Tutorial-4 Electric Field due to continuous charge distributions (Line & surface charge)

By Mohammed Aldosari September 2020 Charge element ⇒ 𝑑𝑄 Total charge ⇒ 𝑄 Charge distribution ⇒ 𝜌 2 ▪ The electric field intensity due to each of the charge distributions 𝜌𝐿 , 𝜌𝑠 , and 𝜌𝑣 may be regarded as the summation of the field contributed by the numerous point charges making up the charge distribution. 3 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 ▪ Calculate the total charge Q contained in a cylindrical tube of charge oriented along the z-axis as shown in the figure. Where the charge density 𝜌𝐿 = 2𝑧 the tube length is 10cm Solution: 𝑑𝑙 = 𝑑𝑧 0.1 𝑄 = න 𝜌𝐿 𝑑𝑙 = න 2𝑧 𝑑𝑧 = 0 0.1 2 𝑧 ቚ 0 = 0.01 𝐶 4 4 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 ▪ A square plate in the x-y plane is situated in the space defined by −3𝑚 ≤ 𝑥 ≤ 3𝑚 and −3𝑚 ≤ 𝑦 ≤ 3𝑚. Find the total charge on the plate if the surface charge density is given by 𝜌𝑠 = 2𝑦 2 𝜇𝐶/𝑚2 Solution: 3 𝑄 = න 𝜌𝑠 𝑑𝑠 = න 3 න 𝑦=−3 𝑥=−3 2𝑦 2 (10−6 ) 𝑑𝑥𝑑𝑦 = 𝑦3 3 3 อ 𝑦=−3 . 𝑥ቚ 3 𝑥=−3 (2 × 10−6 ) = 216 𝜇𝐶 5 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 ▪ A spherical shell centered at the origin extends between 𝑟 = 2𝑐𝑚 𝑎𝑛𝑑 𝑟 = 3𝑐𝑚. If the volume charge density is given by 𝜌𝑣 = 6𝑟 × 10−4 C/𝑚3 . Find the total charge contained in the shell. Solution: 𝑄 = ‫= 𝑣𝑑 𝑣𝜌׬‬ 0.03 𝜋 2𝜋 ‫=𝑟׬‬0.02 ‫=𝜃׬‬0 ‫=∅׬‬0 6𝑟(10−4 ) 𝑟 2 𝑠𝑖𝑛𝜃 𝑑𝑟𝑑𝜃𝑑∅ = 0.03 𝑟4 ฬ . (−𝑐𝑜𝑠𝜃)ȁ𝜋𝜃=0 (2𝜋)(6 × 4 𝑟=0.02 10−4 ) = 1.22 𝑛𝐶 6 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 ▪ A line charge with uniform charge density 𝜌𝐿 is placed on the z-axis from point −ℎ to point +ℎ ① Find the electric field intensity 𝐸ത at point 𝑝 on the y-axis at distance L from the z-axis ② If 𝜌𝐿 = 10 𝜇𝐶/𝑚 ℎ = 4𝑚 𝑎𝑛𝑑 𝐿 = 3𝑚, a. Find the total charge on the line, b. Find 𝐸ത at point 𝑝 Hint ‫׬‬ 𝑑𝑥 𝑎2 +𝑥 2 3/2 = 𝑥 𝑎2 𝑎2 +𝑥 2 Solution ① 𝑑𝑙 = 𝑑𝑧 𝜌 𝑑𝑙 𝐸ത = 𝐾 ‫ 𝐿 ׬‬2 𝑎ො𝑅 𝑅 𝑅ത = 0, 𝐿, 0 − 0,0, 𝑧 = 𝐿𝑎ො 𝑦 − 𝑧𝑎ො𝑧 𝑅ത = 𝐿2 + 𝑧 2 𝑎ො𝑅 = 𝐿 𝑎ො 𝑦 −𝑧𝑎ො 𝑧 𝐿2 +𝑧 2 However the component of 𝐸ത in the z direction will be always zero, because of the symmetry 𝐿𝑎ො𝑦 𝑎ො𝑅 = 𝐿2 + 𝑧 2 ℎ ℎ 𝐿𝑎ො𝑦 𝜌𝐿 𝑑𝑙 𝑑𝑍 𝑑𝑍 𝐸ത = 𝐾 න 2 𝑎ො𝑅 = 𝐾ρ𝐿 න = 𝐿𝐾𝜌 න ො𝑦 ) 𝐿 2 3 (𝑎 2 2 𝑅 2 2 𝐿 +𝑧 2 2 𝐿 +𝑧 −ℎ −ℎ 𝐿 + 𝑧 2 ℎ 𝑧 ℎ −ℎ ℎ ℎ 𝐾𝐿𝜌𝐿 𝑎ො𝑦 = 𝐾𝐿𝜌𝐿 − 𝑎ො𝑦 = 2𝐾𝐿𝜌𝐿 𝑎ො𝑦 = 2𝐾𝜌𝐿 𝑎ො𝑦 2 2 2 2 2 2 2 2 2 2 2 2 2 2 𝐿 𝐿 + 𝑧 −ℎ 𝐿 𝐿 +ℎ 𝐿 𝐿 +ℎ 𝐿 𝐿 +ℎ 𝐿 𝐿 +ℎ ② 4 𝑄 = න 𝜌𝐿 𝑑𝑙 = න 10 × 10−6 𝑑𝑧 = 80 𝜇𝐶 −4 𝐸ത = 2𝐾𝜌𝐿 ℎ 𝐿 𝐿2 +ℎ2 𝑎ො𝑦 = 2(9 × 109 )(10 × 10−6 ) 4 3 32 +42 𝑎ො 𝑦 = 47.9𝑎ො𝑦 𝑘𝑉/𝑚 7 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 ▪ A rod of length L has a uniform charge per unit length λ and a total charge Q. Calculate the electric field at a point P along the axis of the rod at a distance d from one end. Solution: 𝜌 𝑑𝑙 𝐸ത = 𝐾 ‫ 𝐿 ׬‬2 𝑎ො𝑅 𝑅 𝜌𝐿 = λ 𝑑𝑙 = 𝑑𝑥 𝑎ො𝑅 = −𝑎ො𝑥 Why? Prove: 𝑅ത = 0,0,0 − 𝑥, 0,0 = −𝑥 𝑎ො𝑥 𝑅ത = 𝑥 𝑎ො𝑅 = −𝑎ො𝑥 (because it is one dimension problem. Don’t do this direct way with two or three 𝑅ത dimensions, because 𝑎ො 𝑅 = ത . In another word, in two dimensions problem Each 𝑅 differential charge dQ at dl contributes to the field in two directions “see Q4”) 𝐿+𝑑 λ𝑑𝑥 𝐿+𝑑 𝜌 𝑑𝑙 1 1 𝐸ത = 𝐾 ‫ 𝐿 ׬‬2 𝑎ො𝑅 = 𝐾 ‫ 𝑑=𝑥׬‬2 −𝑎ො𝑥 = 𝐾λ ‫ 𝑥 𝑑=𝑥׬‬−2 𝑑𝑥 −𝑎ො𝑥 = 𝐾λ −𝑥 −1 𝐿+𝑑 −𝑎ො𝑥 = 𝐾λ − + −𝑎ො𝑥 𝑑 𝑅 𝑥 𝐿+𝑑 𝑑 𝐿 = 𝐾λ −𝑎ො𝑥 𝑑(𝐿 + 𝑑) But λL = Q So, 𝐸ത = 𝐾 Note if L ≈ 0 𝑄 𝑑(𝐿+𝑑) ⇒ 𝐸ത = 𝐾 −𝑎ො𝑥 𝑄 𝑑2 −𝑎ො𝑥 Which is the electric field of a point charge 8 8 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 ▪ A circular ring of radius a carries a uniform charge 𝜌𝐿 C/m and is placed on the xy-plane. Find the electric field intensity at point P along the z-axis Solution This problem can be solved in different ways 𝜌 𝑑𝑙 𝐸ത = 𝐾 ‫ 𝐿 ׬‬2 𝑎ො𝑅 ① 𝑅 It is easier to deal with cylindrical coordinate system because the distance R(between the ring and point P) is always constant 𝑑𝑙 = 𝑎𝑑∅ 𝑎ො𝑅 = −𝑎𝑎ො 𝜌 +ℎ𝑎ො 𝑧 𝑎2 +ℎ2 𝑅ത = 0,0, ℎ − 𝑎, 0,0 = −𝑎𝑎ො𝜌 + ℎ𝑎ො𝑧 𝑅 = 𝑎 2 + ℎ2 But because of symmetry 𝑎ො𝜌 = 0 ⇒ 𝑎ො𝑅 = 2𝜋 𝜌𝐿 𝑑𝑙 𝐸ത = 𝐾 න 2 𝑎ො𝑅 = 𝐾 න 𝑅 𝜌𝐿 𝑎𝑑∅ 𝑎 2 + ℎ2 ∅=0 ℎ𝑎ො𝑧 2 𝑎 2 + ℎ2 ℎ𝑎ො 𝑧 𝑎2 +ℎ2 ∅=0 𝑘𝑄ℎ ⇒ 𝐸ത = But Q = 2π𝑎 𝜌𝐿 𝑊ℎ𝑒𝑟𝑒 2𝜋𝑎 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑎 𝑐𝑖𝑟𝑐𝑙𝑒 2𝜋 𝑘𝜌𝐿 𝑎ℎ 𝑘𝜌𝐿 𝑎ℎ = 2 න 𝑑∅ 𝑎 ො = 2𝜋 𝑎ො𝑧 𝑧 𝑎 + ℎ2 3/2 𝑎2 + ℎ2 3/2 3 𝑎2 +ℎ2 2 𝑎ො𝑧 ② The electric field because of charge element dQ can be written as 𝑑𝐸 = 𝑘𝑑𝑄 𝑅2 (point charge) 𝐹𝑟𝑜𝑚 𝑠𝑦𝑚𝑒𝑡𝑟𝑦 𝑡ℎ𝑒 𝑒𝑙𝑒𝑐𝑐𝑡𝑟𝑖𝑐 𝑓𝑖𝑒𝑙𝑑 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑧 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛. Then, 𝑑𝐸𝑧 = 𝑘𝑑𝑄 𝑐𝑜𝑠𝛼 𝑅2 𝑑𝐸𝑧 = 𝑘𝑑𝑄 ℎ 2 2 2 𝑎 +ℎ 𝑎 +ℎ2 𝑏𝑢𝑡 𝑅 = 𝑎2 + ℎ2 = 𝑘𝑑𝑄ℎ 3 𝑎2 +ℎ2 2 ⇒ 𝐸𝑧 = 𝑘ℎ 3 𝑎2 +ℎ2 2 𝑎𝑛𝑑 𝑐𝑜𝑠𝛼 = ‫= 𝑄𝑑 ׬‬ ℎ 𝑅 𝑘𝑄ℎ 3 𝑎2 +ℎ2 2 = ℎ 𝑎2 +ℎ2 9 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 ▪ A ring of radius a has a uniform charge per unit length and a total positive charge Q. Calculate the electric field at a point P along the axis of the ring at a distance 𝑥0 from its center. Solution: This problem can be solved in different ways ① We can use the previous result in Question 6 𝐸ത = 𝑘𝑄ℎ 3 𝑎2 +ℎ2 2 𝑎ො𝑧 So, 𝐸ത = However, the electric field here will be in x-direction only (because of symmetry). 𝑘𝑄𝑥0 3 𝑎2 +𝑥0 2 2 𝑎ො𝑥 ② The electric field because of charge element dQ can be written as 𝑑𝐸 = 𝐹𝑟𝑜𝑚 𝑠𝑦𝑚𝑒𝑡𝑟𝑦 𝑡ℎ𝑒 𝑒𝑙𝑒𝑐𝑐𝑡𝑟𝑖𝑐 𝑓𝑖𝑒𝑙𝑑 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑥 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛. Then, 𝑘𝑑𝑄 𝑑𝐸𝑥 = 2 𝑐𝑜𝑠𝜃 𝑏𝑢𝑡 𝑅 = 𝑎2 + 𝑥0 2 𝑅 𝑑𝐸𝑥 = 𝑘𝑑𝑄 𝑎2 +𝑥0 2 𝑥0 𝑎2 +𝑥0 2 = 𝑘𝑑𝑄ℎ 3 𝑎2 +𝑥0 2 2 ⇒ 𝐸𝑥 = 𝑘𝑥0 3 𝑎2 +𝑥0 2 2 ‫= 𝑄𝑑 ׬‬ 𝑘𝑑𝑄 𝑅2 𝑎𝑛𝑑 𝑐𝑜𝑠𝜃 = 𝑥0 = 𝑅 𝑥0 𝑎2 + 𝑥0 2 𝑘𝑄𝑥0 3 𝑎2 +𝑥0 2 2 10 Q9 Q1 Q2 Q3 Q4 Q6 Q5 Q7 Q8 ▪ A disc of radius R has a uniform charge per unit area σ. Calculate the electric field at a point P along the central axis of the disc at a distance 𝑥0 from its center. Hint ‫׬‬ 𝑥 𝑑𝑥 𝑎 2 +𝑥 2 3/2 = −1 𝑎 2 +𝑥 2 Solution We will solve the question by two ways ① -We will select the cylindrical coordinate -To simplify the problem we will place the x-axis by z-axis -𝜌 is a variable (on the xy-plane) change from 0 to R 𝑅ത = 0,0, ℎ − 𝜌, 0,0 = −𝜌𝑎ො𝜌 + ℎ𝑎ො𝑧 𝑑𝑠 = 𝜌𝑑𝜌𝑑∅ 𝑅 𝐸ത = 𝐾𝜌𝑠 න 2𝜋 𝑑𝑠 𝑎ො = 𝐾𝜌𝑠 න න 𝑑∅ 𝑅2 𝑅 𝑅= 𝜌2 + ℎ2 𝜌𝑑𝜌 ℎ𝑎ො𝑧 2 𝜌2 + ℎ2 𝜌=0 ∅=0 𝜌2 + ℎ2 𝑅 = 𝑘𝜌𝑠 ℎ න 𝜌=0 𝑎ො𝑅 = 𝜌 𝜌2 + ℎ2 So by looking to the origin position we should replace 𝑎ො𝑧 = 𝑎ො𝑥 & ℎ = 𝑥0 & 𝜌𝑠 = σ 𝐸ത = 𝑘σ𝑥0 2𝜋 −1 𝑅2 +𝑥0 2 −𝜌𝑎ො 𝜌 +ℎ 𝑎ො 𝑧 𝜌2 +ℎ2 2𝜋 because of symmetry 𝑎ො𝜌 = 0 𝑑𝜌 න 𝑑∅ 𝑎ො𝑧 = 𝑘𝜌𝑠 ℎ 3/2 ∅=0 −1 𝜌2 + ℎ2 ⇒ 𝑎ො𝑅 = 𝑅 2𝜋 𝑎ො𝑧 = 𝑘𝜌𝑠 ℎ 2𝜋 0 ℎ 𝑎ො 𝑧 𝜌2 +ℎ2 −1 𝑅 2 + ℎ2 − −1 𝑎ො𝑧 ℎ So 1 + 𝑥 𝑎ො𝑥 0 ② The electric field at a point P due to a ring has been calculated in Question 7 as 𝐸𝑥 = 𝑘𝑄𝑥0 3 2 𝑎 +𝑥0 2 2 here 𝑎 = 𝑟 ⇒ 𝐸𝑥 = 𝑘𝑥0 𝑄 3 2 𝑟 +𝑥0 2 2 By considering that the disc is made of concentric rings. By presenting 2πr as the length of the ring, and dr as the thickness. Then, 2πr.dr is the area And the charge can be written as 𝑄 = 𝐴𝑟𝑒𝑎 σ = 2πr𝑑𝑟 σ 𝑘𝑥 (2πr𝑑𝑟)σ 𝑑𝐸𝑥 = 0 3 𝑟 2 +𝑥02 2 𝑅 𝐸𝑥 = 2πσ𝑘𝑥0 න 𝑟=0 r 𝑟2 + 3 𝑑𝑟 2 𝑥0 2 = 2πσ𝑘𝑥0 −1 𝑟 2 + 𝑥0 2 𝑅 = 2πσ𝑘𝑥0 𝑟=0 −1 𝑅 2 + 𝑥0 2 − −1 𝑥0 2 = 2πσ𝑘𝑥0 −1 𝑅 2 + 𝑥0 2 + 1 𝑥0 11 Q9 Calculate the electric field at a distance 𝑥0 from an infinite plane sheet with a uniform charge density 𝜎. 12

Tutorial-4 Electric Field due to continuous charge distributions (Line & surface charge)

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