclockwise direction.
Therefore, the force is
F~
= F~(x,0,0) = F~1 + F~2 + F~3 + F~4 =
(7)
!
(x + a)~i − aĵ
(x + a)~i + aĵ
(x − a)~i + aĵ
(x − a)~i − aĵ
+
−
+
3/2
3/2
3/2
2
2
2
2
2
2
((x − a) + a )
((x + a) + a )
((x + a) + a )
((x − a)2 + a2 )3/2
1
1
ĵ
(8)
= 2akqQ −
+
((x + a)2 + a2 )3/2 ((x − a)2 + a2 )3/2
= kqQ −
For |x| a
1
((x ± a)2 + a2 )3/2
=
'
1
1
=
2a
3
(x2 ± 2ax + 2a2 )3/2
|x| (1 ± x +
3a
1
1
'
1∓
3/2
|x|3
x
|x|3 (1 ± 2a
x)
1
2a2 3/2
)
x2
(9)
(10)
Then
3Qqa2 sign(x)
·
ŷ
F~ =
πε0
x4
(11)
which is the the force acting by a quadrupole.
2
Electric force
Submitted by: I.D. 061110185
The problem:
An infinite wire charged uniformly with the charge density λ is bent as shown on the picture. What
is the force acting on a charge q which is put into the center of the half-circle?
The solution:
We divide the problems to 2 parts: (1) finding the force acting on the straight line and (2) finding
the force acting on the half circle.
The infinidecimal charge is dq = λdl.
For the straight line:
~r = xî + Rĵ
kQdq
kQλxdx
dFx = −
3 x = −
3
2
2
(x + R ) 2
(x2 + R2 ) 2
kλRdx
dFy =
3
(x2 + R2 ) 2
Z∞
kQλxdx
2λQk
Fx = −2
3 = −
R
(x2 + R2 ) 2
(1)
(2)
(3)
(4)
0
The y-components of the force are canceled since there are 2 wires and the factor of 2 in the last
equation also comes from this fact.
For the half circle:
dF~
=
Fx =
kQλR2 dθ
sin θ
R3
2λQk
R
(5)
(6)
Therefore,
X
F~ = 0
(7)
1
Gauss’ law - plane symmetry
Submitted by: I.D. 039023262
The problem:
Find the electric field for:
1. infinite uniformly charged plane (charge density σ)
2. infinite uniformly charged layer (charge density ρ and width h)
3. two adjacent infinite layers of width h charged ρ+ and ρ− uniformly
4. two infinite planes perpendicular to each other, both charged σ uniformly.
The solution:
H
R
~ · d~s = 4πk ρdv, and knowing that for plane symmetry E
~ = Ez ẑ
1. According to Gauss’ law E
we construct a Gauss shell enclosing a piece of the plane of area S:
I
Z
~
E · d~s = 4πk ρdv
(1)
2Ez · S = 4πkσS
(2)
Ez = 2πkσ · sign(z)ẑ
(3)
2. For infinite layer of the width h (0 < z < h), if z > h or z < 0 the solution is identical to the one
for the infinite plane. In the case 0 < z < h we construct a shell of area S symmetrical relatively
to the middle of the layer - its its bottom plate is at z, the top plate is at h − z and the height,
therefore, is h − 2z.
2E · S = 4πk(h − 2z)Sρ
(4)
E = 2πkρ(h − 2z)
(5)


2πkρh,
z>h



2πkρ(2z − h), h/2 < z < h
=

2πkρ(h − 2z), 0 < z < h/2



−2πkρh,
z<0
(6)
Then
~z
E
3. Two infinite layers of thickness h charged ρ+ and ρ− uniformly. We choose the ρ+ charged layer
to be at 0 < z < h and the ρ− charged layer to be at −h < z < 0. Then using the result of for the
one layer and the superposition principle we obtain:


2πk(ρ+ + ρ− )h,
z>h




2πkρ+ (2z − h) + 2πkρ− h,
h/2 < z < h

~
(7)
Ez =
2πkρ+ (h − 2z) + 2πkρ− h,
0 < z < h/2



−2πkρ+ h + 2πkρ− (h + 2z), −h < z < 0




−2πk(ρ+ + ρ− )h,
z < −h
4. Two infinite plains perpendicular to each other, both charged σ uniformly.
By using the principle of superposition, we simply calculate the field intensity of each plane, and
~ = 2πkσ(sign(z)ẑ + sign(y)ŷ) (assuming that the plains in
combine them. For tow plane we get E
question are the XY and XZ)
1
Gauss’s law - plane symmetry
Submitted by: I.D. 039023262
The problem:
Find the electric field along the z-axis of an infinite uniformely charged plane at the x − y plane
(charge density σ) with a hole at the origin of a radius r0 .
The solution:
Using the principle of superposition we shall calculate the electric field induced by the missing hole
as being of a disk charged with −σ, then combine it with an infinite plane:
!
z
Ezdisk = −2πkσ sign(z) − p
(1)
z 2 + r02
Ezplane = 2πkσ sign(z)
z
Ez = 2πkσ p
2
z + r02
(2)
(3)
1