Answer :
Answer:
ρ₀ = 3λ/(2πR⁴)
Explanation:
If we take an elemental part of the cross sectional Area of the infinitely long cylinder.
The small elemental part has a small length dr in the cylinder's radial direction such that the cross sectional Area of the elemental part = 2πr dr
The linear density over the entire radial length of the cylinder will be equal to the sum of (volume charge density × elemental cross sectional Area)
That is,
λ = Σ ρ₀rR × 2πr dr
summing from 0 to R
λ = ∫ᴿ₀ (ρ₀rR × 2πr) dr
λ = ∫ᴿ₀ ρ₀R × 2πr² dr
λ = [ρ₀R × 2πr³/3 ]ᴿ₀
λ = (2π ρ₀R/3) [ r³ ]ᴿ₀
λ = (2π ρ₀R/3) [ R³ ]
λ = 2π ρ₀R⁴/3
ρ₀ = 3λ/(2πR⁴)
The Answer is: ρ₀ = 3λ/(2πR⁴)
- When If we take an elemental part of the cross-sectional then Area of the infinitely long cylinder.
- After that The small elemental part has a small length dr in the cylinder's radial direction such that the cross-sectional then Area of the elemental part = 2πr dr
- When The linear density over the entire radial length of the cylinder will be equal to the sum of (volume charge density × elemental cross-sectional Area)
That is,
- Then λ = Σ ρ₀rR × 2πr dr
- Then summing from 0 to R
- After that λ = ∫ᴿ₀ (ρ₀rR × 2πr) dr
- Then λ = ∫ᴿ₀ ρ₀R × 2πr² dr
- Then λ = [ρ₀R × 2πr³/3 ]ᴿ₀
- Nowλ = (2π ρ₀R/3) [ r³ ]ᴿ₀
- Then λ = (2π ρ₀R/3) [ R³ ]
- Now λ = 2π ρ₀R⁴/3
- Thus, ρ₀ = 3λ/(2πR⁴)
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