Download Electric field

Electric field
Submitted by: I.D. 310159025
The problem:
Given a thin half-sphere, uniformly charged with density σ find the electric field at the center of
the sphere.
The solution:
Because the symmetry of the half-sphere, the only field contributing component is the ẑ component (coinciding with the symmetry axis of the half sphere). Every point-area on the half-sphere
contributes an electric field dE:
ds = R2 sin θdθdϕ
(1)
dq = σds
kdq cos θ
kR2 σ sin θ cos θdθdϕ
dE =
=
= kσ sin θ cos θdθdϕ
R2
R2
(2)
(3)
where θ is the angle between the radius-vector and the z axis.
Integration of the both sides gives:
π
Z2π
Z2
E = kσ
sin θ cos θdθ
0
dϕ = πkσ
(4)
0
~ = πkσẑ
E
(5)
1
Electric field
Submitted by: I.D. 066370016
The problem:
A thin wire of a length L is charged with an uniform electrical charge q. What is the electric field
in points P,R and P’ ? The wire is on the X-axis (one edge is located at the origin and the other
at point L) , point P is located at (L1 , h) , point R is at (L + L3 , 0) and point P’ is located at
(L + h tan θ1 , h)
The solution:
1. It is given that the thin wire is charged with an uniform electrical charge q so
λ=
q
L
(1)
The vector to the point P is
~r = (L1 , h)
(2)
The vector to the element charge dq is
~r0 = (x0 , 0)
(3)
Subtracting ~r0 from ~r and calculating the norm
~r − ~r0 = (L1 − x0 , h)
p
~r − ~r0 =
(L1 − x0 )2 + h2
(4)
(5)
The element charge is
dq = λdx
(6)
By using the electric field equation
kdq
L1 − x0
~
dE =
3
h
((L1 − x0 )2 + h2 ) 2
Z
Z L
0
kdq
E
L
−
x
x
1
~ =
~ =
E
= dE
3
Ey
h
0 ((L1 − x0 )2 + h2 ) 2
So to find Ex
Z L
Ex =
0
kdq(L1 − x0 )
((L1 − x0 )2 + h2 )
3
2
1
= kλ( p
h2 +
L22
−p
1
h2 + L21
)
(7)
(8)
(9)
Finding Ey
Z
Ey = kλh
0
L
dx0
((L1 − x0 )2 + h2 )
3
2
=
L2
L1
kλ
(q
+q
)
h
h2 + L22
h2 + L21
(10)
2. It is given that the thin wire is charged with an uniform electrical charge q so
λ=
q
L
(11)
1
The vector to the point R is
~r = (L3 + L, 0)
(12)
The vector to the element charge dq is
~r0 = (x0 , 0)
(13)
Subtracting ~r0 from ~r and calculating the norm
~r − ~r0 = (L3 + L − x0 , 0)
p
~r − ~r0 =
(L3 + L − x0 )2 + 02 = {0 ≤ x ≤ L} = L − x0 + L3
(14)
(15)
The element charge is
dq = λdx
(16)
By using the electric field equation
kdq
L − x0 + L3
~
dE =
0
(L − x0 + L3 )3
Z
Z L
kdq
Ex
L − x0 + L3
~ =
~ =
E
= dE
3
0
Ey
0
0 (L − x + L3 )
(17)
(18)
Therefore Ey = 0 and Ex is:
Z L
dx(L − x0 + L3 )
Ex = kλ
(L − x0 + L3 )3
0
q
= k
L3 (L + L3 )
(19)
3. It is given that the thin wire is charged with an uniform electrical charge q so
q
λ=
L
The vector to the point P is
(21)
(20)
~r = (L + h tan θ1 , h)
(22)
The vector to the element charge dq is
~r0 = (x0 , 0)
(23)
Subtracting ~r0 from ~r and calculating the norm
~r − ~r0 = (L + h tan θ1 − x0 , h)
p
~r − ~r0 =
(L + h tan θ1 − x0 )2 + h2
(24)
(25)
The element charge is
dq = λdx
(26)
By using the electric field equation
~ =
dE
kdq
3
L + h tan θ1 − x0
h
((L + h tan θ1 − x0 )2 + h2 ) 2
Z
Ex
~
~
E =
= dE
Ey
Z L
kdq
L + h tan θ1 − x0
~
E =
3
h
0 ((L + h tan θ1 − x0 )2 + h2 ) 2
2
(27)
(28)
(29)
So to find Ex
Z
kλdx(L + h tan θ1 − x0 )
L
Ex =
3
((L + h tan θ1 − x0 )2 + h2 ) 2
1
1
= kλ( p
−p
)
h2 + (h tan θ1 )2
h2 + (L + h tan θ1 )2
(30)
0
(31)
Finding Ey
L
Z
dx
Ey = kλh
3
((L + h tan θ1 − x0 )2 + h2 ) 2
kλ
L + h tan θ1
L − h tan θ1
+p
)
(p
h
((L − h tan θ1 )2 + h2 )
((L + h tan θ1 )2 + h2 )
(32)
0
=
Appendix: calculating the integrals
Part One
Z L
Z L
kdq(L1 − x0 )
dx(L1 − x0 )
Ex =
=
kλ
3
3
0 ((L1 − x0 )2 + h2 ) 2
0 ((L1 − x0 )2 + h2 ) 2

1 

t = ((L1 − x0 )2 + h2 )− 2 






(L1 −x0 )


Z √ 1
√ 21 2
dx

 dt =
3
h +L2
h2 +L2
((L1 −x0 )2 +h2 ) 2
2 dt = kλ[t]
=
= kλ
1
1
√
√
x
=
0
=⇒
t
=
1


√2 2

h2 +L2
h2 +L21 


1
h
+L


1


 x = L =⇒ t = √ 1 2 
2
(33)
(34)
(35)
h +L2
1
Ex = kλ( p
h2
Z
+
L22
1
−p
h2
+ L21
)
(36)
dx0
L
Ey = kλh
3
((L1 − x0 )2 + h2 ) 2


0 = h tan t + L
x
1




Z arctan L2


hdt
h
hdt
1
dx0 = cos
2t
=
= kλh
−L1
3
0
−L1 cos2 t
x
=
0
=⇒
t
=
arctan


2
2 tan t + h2 ) 2
arctan h


h
(h

 0
x = L =⇒ t = arctan Lh2
Z arctan L2
h
hdt
1
= kλh
3
−L1 h3 cos2 t
arctan h
(tan2 t + 1) 2
Z arctan L2
h
dt
= kλh
cos3 t
2
−L1 h cos2 t
arctan
(37)
0
(38)
(39)
(40)
h
=
L
kλ
kλ
L2
L1
arctan h2
[sin t]
(q
+q
−L1 =
)
arctan h
h
h
h2 + L22
h2 + L21
Part Two
3
(41)
L
Z
Ex = kλ
0
Z
dx(L − x0 + L3 )
(L − x0 + L3 )3
(42)
L
1
dx
L
2 = kλ[ (L − x0 + L ) ]0
0
3
0 (L − x + L3 )
1
1
λ
= kλ(
−
) = Lk
(L − L + L3 ) (L − 0 + L3 )
L3 (L + L3 )
kqL
q
=
=k
LL3 (L + L3 )
L3 (L + L3 )
= kλ
(43)
(44)
(45)
Part Three
Z
kλdx(L + h tan θ1 − x0 )
L
Ex =
((L + h tan θ1 − x0 )2 + h2 )
0
3
2
=

1

t = ((L + h tan θ1 − x0 )2 + h2 )− 2



(L+h tan θ1 −x0 )

dt =

3 dx
0 2
2







1
 x = 0 =⇒ t = √ 2

h +(L+h tan θ1 )2



1
 x = L =⇒ t = √






((L+h tan θ1 −x ) +h ) 2
h2 +(h tan θ1 )2
Z √
= kλ
√
1
h2 +(h tan θ1 )2
1
h2 +(L+h tan θ1 )2
√
dt = λ[t] √
1
Ex = kλ( p
h2 + (h tan θ1 )2
Z
Ey = kλh
0
Z
L
arctan
=
=
L−h tan θ1
h
−(L+h tan θ1 )
h
arctan
= kλh
= kλh
h2 +(h tan θ1 )2
1
h2 +(L+h tan θ1 )2
1
h2 + (L + h tan θ1 )2
(47)
)




arctan
arctan
Z
1
x = h tan t + L + h tan θ1
hdt
dx = cos
dx
2t
−(L+h tan θ1 )
3 =

x
=
0
=⇒
t
=
arctan
0
2
2

((L + h tan θ1 − x ) + h ) 2
h

x = L =⇒ t = arctan L−h htan θ1
= kλh
Z
−p
L−h tan θ1
h
−(L+h tan θ1 )
h
arctan
(46)
(48)




(49)



hdt
1
2
cos t (h2 tan2 t + h2 ) 32
(50)
hdt
1
h3 cos2 t (tan2 t + 1) 32
(51)
L−h tan θ1
h
dt
h2 cos2 t
−(L+h tan θ1 )
arctan
h
L−h tan θ1
arctan
h
−(L+h tan θ1 )
arctan
h
cos3 t
kλ
[sin t]
h
kλ
L − h tan θ1
L + h tan θ1
(p
+p
)
2
2
h
((L − h tan θ1 ) + h )
((L + h tan θ1 )2 + h2 )
4
(52)
(53)
(54)
The electric field
Submitted by: I.D. 040460479
The problem:
An isolated electrical wire is charged uniformly with charge q and bent into a circular shape (radius
R) with a small hole b R (where b is the arc length).
What is the electrical field in the middle of the circle?
The solution:
The simple solution is to use superposition.
The electric field in the middle of a complete ring is, of course, zero.
Now we’ll sum up the field of a complete ring with the field of a very small wire of the same size
and shape as the hole but with a negative charge.
~ = kλb
x̂
Because b R the wire can be taken as a negative point charge and, therefore, the field is E
R2
q
(when we take the hole to be on the X axis and λ = 2πR ).
It is also possible to calculate the field directly. Taking ~r = (0, 0, 0) and ~r0 = (R cos θ, R sin θ, 0) we
have
kdq
(−R cos θ, −R sin θ, 0)
R3
dq = λRdθ
Z −α
kλ
kλRdθ0
~
(−R cos θ0 , −R sin θ0 , 0) =
(2 sin α, 0, 0)
E =
3
R
R
α
~ =
dE
(1)
(2)
(3)
where ±α are the angles of the edges of the bent wire (or, the upper and lower limits of the hole).
Since b R we can approximate tan α ' sin α = b/2
R .
Substituting into the expression for the field we obtain
~ = kλb x̂
E
R2
(4)
1
Electric Field
The problem:
A thin ring is consists of two semi-rings, each of which is charged homogeneously but with the
opposite charges, q and −q. Find the electric field on the ring axis.
The solution:
Let us choose the coordinates such that the z axis consides with the ring axis, and the ring lies in
the x − y plane. Let ϕ be the angle in the x − y plane measured from the positive direction of the
x axis.
The charge desity is
(
q
,
0≤ϕ≤π
λ = πR q
(1)
− πR , π ≤ ϕ ≤ 2π
where R is the radius of the ring.
Then
Z π
Z 2π
kqR
0
0
0
0
Ex = −
cos ϕ dϕ −
cos ϕ dϕ = 0,
π(R2 + z 2 )3/2
0
π
Z π
Z 2π
kqR
4kqR
0
0
0
0
Ey = −
sin ϕ dϕ −
sin ϕ dϕ = −
,
3/2
2
2
2
π(R + z )
π(R + z 2 )3/2
0
π
Z π
Z 2π
kqz
0
0
dϕ −
dϕ = 0
Ez =
π(R2 + z 2 )3/2
0
π
1
(2)
(3)
(4)