Electromagnetic Theory
2018-2019
Prof. dr. Ali Hadi Hassan Al-Batat 1-7
Problems of Chapter Two
Problem.1: A circular ring of radius a carries a uniform charge λ(C/m) is placed
on the x-y plane with axis the same as the z-axis.
1- Find the electrostatic field at (0,0,h).
2- What values of h that gives the maximum value of the electric field?
3- If the total charge on the ring is Q, find the electric field as 𝑎 → 0.
Solution: 1) consider the system as shown in the fig.(1) below.
𝑑𝑙 = 𝑎𝑑𝜃
(cylindrical coordinates)
𝑅⃗ = (0 − 𝑎)𝑎̂𝑅 + (𝜃 − 𝜃)𝑎̂𝜃 + (ℎ − 0)𝑎̂𝑧
𝑅⃗ = ℎ𝑎̂𝑧 − 𝑎𝑎̂𝑅
The scalar value of 𝑅⃗ = ℎ𝑎̂𝑧 − 𝑎𝑎̂𝑅 is |𝑅⃗| = √ℎ2 + 𝑎2 , and 𝑎̂𝑅 =
⃗
𝑅
⃗|
|𝑅
For line charge, the electric field is written as
𝐸⃗ (0,0, ℎ) = ∫
𝜆𝑑𝑙
4𝜋𝜀0 𝑅
𝑎̂ = ∫
2 𝑅
⃗
𝜆𝑑𝑙 𝑅
4𝜋𝜀0 𝑅 3
=∫
𝜆 𝑎 𝑑𝜃 (ℎ𝑎̂𝑧 −𝑎𝑎̂𝑅 )
4𝜋𝜀0 (ℎ2 +𝑎2 )3/2
2
𝜆𝑎ℎ
𝑑𝜃
𝜆𝑎
𝑑𝜃
𝐸⃗ (0,0, ℎ) =
𝑎
̂
−
𝑎̂ , this means that 𝐸⃗ =
∫
∫
𝑧
2
2
3/2
2
(ℎ +𝑎 )
(ℎ +𝑎2 )3/2 𝑅
4𝜋𝜀
4𝜋𝜀
0
0
𝐸𝑧 𝑎̂𝑧 + 𝐸𝑅 𝑎̂𝑅
Since there is symmetry, the radial components of the electric field cancel each
other, thus, the electric field is directed along z-axis only.
𝜆𝑎ℎ
𝐸⃗ (0,0, ℎ) =
∫
4𝜋𝜀
0
𝜆𝑎ℎ
𝐸⃗ (0,0, ℎ) =
𝑑𝜃
3
(ℎ2 +𝑎2 )2
2𝜋
4𝜋𝜀0 (ℎ2 2 )32
+𝑎
3
2)
𝑑|𝐸⃗ |
𝑑ℎ
=
𝜆𝑎
2𝜀0
[
𝑎̂𝑧 =
𝜆𝑎ℎ
𝜆𝑎ℎ
3
2𝜀0 (ℎ2 +𝑎2 )2
1
3
2
(ℎ2 +𝑎2 )3
𝑑|𝐸⃗ |
𝑑ℎ
1
4𝜋𝜀0 (ℎ2 2 )32
+𝑎
(ℎ2 +𝑎2 )2 −ℎ( )(ℎ2 +𝑎2 )2 2ℎ
For maximum 𝐸⃗ ,
3
𝑎̂𝑧 =
]
= 0, which implies
1
(ℎ2 + 𝑎2 )2 − 3ℎ2 (ℎ2 + 𝑎2 )2 = 0
1
𝑎̂𝑧
2𝜋
∫0 𝑑𝜃
Electromagnetic Theory
2018-2019
Prof. dr. Ali Hadi Hassan Al-Batat 1-7
1
(ℎ2 + 𝑎2 )2 [ℎ2 + 𝑎2 − 3ℎ2 ] = 0 , thus, ℎ2 + 𝑎2 − 3ℎ2 = 0 , 𝑎2 − 2ℎ2 = 0 ,
ℎ = ∓√
𝑎2
2
=∓
𝑎
√2
,ℎ=∓
𝑎
√2
3) Since the charge is uniformly distributed, the line charge density is;
𝜆=
𝐸⃗ =
𝐸⃗ =
𝑄
2𝜋𝑎
, 2𝜋𝑎𝜆 = 𝑄, 𝜆𝑎 =
𝜆𝑎ℎ
3
2𝜀0 (ℎ2 +𝑎2 )2
𝑄ℎ
4𝜋𝜀0
ℎ3
𝑎̂𝑧 =
𝑎̂𝑧 , 𝐸⃗ =
𝑄
2𝜋
, so that
𝑄ℎ
3
2𝜀0 2𝜋(ℎ2 +𝑎2 )2
𝑄
4𝜋𝜀0 ℎ2
𝑎̂𝑧 , 𝐸⃗ =
𝑎̂𝑧 or in general
𝑄ℎ
3
4𝜋𝜀0 (ℎ2 +𝑎2 )2
𝑄
𝐸⃗ =
𝑎̂
4𝜋𝜀 𝑟 2 𝑟
0
𝑎̂𝑧 as 𝑎 → 0
which is the same as
that of a point charge.
Fig.(1): A circular ring of radius a carrying charge density λ.
2
Electromagnetic Theory
2018-2019
Prof. dr. Ali Hadi Hassan Al-Batat 1-7
Problem.2(H.W): A circular ring placed in the x-y plane, where its inner radius is
1m and the exterior radius is 2m. The ring carrying a surface charge density
𝜎(𝑟) =
100 𝜇𝐶
(
𝑟
𝑚2
). Find the electrostatic field on the axis of the ring at 10m from the
center.
Problem.3: A circular disk of radius a is uniformly charged with surface charge
density 𝜎(C/m2). If the disk lies on the z=0 plane (i.e., x-y plane) with its axis along
the z-axis. Show that the electric field at the point (0, 0, h) is given by:
𝐸⃗ =
𝜎
2𝜀0
[1 −
ℎ
√ℎ2 +𝑎2
] 𝑎̂𝑧
Solution: The electric field may be determined using two methods:
Method 1: In this method the electric field will be determined directly from the
conventional definition of the electrostatic field due to surface charge distribution,
see fig.(2).
The position vector of the field point is
𝑅⃗1 = 0𝑎̂𝑅 + 𝜃𝑎̂𝜃 + ℎ𝑎̂𝑧
The position vector of the source point is
𝑅⃗2 = 𝑟𝑎̂𝑅 + 𝜃𝑎̂𝜃 + 0𝑎̂𝑧
Thus, 𝑅⃗ = 𝑅⃗1 − 𝑅⃗2 = (0 − 𝑟)𝑎̂𝑅 + (𝜃 − 𝜃)𝑎̂𝜃 + (ℎ − 0)𝑎̂𝑧
𝑅⃗ = ℎ𝑎̂𝑧 − 𝑟𝑎̂𝑟 and |𝑅⃗| = √ℎ2 + 𝑟 2 , the surface element is 𝑑𝑠 = 𝑟𝑑𝑟𝑑𝜃 and
𝑎̂𝑅 =
𝑑𝐸⃗ =
𝑑𝐸⃗ =
𝐸⃗ =
⃗
𝑅
⃗|
|𝑅
𝑑𝑄
4𝜋𝜀0 𝑅
𝑎̂ =
2 𝑅
⃗
𝜎𝑑𝑠 𝑅
⃗|
4𝜋𝜀0 𝑅 2 |𝑅
1
𝜎𝑑𝑠
4𝜋𝜀0
(ℎ2 +𝑟 2 )3/2
1
4𝜋𝜀0
∫𝑆
𝜎𝑑𝑠
3
(ℎ2 +𝑟 2 )2
1
=
1
𝜎𝑑𝑠
4𝜋𝜀0 𝑅 3
𝑅⃗
(ℎ𝑎̂𝑧 − 𝑟𝑎̂𝑟 )
ℎ𝑎̂𝑧 −
1
4𝜋𝜀0
∫𝑆
𝜎𝑑𝑠
3
(ℎ2 +𝑟 2 )2
3
𝑟𝑎̂𝑟 = 𝐸𝑟 + 𝐸𝑧
Electromagnetic Theory
2018-2019
Prof. dr. Ali Hadi Hassan Al-Batat 1-7
From the symmetry of the problem the radial components of the field are cancel each
other, thus, the only remaining component is the z-component:
𝐸⃗ =
𝜎ℎ
4𝜋𝜀0
2𝜋
𝑎 𝑟 𝑑𝑟 𝑑𝜃
̂𝑧 =
3𝑎
∫0 ∫0
(ℎ2 +𝑟 2 )2
1
=
𝐸⃗ =
−
𝜎ℎ (ℎ2 +𝑟 2 ) 2
4𝜀0
𝜎
2𝜀0
−1/2
[1 −
2𝜋𝜎ℎ
4𝜋𝜀0
𝑎
∫0
𝑎
𝜎ℎ
⌋ 𝑎̂𝑧 = − 2𝜀
0
ℎ
√ℎ2 +𝑎2
0
1
√ℎ2 +𝑟
𝑟 𝑑𝑟
̂𝑧 =
3𝑎
(ℎ2 +𝑟 2 )2
2𝜋𝜎ℎ
8𝜋𝜀0
𝑎
𝜎ℎ
0
2𝜀0 √ℎ2 +𝑎
| 𝑎̂𝑧 = −
2
[
3
𝑎
−
∫0 (ℎ2 + 𝑟 2 ) 2 2𝑟 𝑑𝑟𝑎̂𝑧
1
1
− ] 𝑎̂𝑧
2
ℎ
] 𝑎̂𝑧
Fig.(2): A uniformly charged disk.
Method 2: To determine the electric field, we firstly evaluate the electrostatic
potential and then determine the electrostatic field using the formula 𝐸⃗ = −∇𝑉.
The potential of the surface charge is 𝑉 =
4
1
4𝜋𝜀0
∫𝑆
𝜎
𝑅
𝑑𝑠
Electromagnetic Theory
2018-2019
Prof. dr. Ali Hadi Hassan Al-Batat 1-7
2𝜋 𝑎
𝜎
𝑟 𝑑𝑟 𝑑𝜃
𝑉=
∫∫ 2
(𝑧 + 𝑟 2 )1/2
4𝜋𝜀0
0
0
𝑎
𝑎
2𝜋𝜎
𝑟 𝑑𝑟
𝜎
=
∫ 2
=
∫ 2(𝑧 2 + 𝑟 2 )−1/2 𝑟𝑑𝑟
4𝜋𝜀0 (𝑧 + 𝑟 2 )1/2 4𝜀0
0
0
2 )1/2 𝑎
(𝑧 2
𝜎
+𝑟
𝜎
𝑎
(𝑧 2 + 𝑟 2 )1/2 |
| =
0
4𝜀0
1/2
2𝜀0
0
𝜎
=
[(𝑧 2 + 𝑎2 )1/2 − 𝑧]
2𝜀0
=
⃗ 𝑉 = − 𝑑𝑉 𝑎̂𝑧
𝐸⃗ = −∇
𝑑𝑧
𝐸⃗ = −𝑎̂𝑧
=
𝜎
2𝜀0
1
1
𝑑 𝜎
𝜎 1 2
[(𝑧 2 + 𝑎2 )2 − 𝑧] = −
[ (𝑧 + 𝑎2 )−2 2𝑧 − 1] 𝑎̂𝑧
𝑑𝑧 2𝜀0
2𝜀0 2
[−
𝑧
√𝑧 2 +𝑎2
When z=h, then 𝐸⃗ =
𝜎
2𝜀0
+ 1] 𝑎̂𝑧 =
[1 −
ℎ
√ℎ2 +𝑎2
𝜎
2𝜀0
[1 −
𝑧
√𝑧 2 +𝑎2
] 𝑎̂𝑧
] 𝑎̂𝑧
Problem.4: Consider an infinite sheet of charge in the x-y plane with uniform
surface charge density σ. Find an expression for the electrostatic field above the
sheet.
Solution: The charge associated with an element area dS is 𝑑𝑄 = 𝜎𝑑𝑆, hence the
total charge is 𝑄 = ∫ 𝜎𝑑𝑆
The contribution to the electric field 𝐸⃗ at point P(0,0,h) by the element surface 1
shown in fig.(3) is
𝑑𝐸⃗ =
𝑑𝑄
4𝜋𝜀0 𝑅 2
𝑎̂𝑅
The position vector of the field point is
𝑅⃗1 = 0𝑎̂𝑅 + 0𝑎̂𝜃 + ℎ𝑎̂𝑧
The position vector of the source point is
𝑅⃗2 = 𝑟𝑎̂𝑅 + 0𝑎̂𝜃 + 0𝑎̂𝑧
5
Electromagnetic Theory
2018-2019
Prof. dr. Ali Hadi Hassan Al-Batat 1-7
Thus, 𝑅⃗ = 𝑅⃗1 − 𝑅⃗2 = (0 − 𝑟)𝑎̂𝑅 + (0 − 0)𝑎̂𝜃 + (ℎ − 0)𝑎̂𝑧
𝑅⃗ = ℎ𝑎̂𝑧 − 𝑟𝑎̂𝑟 and |𝑅⃗| = √ℎ2 + 𝑟 2 , the surface element is 𝑑𝑠 = 𝑟𝑑𝑟𝑑𝜃 and
⃗
𝑅
⃗|
|𝑅
𝑎̂𝑅 =
𝑑𝑄
𝑑𝐸⃗ =
𝐸⃗ =
, 𝑑𝑄 = 𝜎𝑑𝑆 = 𝜎 𝑟 𝑑𝑟 𝑑𝜃
4𝜋𝜀0
1
4𝜋𝜀0
𝑅2
𝑎̂𝑅 =
1
𝜎 𝑟 𝑑𝑟 𝑑𝜃
4𝜋𝜀0
𝑅3
𝜎 𝑟 𝑑𝑟 𝑑𝜃
∫𝑆
3
(ℎ2 +𝑟 2 )2
ℎ𝑎̂𝑧 −
𝑅⃗ =
1
4𝜋𝜀0
∫𝑆
1
𝜎 𝑟 𝑑𝑟 𝑑𝜃
4𝜋𝜀0 (ℎ2 +𝑟 2 )3/2
𝜎 𝑟 𝑑𝑟 𝑑𝜃
3
(ℎ2 +𝑟 2 )2
(ℎ𝑎̂𝑧 − 𝑟𝑎̂𝑟 )
𝑟𝑎̂𝑟 = 𝐸𝑟 + 𝐸𝑧
Due to the symmetry of the charge distribution, for every element 1, there is
corresponding element 2 whose contribution along 𝑎̂𝑟 cancels that of element 1, as
illustrated in fig.(3). Thus, the contribution to 𝐸𝑟 add up to zero so that 𝐸⃗ has only
z-component. Therefore,
𝜎ℎ
𝐸⃗ =
∫
4𝜋𝜀0
𝑆
−
=
∞
∫
2𝜀0 0
𝜎ℎ
1
𝑟 𝑑𝑟 𝑑𝜃
(ℎ2
𝜎ℎ
+
𝑟 𝑑𝑟
3
(ℎ2 +𝑟 2 )2
∞
𝑟2)
𝑎̂𝑧 =
| 𝑎̂𝑧 = −
2𝜀0 √ℎ2 +𝑟 2 0
2𝜋 ∞
𝜎ℎ
̂𝑧 =
∫∫
3𝑎
4𝜋𝜀
0
2
𝜎ℎ
0
∞ 2
(ℎ
∫
4𝜀0 0
𝜎ℎ
[
1
2𝜀0 √ℎ2 +∞2
0
+𝑟
𝑟 𝑑𝑟 𝑑𝜃
(ℎ2
3
2 )−2
+
𝑟2)
∞
2𝜋𝜎ℎ
̂𝑧 =
∫
3𝑎
4𝜋𝜀
0
2
0
𝑟 𝑑𝑟
(ℎ2
∞
1
2𝑟 𝑑𝑟 𝑎̂𝑧 =
1
𝜎ℎ
ℎ
2𝜀0
− ] 𝑎̂𝑧 = −
−
𝜎ℎ (ℎ2 +𝑟 2 ) 2
4𝜀0
−1/2
1
𝜎
ℎ
2𝜀0
[0 − ] 𝑎̂𝑧 =
+
̂𝑧
3𝑎
2
𝑟 )2
| 𝑎̂𝑧 =
0
𝑎̂𝑧
That is, 𝐸⃗ has only z-component if the charge is in the x-y plane. In general, for an
infinite sheet of charges, the electric field is
𝐸⃗ =
𝜎
2𝜀0
𝑎̂𝑛
Where 𝑎̂𝑛 is a unit vector normal to the sheet. From the latter two equations, we
note that the electric field is normal to the sheet and it is surprising independent of
the distance between the sheet and the point of observation P. In a parallel plate
capacitor, the electric field existing between the two plates having normal and
opposite charges is given by
𝐸⃗ =
𝜎
2𝜀0
𝑎̂𝑛 +
−𝜎
2𝜀0
(−𝑎̂𝑛 ) =
𝜎
𝜀0
𝑎̂𝑛
6
Electromagnetic Theory
2018-2019
Prof. dr. Ali Hadi Hassan Al-Batat 1-7
Fig.(3): Evaluation of the electrostatic field due to an infinite sheet of charge.
7