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

Result:
a. The electric field strength outside the cylinder (r ≥ R) is $E=\frac{\lambda }{2\pi {ϵ}_{0}r}$.
b. The electric field strength inside the cylinder (r ≤ R) is $E=\frac{\lambda }{2\pi {ϵ}_{0}R}$.
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:
${\Phi }_E=E·A$
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:
${\Phi }_E=E·A=E·2\pi rL$
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):
${\Phi }_E=\frac{Q_{enc}}{{ϵ}_0}$
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:
$Q_{enc}=\lambda L$
By equating the two expressions for the electric flux, we can solve for E:
$E·2\pi rL=\frac{\lambda L}{{ϵ}_0}$
Simplifying and solving for E, we have:
$E=\frac{\lambda }{2\pi {ϵ}_{0}r}$
To express the answer in terms of R, r, and π, we substitute L with the length of the cylinder:
$E=\frac{\lambda }{2\pi {ϵ}_{0}r}=\frac{\lambda }{2\pi {ϵ}_{0}R}$
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:
${\Phi }_E=E·A$
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:
${\Phi }_E=E·A=E·2\pi rL$
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):
${\Phi }_E=\frac{Q_{enc}}{{ϵ}_0}$
The charge enclosed by the Gaussian surface is now determined by the length of the Gaussian surface within the cylinder:
$Q_{enc}=\lambda L_{enc}$
The length of the Gaussian surface within the cylinder is equal to the length of the Gaussian cylinder itself:
$L_{enc}=L$
By equating the two expressions for the electric flux, we can solve for E:
$E·2\pi rL=\frac{\lambda L}{{ϵ}_0}$
Simplifying and solving for E, we have:
$E=\frac{<}{b}r>\lambda 2\pi {ϵ}_{0}r$
To express the answer in terms of R, r, and π, we substitute L with 2πR:
$E=\frac{\lambda }{2\pi {ϵ}_{0}r}=\frac{\lambda }{2\pi {ϵ}_{0}R}$

a. The electric field strength outside the cylinder ($r\ge R$) is given by:
$E=\frac{\lambda }{2\pi {\xi }_0}\frac{1}{r}$
b. The electric field strength inside the cylinder ($r\le R$) is given by:
$E=\frac{\lambda r}{2\pi {\xi }_0{R}^2}$

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:
${\displaystyle {\Phi }_E=E·A=E·2\pi rL}$
${\displaystyle Q_{\text{enclosed}}=\lambda L}$
Applying Gauss's law, we get:
${\displaystyle {\Phi }_E=\frac{Q_{\text{enclosed}}}{{\epsilon }_0}}$
${\displaystyle E·2\pi rL=\frac{\lambda L}{{\epsilon }_0}}$
Simplifying the equation:
${\displaystyle E=\frac{\lambda }{2\pi {\epsilon }_{0}r}}$
Expressing the answer in terms of R and π:
${\displaystyle \boxed{E=\frac{\lambda }{2\pi {\epsilon }_{0}R}}}$
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:
${\displaystyle Q_{\text{total}}=\lambda L}$
Applying Gauss's law, we get:
${\displaystyle {\Phi }_E=\frac{Q_{\text{total}}}{{\epsilon }_0}}$
${\displaystyle E·2\pi rL=\frac{\lambda L}{{\epsilon }_0}}$
Simplifying the equation:
${\displaystyle E=\frac{\lambda r}{2\pi {\epsilon }_0}}$
Expressing the answer in terms of R and π:
${\displaystyle \boxed{E=\frac{\lambda r}{2\pi {\epsilon }_0}}}$