Continuous Charge Distributions, Example

1. 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 = 𝜆0(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. 𝑑𝑧 𝑧 𝑦 𝑥 𝑑𝑠 = 𝑅 𝑑𝜑 𝜑 𝑧 𝑟 𝑑E 𝑃 𝐴 FIG. 1. Points 𝐴 and 𝑃 are at the coordinates where the differentials 𝑑𝑠 and 𝑑𝑧 are respectively. II. SOLUTION A. Part I The amount of charge contained within 𝑑𝑠 is 𝑑𝑞 = 𝑞 2𝜋𝑅 𝑑𝑠 = 𝑞 2𝜋𝑅 𝑅 𝑑𝜑, where 𝑞 2𝜋𝑅 is the charge density. Its contribution to the electric field at the point 𝑃 is 𝑑E = 1 4𝜋𝜖0 𝑑𝑞 𝑟2 r̂ = 1 4𝜋𝜖0 𝑞 𝑑𝜑 2𝜋𝑟2 r̂. (1) We obtain r and its length in cylindrical coordinates r = −𝑅ˆ 𝛒 + 𝑧ẑ krk = 𝑟 = p 𝑅2 + 𝑧2.

2. 2 Replacing in (1), we have the electric field at 𝑃. 𝑑E = 1 4𝜋𝜖0 𝑞 𝑑𝜑 2𝜋(𝑅2 + 𝑧2)3/2 (−𝑅ˆ 𝛒 + 𝑧ẑ). (2) Upon integrating along the ring, we obtain E = − 1 4𝜋𝜖0 𝑞𝑅 2𝜋(𝑅2 + 𝑧2)3/2 ∮ ˆ 𝛒 𝑑𝜑 + ẑ 1 4𝜋𝜖0 𝑞𝑧 2𝜋(𝑅2 + 𝑧2)3/2 ∮ 𝑑𝜑, if we replace the equation ˆ 𝛒 = cos 𝜑ˆ 𝚤 − sin 𝜑ˆ 𝚥, we have E = − 1 4𝜋𝜖0 𝑞𝑅 2𝜋(𝑅2 + 𝑧2)3/2 ˆ 𝚤 ∫ 2𝜋 0 cos 𝜑 𝑑𝜑 − ˆ 𝚥 ∫ 2𝜋 0 sin 𝜑 𝑑𝜑 +ẑ 1 4𝜋𝜖0 𝑞𝑧 2𝜋(𝑅2 + 𝑧2)3/2 ∫ 2𝜋 0 𝑑𝜑. And finally, we have E = 1 4𝜋𝜖0 2𝜋𝑞𝑧 2𝜋(𝑅2 + 𝑧2)3/2 ẑ = 1 4𝜋𝜖0 𝑞𝑧 (𝑅2 + 𝑧2)3/2 ẑ. (3) 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 = 𝜆0(𝐿 − 𝑧) 𝑑𝑧. So, the force at any point of the wire is 𝑑F = 1 4𝜋𝜖0 𝑞𝑧 (𝑅2 + 𝑧2)3/2 𝜆0(𝐿 − 𝑧) 𝑑𝑧ẑ. (4) Integrating from 𝑧 = 0 to 𝑧 = 𝐿, the total electric force on the wire is F = 𝜆0𝑞 4𝜋𝜖0 𝐿 ∫ 𝐿 0 𝑧 (𝑅2 + 𝑧2)3/2 𝑑𝑧 − ∫ 𝐿 0 𝑧2 (𝑅2 + 𝑧2)3/2 𝑑𝑧 ẑ. For the first integral, we can replace 𝑡 = 𝑅2 + 𝑧2 and it will become ∫ 𝑙+𝑅2 𝑅2 1 2𝑡3/2 𝑑𝑧 = − 1 √ 𝑅2 + 𝑧2

6. 𝑙+𝑅2 𝑅2 = 1 𝑅 − 1 √ 𝑅2 + 𝐿2 . (5) And for the second integral, integrating by parts with 𝑢 = 𝑧, 𝑑𝑣 = 𝑧 (𝑅2 + 𝑧2)3/2 𝑑𝑧 −→ 𝑑𝑢 = 𝑑𝑧, 𝑣 = − 1 √ 𝑅2 + 𝑧2 ,

7. 3 the integral becomes 𝑧 √ 𝑅2 + 𝑧2

11. 𝐿 0 + ∫ 𝐿 0 1 √ 𝑅2 + 𝑧2 𝑑𝑧 = 𝑧 √ 𝑅2 + 𝑧2 + ln

14. 𝑧 + p 𝑅2 + 𝑧2

20. 𝐿 0 = 𝐿 √ 𝑅2 + 𝐿2 + ln 𝐿 + √ 𝑅2 + 𝐿2 − ln 𝑅. (6) Finally, the electric force is F = 𝜆0𝑞 4𝜋𝜖0 𝐿 √ 𝑅2 + 𝐿2 + ln 𝐿 + √ 𝑅2 + 𝐿2 𝑅 ! + 𝐿 𝑅 − 𝐿 √ 𝑅2 + 𝐿2 ! ẑ = 𝜆0𝑞 4𝜋𝜖0 𝐿 𝑅 + ln 𝐿 + √ 𝑅2 + 𝐿2 𝑅 !# ẑ. (7) B. Part II The distance from 𝑃 to 𝐴 is given by r = 𝑅ˆ 𝛒−𝑧ẑ and krk = 𝑟 = √ 𝑅2 + 𝑧2. The contribution to the electric force at 𝐴 is 𝑑E = 1 4𝜋𝜖0 𝑑𝑞 𝑟2 r̂ = 1 4𝜋𝜖0 𝜆0(𝐿 − 𝑧) 𝑑𝑧 (𝑅2 + 𝑧2)3/2 (𝑅ˆ 𝛒 − 𝑧ẑ). (8) Upon integrating from 0 to 𝐿, the equation results E = 1 4𝜋𝜖0 ∫ 𝐿 0 𝜆0(𝐿 − 𝑧)(𝑅ˆ 𝛒 − 𝑧ẑ) (𝑅2 + 𝑧)3/2 𝑑𝑧 = 𝜆0 4𝜋𝜖0 𝐿𝑅ˆ 𝛒 ∫ 𝐿 0 1 (𝑅2 + 𝑧2)3/2 𝑑𝑧 − 𝑅ˆ 𝛒 ∫ 𝐿 0 𝑧 (𝑅2 + 𝑧2)3/2 𝑑𝑧 + 𝜆0 4𝜋𝜖0 −𝐿ẑ ∫ 𝐿 0 𝑧 (𝑅2 + 𝑧2)3/2 𝑑𝑧 + ẑ ∫ 𝐿 0 𝑧2 (𝑅2 + 𝑧2)3/2 𝑑𝑧 . To integrate the first integral, let’s treat the equation as a differential binomial, in order to use the Chebyshev theorem on the integration of binomial differentials. The differential equation takes the form 𝑧0(𝑅2 + 𝑧2)−3/2𝑑𝑧 and as (0 + 1)/2 + (−3/2) = −1 is an integer, we will make change of variable: 𝑢 = 𝑅2 + 𝑧2 𝑧2 1/2 = (𝑅2 𝑧−2 + 1)1/2 , which gives (𝑢𝑧)−3/2 = (𝑅2 + 𝑧2 )−3/2 and 𝑧 = 𝑢2 − 1 𝑅2 −1/2 −→ 𝑑𝑧 = − 𝑅𝑢 (𝑢2 − 1)3/2 𝑑𝑢,

21. 4 the above integral becomes ∫ 𝐿 0 1 (𝑅2 + 𝑧2)3/2 𝑑𝑧 = ∫ √ 𝑅2/𝐿2+1 +∞ (𝑢𝑧)−3 −𝑅𝑢 (𝑢2 − 1)3/2 𝑑𝑢 = − ∫ √ 𝑅2/𝐿2+1 +∞ 𝑢2 − 1 𝑅2 3/2 𝑅𝑢−2 (𝑢2 − 1)3/2 𝑑𝑢 = −𝑅2 ∫ √ 𝑅2/𝐿2+1 +∞ 𝑢−2 𝑑𝑢 = lim 𝑎→+∞ 1 𝑅2𝑢

25. √ 𝑅2/𝐿2+1 𝑎 = 𝐿 𝑅2 √ 𝑅2 + 𝐿2 . (9) The second and third integral are ∫ 𝐿 0 𝑧 (𝑅2 + 𝑧2)3/2 𝑑𝑧 = 1 2 ∫ 𝑅2+𝐿2 𝑅2 1 𝑡3/2 𝑑𝑡 = − 1 √ 𝑡

29. 𝑅2+𝐿2 𝑅2 = 1 𝑅 − 1 √ 𝑅2 + 𝐿2 . (10) The last integral can be solved using the differential binomial as in the first integral. So, as the differential binomial is 𝑧2(𝑅2 + 𝑧2)−3/2 𝑑𝑧 and (2 + 1)/2 − 3/2 = 0 is an integer, we will use the same substitution variables of the first integral. Thus, we obtain ∫ 𝐿 0 𝑧2 (𝑅2 + 𝑧2)3/2 𝑑𝑧 = ∫ √ 𝑅2/𝐿2+1 +∞ 𝑧2 (𝑢𝑧)−3 −𝑅𝑢 (𝑢2 − 1)3/2 𝑑𝑢 = − ∫ √ 𝑅2/𝐿2+1 +∞ 𝑧−1 𝑅𝑢−2 (𝑢2 − 1)3/2 𝑑𝑢 = − ∫ √ 𝑅2/𝐿2+1 +∞ 𝑢2 − 1 𝑅2 𝑅𝑢−2 (𝑢2 − 1)3/2 𝑑𝑢 = − ∫ √ 𝑅2/𝐿2+1 +∞ 1 𝑢2(𝑢2 − 1) 𝑑𝑢 = ∫ √ 𝑅2/𝐿2+1 +∞ 1 𝑢2 𝑑𝑢 − ∫ √ 𝑅2/𝐿2+1 +∞ 1 𝑢2 − 1 𝑑𝑢 = − lim 𝑎→+∞ 1 𝑢 + 1 2 ln

33. 𝑢 − 1 𝑢 + 1

40. √ 𝑅2/𝐿2+1 𝑎 = − 𝐿 √ 𝑅2 + 𝐿2 − ln √ 𝑅2 + 𝐿2 − 𝑧 𝑅 ! . (11) The resulting electric field at the point 𝐴 is E = 𝜆0 4𝜋𝜖0 𝐿2 + 𝑅2 𝑅 √ 𝑅2 + 𝐿2 − 1 ˆ 𝛒 − 𝜆0 4𝜋𝜖0 𝐿 𝑅 + ln √ 𝑅2 + 𝐿2 − 𝑧 𝑅 !# ẑ. (12) Now, the electric force applied to the ring at the pint 𝐴 is 𝑑F = E𝑑𝑞, where 𝑑𝑞 = 𝑞 2𝜋𝑅 𝑑𝑠 = 𝑞 2𝜋 𝑑𝜑. Upon integrating along the ring, we have F = ∮ E 𝑑𝑞 = 𝑞𝐸𝑟 2𝜋 ∫ 2𝜋 0 ˆ 𝛒 𝑑𝜑 + 𝑞𝐸𝑧 2𝜋 ẑ ∫ 2𝜋 0 𝑑𝜑 = 𝑞𝐸𝑟 2𝜋 ˆ 𝚤 ∫ 2𝜋 0 cos 𝜑 𝑑𝜑 + 𝑞𝐸𝑟 2𝜋 ˆ 𝚥 ∫ 2𝜋 0 sin 𝜑 𝑑𝜑 + 2𝜋𝑞𝐸𝑧 2𝜋 ẑ = − 𝑞𝜆0 4𝜋𝜀0 𝐿 𝑅 + ln √ 𝑅2 + 𝐿2 − 𝑧 𝑅 !# ẑ. (13)

41. 5 III. CONCLUSION As we saw in equations (7) and (13), the electric force is Newtonian even if the object is a body.

Continuous Charge Distributions, Example

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