A very long, uniformly charged cylinder has radius R and lin

tripiverded9

tripiverded9

Answered question

2022-01-15

A very long, uniformly charged cylinder has radius R and linear charge density λ.
a. Find the cylinder's electric field strength outside the cylinder, rR. Give your answer as a multiple of λξ0.
Express your answer in terms of some or all of the variables R, r, and the constant π.
b. Find the cylinder's electric field strength inside the cylinder, rR. Give your answer as a multiple of λξ0.
Express your answer in terms of some or all of the variables R, r, and the constant π.

Answer & Explanation

Mary Herrera

Mary Herrera

Beginner2022-01-16Added 37 answers

Step 1
Given:
Radius of the cylinder =R
Linear Charge density of the cylinder =λ
Step 2
Calculating the electric field outside the cylinder:
Let’s assume a Cylindrical Gaussian surface of radius ‘r’ (r>R) and length ‘L’ around the given charged cylinder.
Charge enclosed by the Guassian surface, qenclosed=λL
Using Gausss
Piosellisf

Piosellisf

Beginner2022-01-17Added 40 answers

Given Data:
- Radius of cylinder is
- Linear charge density is λ
Part a
We have to find out the electric field strength outside the cylinder rR
Let P be the external point, where we have to find the electric field.
So, we consider a gaussian surface of length L, at a distance r from the axis of the cylinder, which is also a coaxial cylinder.
From Gauss law,
SE.dS=λL0
Where λL is the charge enclosed by the surface.
Thus,
SE.dS=EdS
=E.2πrL
=λL0
So, E=14π02λrr^
Where r^ is the direction of electric field and it is normal to the curved portion.
The cylinder's electric field strength outside the cylinder is E=14π02λrr^
Part b
Now, we have to find out the electric field strength inside the cylinder rR
Let P be any internal point, where we have to find the electric field.
Similarly, we consider a gaussian surface of length L at a distance from the axis of the cylinder, which is also a coaxial cylinder.
Charge enclosed by it is,
Qenclosed=πr2Lρ
Where ρ is the volume charge density.
Now, volume charge density is related to linear charge density as,
πR21ρ=λρ=λπR2
From Gauss law,
SE^.dS^=Qenclosed0
So, SE^.dS^=EdS
=E2πr L
=πr2Lρ0
Now, E2πr L=πr2L0λπR2
So, E=14π02λrR2r^
Where r^ is the direction of electric field and it is normal to the curved portion.
The cylinder's electric field strength inside the cylinder is
E=14π02λrR2r^
user_27qwe

user_27qwe

Skilled2023-06-19Added 375 answers

Result:
a. The electric field strength outside the cylinder (r ≥ R) is E=λ2πϵ0r.
b. The electric field strength inside the cylinder (r ≤ R) is E=λ2πϵ0R.
Solution:
a. Outside the cylinder (r ≥ R):
To find the electric field strength outside the cylinder, we'll consider a Gaussian surface in the form of a cylinder of radius r and length L, with the axis coinciding with the axis of the charged cylinder. The Gaussian surface encloses the entire charged cylinder.
The electric field strength, E, outside the cylinder will be constant and perpendicular to the surface of the Gaussian cylinder due to its symmetry. The electric flux through the Gaussian surface is given by:
ΦE=E·A
where E is the electric field strength and A is the area of the cylindrical surface.
Since the electric field is constant over the surface and perpendicular to it, the flux becomes:
ΦE=E·A=E·2πrL
According to Gauss's Law, the electric flux through the Gaussian surface is also equal to the charge enclosed by the surface divided by the electric constant (ε₀):
ΦE=Qencϵ0
The linear charge density, λ, is given by the charge per unit length of the cylinder, so the total charge enclosed by the Gaussian surface is:
Qenc=λL
By equating the two expressions for the electric flux, we can solve for E:
E·2πrL=λLϵ0
Simplifying and solving for E, we have:
E=λ2πϵ0r
To express the answer in terms of R, r, and π, we substitute L with the length of the cylinder:
E=λ2πϵ0r=λ2πϵ0R
b. Inside the cylinder (r ≤ R):
To find the electric field strength inside the cylinder, we'll again consider a Gaussian surface in the form of a cylinder of radius r and length L, with the axis coinciding with the axis of the charged cylinder. The Gaussian surface is completely within the charged cylinder.
The electric field strength, E, inside the cylinder will be constant and perpendicular to the surface of the Gaussian cylinder due to its symmetry. The electric flux through the Gaussian surface is given by:
ΦE=E·A
where E is the electric field strength and A is the area of the cylindrical surface.
Again, since the electric field is constant over the surface and perpendicular to it, the flux becomes:
ΦE=E·A=E·2πrL
Using Gauss's Law, the electric flux through the Gaussian surface is equal to the charge enclosed by the surface divided by the electric constant (ε₀):
ΦE=Qencϵ0
The charge enclosed by the Gaussian surface is now determined by the length of the Gaussian surface within the cylinder:
Qenc=λLenc
The length of the Gaussian surface within the cylinder is equal to the length of the Gaussian cylinder itself:
Lenc=L
By equating the two expressions for the electric flux, we can solve for E:
E·2πrL=λLϵ0
Simplifying and solving for E, we have:
E=<br>λ2πϵ0r
To express the answer in terms of R, r, and π, we substitute L with 2πR:
E=λ2πϵ0r=λ2πϵ0R
karton

karton

Expert2023-06-19Added 613 answers

a. The electric field strength outside the cylinder (rR) is given by:
E=λ2πξ01r
b. The electric field strength inside the cylinder (rR) is given by:
E=λr2πξ0R2
alenahelenash

alenahelenash

Expert2023-06-19Added 556 answers

Step 1: To solve the problem, let's consider both cases separately:
a. Outside the cylinder (r ≥ R):
We can use Gauss's law to find the electric field strength outside the cylinder. Consider a Gaussian surface in the form of a cylindrical surface with radius r and length L, enclosing the cylinder.
According to Gauss's law, the total electric flux through the Gaussian surface is equal to the charge enclosed divided by the permittivity of free space (ε₀).
Since the cylinder is uniformly charged, the charge enclosed within the Gaussian surface is equal to the linear charge density (λ) multiplied by the length of the cylinder (L).
The electric field is uniform and perpendicular to the Gaussian surface, so its magnitude is constant over the entire surface.
Using these considerations, we have:
ΦE=E·A=E·2πrL
Qenclosed=λL
Applying Gauss's law, we get:
ΦE=Qenclosedε0
E·2πrL=λLε0
Simplifying the equation:
E=λ2πε0r
Expressing the answer in terms of R and π:
E=λ2πε0R
Step 2: b. Inside the cylinder (r ≤ R):
Inside the cylinder, we can assume a Gaussian surface in the form of a cylindrical surface with radius r and length L, still entirely inside the cylinder.
Applying Gauss's law, the total electric flux through this Gaussian surface is equal to the charge enclosed divided by the permittivity of free space (ε₀).
Since the entire charge of the cylinder is enclosed within the Gaussian surface, the charge enclosed is equal to the total charge of the cylinder.
Using the linear charge density (λ) and the length of the cylinder (L), the total charge can be calculated as:
Qtotal=λL
Applying Gauss's law, we get:
ΦE=Qtotalε0
E·2πrL=λLε0
Simplifying the equation:
E=λr2πε0
Expressing the answer in terms of R and π:
E=λr2πε0

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