Continuous Charge Distributions, Example
I. EXAMPLE
A system is composed of a circular ring of radius 𝑅 uniformly charged with total charge
𝑞 and a straight wire of length 𝐿 with linear charge density 𝜆 = 𝜆 0 (𝐿 − 𝑧). One of the ends of
the charged wire coincides with the center of the ring, as shown in the figure 1. Determinate
the electric force that the ring exerts on the wire and the electric force that the wire exerts
on the ring.
𝑧
𝑑E 𝑑𝑧
𝑃
𝑧
𝑟
𝜑 𝑦
𝑑𝑠 = 𝑅 𝑑 𝜑
𝑥 𝐴
FIG. 1. Points 𝐴 and 𝑃 are at the coordinates where the differentials 𝑑𝑠 and 𝑑𝑧 are respectively.
II. SOLUTION
A. Part I
𝑞 𝑞 𝑞
The amount of charge contained within 𝑑𝑠 is 𝑑𝑞 = 2𝜋𝑅 𝑑𝑠 = 2𝜋𝑅 𝑅 𝑑𝜑, where 2𝜋𝑅 is the
charge density. Its contribution to the electric field at the point 𝑃 is
1 𝑑𝑞 1 𝑞 𝑑𝜑
𝑑E = 2
r̂ = r̂. (1)
4𝜋𝜖 0 𝑟 4𝜋𝜖 0 2𝜋𝑟 2
We obtain r and its length in cylindrical coordinates
p
r = −𝑅ˆ
𝛒 + 𝑧ẑ krk = 𝑟 = 𝑅2 + 𝑧2 .
, 2
Replacing in (1), we have the electric field at 𝑃.
1 𝑞 𝑑𝜑
𝑑E = (−𝑅ˆ
𝛒 + 𝑧ẑ). (2)
4𝜋𝜖 0 2𝜋(𝑅 2 + 𝑧2 ) 3/2
Upon integrating along the ring, we obtain
∮ ∮
1 𝑞𝑅 1 𝑞𝑧
E=− ˆ
𝛒 𝑑𝜑 + ẑ 𝑑𝜑,
4𝜋𝜖 0 2𝜋(𝑅 2 + 𝑧2 ) 3/2 4𝜋𝜖 0 2𝜋(𝑅 2 + 𝑧2 ) 3/2
if we replace the equation ˆ 𝛒 = cos 𝜑ˆ𝚤 − sin 𝜑 ˆ𝚥 , we have
∫ 2𝜋 ∫ 2𝜋 ∫ 2𝜋
1 𝑞𝑅 1 𝑞𝑧
E=− ˆ𝚤 cos 𝜑 𝑑𝜑 − ˆ𝚥 sin 𝜑 𝑑𝜑 +ẑ 𝑑𝜑.
4𝜋𝜖 0 2𝜋(𝑅 2 + 𝑧2 ) 3/2 0 0 4𝜋𝜖 0 2𝜋(𝑅 2 + 𝑧2 ) 3/2 0
And finally, we have
1 2𝜋𝑞𝑧 1 𝑞𝑧
E= 2 2 3 2
ẑ = ẑ. (3)
4𝜋𝜖 0 2𝜋(𝑅 + 𝑧 ) / 4𝜋𝜖 0 (𝑅 + 𝑧2 ) 3/2
2
Now, to obtain the electric force that the rings exerts on the wire we make use of the
differential form of the electric force
𝑑F = E 𝑑𝑞,
and as we know, the charge of the wire is
𝑑𝑞 = 𝜆 0 (𝐿 − 𝑧) 𝑑𝑧.
So, the force at any point of the wire is
1 𝑞𝑧
𝑑F = 𝜆 0 (𝐿 − 𝑧) 𝑑𝑧ẑ. (4)
4𝜋𝜖 0 (𝑅 + 𝑧2 ) 3/2
2
Integrating from 𝑧 = 0 to 𝑧 = 𝐿, the total electric force on the wire is
𝑧2
∫ 𝐿 ∫ 𝐿
𝜆0 𝑞 𝑧
F= 𝐿 2 2 3/2
𝑑𝑧 − 2 2 3/2
𝑑𝑧 ẑ.
4𝜋𝜖 0 0 (𝑅 + 𝑧 ) 0 (𝑅 + 𝑧 )
For the first integral, we can replace 𝑡 = 𝑅 2 + 𝑧 2 and it will become
∫ 𝐿+𝑅 2 𝐿+𝑅 2
1 1 1 1
𝑑𝑧 = − √ = −√ . (5)
𝑅2 2𝑡 𝑡 𝑅2 𝑅 𝑅2 + 𝐿 2
And for the second integral, integrating by parts with
𝑧 1
𝑢 = 𝑧, 𝑑𝑣 = 𝑑𝑧 −→ 𝑑𝑢 = 𝑑𝑧, 𝑣 = − √ ,
(𝑅 2 + 𝑧2 ) 3/2 𝑅2 + 𝑧2
I. EXAMPLE
A system is composed of a circular ring of radius 𝑅 uniformly charged with total charge
𝑞 and a straight wire of length 𝐿 with linear charge density 𝜆 = 𝜆 0 (𝐿 − 𝑧). One of the ends of
the charged wire coincides with the center of the ring, as shown in the figure 1. Determinate
the electric force that the ring exerts on the wire and the electric force that the wire exerts
on the ring.
𝑧
𝑑E 𝑑𝑧
𝑃
𝑧
𝑟
𝜑 𝑦
𝑑𝑠 = 𝑅 𝑑 𝜑
𝑥 𝐴
FIG. 1. Points 𝐴 and 𝑃 are at the coordinates where the differentials 𝑑𝑠 and 𝑑𝑧 are respectively.
II. SOLUTION
A. Part I
𝑞 𝑞 𝑞
The amount of charge contained within 𝑑𝑠 is 𝑑𝑞 = 2𝜋𝑅 𝑑𝑠 = 2𝜋𝑅 𝑅 𝑑𝜑, where 2𝜋𝑅 is the
charge density. Its contribution to the electric field at the point 𝑃 is
1 𝑑𝑞 1 𝑞 𝑑𝜑
𝑑E = 2
r̂ = r̂. (1)
4𝜋𝜖 0 𝑟 4𝜋𝜖 0 2𝜋𝑟 2
We obtain r and its length in cylindrical coordinates
p
r = −𝑅ˆ
𝛒 + 𝑧ẑ krk = 𝑟 = 𝑅2 + 𝑧2 .
, 2
Replacing in (1), we have the electric field at 𝑃.
1 𝑞 𝑑𝜑
𝑑E = (−𝑅ˆ
𝛒 + 𝑧ẑ). (2)
4𝜋𝜖 0 2𝜋(𝑅 2 + 𝑧2 ) 3/2
Upon integrating along the ring, we obtain
∮ ∮
1 𝑞𝑅 1 𝑞𝑧
E=− ˆ
𝛒 𝑑𝜑 + ẑ 𝑑𝜑,
4𝜋𝜖 0 2𝜋(𝑅 2 + 𝑧2 ) 3/2 4𝜋𝜖 0 2𝜋(𝑅 2 + 𝑧2 ) 3/2
if we replace the equation ˆ 𝛒 = cos 𝜑ˆ𝚤 − sin 𝜑 ˆ𝚥 , we have
∫ 2𝜋 ∫ 2𝜋 ∫ 2𝜋
1 𝑞𝑅 1 𝑞𝑧
E=− ˆ𝚤 cos 𝜑 𝑑𝜑 − ˆ𝚥 sin 𝜑 𝑑𝜑 +ẑ 𝑑𝜑.
4𝜋𝜖 0 2𝜋(𝑅 2 + 𝑧2 ) 3/2 0 0 4𝜋𝜖 0 2𝜋(𝑅 2 + 𝑧2 ) 3/2 0
And finally, we have
1 2𝜋𝑞𝑧 1 𝑞𝑧
E= 2 2 3 2
ẑ = ẑ. (3)
4𝜋𝜖 0 2𝜋(𝑅 + 𝑧 ) / 4𝜋𝜖 0 (𝑅 + 𝑧2 ) 3/2
2
Now, to obtain the electric force that the rings exerts on the wire we make use of the
differential form of the electric force
𝑑F = E 𝑑𝑞,
and as we know, the charge of the wire is
𝑑𝑞 = 𝜆 0 (𝐿 − 𝑧) 𝑑𝑧.
So, the force at any point of the wire is
1 𝑞𝑧
𝑑F = 𝜆 0 (𝐿 − 𝑧) 𝑑𝑧ẑ. (4)
4𝜋𝜖 0 (𝑅 + 𝑧2 ) 3/2
2
Integrating from 𝑧 = 0 to 𝑧 = 𝐿, the total electric force on the wire is
𝑧2
∫ 𝐿 ∫ 𝐿
𝜆0 𝑞 𝑧
F= 𝐿 2 2 3/2
𝑑𝑧 − 2 2 3/2
𝑑𝑧 ẑ.
4𝜋𝜖 0 0 (𝑅 + 𝑧 ) 0 (𝑅 + 𝑧 )
For the first integral, we can replace 𝑡 = 𝑅 2 + 𝑧 2 and it will become
∫ 𝐿+𝑅 2 𝐿+𝑅 2
1 1 1 1
𝑑𝑧 = − √ = −√ . (5)
𝑅2 2𝑡 𝑡 𝑅2 𝑅 𝑅2 + 𝐿 2
And for the second integral, integrating by parts with
𝑧 1
𝑢 = 𝑧, 𝑑𝑣 = 𝑑𝑧 −→ 𝑑𝑢 = 𝑑𝑧, 𝑣 = − √ ,
(𝑅 2 + 𝑧2 ) 3/2 𝑅2 + 𝑧2
