A very long insulating cylinder has radius R and carries positive charge distributed throughout its...

Danelle Albright

Danelle Albright

Answered

2022-01-16

A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribution has cylindrical symmetry but varies with perpendicular distance from the axis of the cylinder. The volume charge density is ρ(r)=α(1rR), where α is a constant with units Cm3 and r is the perpendicular distance from the center line of the cylinder.

Answer & Explanation

Stuart Rountree

Stuart Rountree

Expert

2022-01-17Added 29 answers

Step 1
Using the Gauss law,
ϕEdϕA=Qencξ0 [1]
Step 2
Use equation (1) to solve for the electric field as a function of r.
E2πrh=Qencξ0
=ρVξ0 [2]
Step 3
The volume charge density of the cylinder is
ρ=α(1rR) [3]
Step 4
Use equation (3) in (2) to find the electric field as a function of r.
E2πrh=α(1rR)πr2hξ0
E=α2ξ0(rr2R)
Alex Sheppard

Alex Sheppard

Expert

2022-01-18Added 36 answers

Answer:
1. E(r)=α4πξ0(2rR)
2. E(r)=14πξ0αRr
3.The results from part 1 and 2 agree when r=R.
Explanation:
The volume charge density is given as
ρ(r)=α(1rR)
We will investigate this question in two parts. First r<R, then r>R. We will show that at r=R, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.
Eda=Qencξ0
The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is
da=2πrh
where ‘h’ is the length of the imaginary Gaussian surface.
Qenc=0rρ(r)h dr=αh0r(1r/R)dr=αh(rr22R){r=rr=0=αh(2Rrr22R)
E2πrh=αh2Rrr22Rξ0
E(r)=α2Rr4πξ0R
E(r)=α4πξ0(2rR)
2. For r>R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,
Qenc=0Rρ(r)h dr=α0R(1r/R)h dr=αh(rr22R){r=Rr=0=α h(RR22R)=αhR2
E2πr h=αRh2ξ0
E(r)=14πξ0αRr
3. At the boundary where r=R:
E(r=R)=α4πξ0(2rR)=α4πξ0

alenahelenash

alenahelenash

Expert

2022-01-23Added 366 answers

Step 1 1. Since the cylinder is very long, Gauss

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