A very long, uniformly charged cylinder has radius R and linear charge densityλ.a. Find the...

tripiverded9

tripiverded9

Answered

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

Expert

2022-01-16Added 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

Expert

2022-01-17Added 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^
alenahelenash

alenahelenash

Expert

2022-01-23Added 366 answers

Step 1 a) Apply Gausss

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