Answer:
[tex]E=\frac{\lambda}{2\pi r\epsilon_0}[/tex]
Explanation:
We are given that
Linear charge density of wire=[tex]\lambda[/tex]
Radius of hollow cylinder=R
Net linear charge density of cylinder=[tex]2\lambda[/tex]
We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R
By Gauss theorem
[tex]\oint E.dS=\frac{q}{\epsilon_0}[/tex]
[tex]q=\lambda L[/tex]
[tex]E(2\pi rL)=\frac{L\lambda}{\epsilon_0}[/tex]
Where surface area of cylinder=[tex]2\pi rL[/tex]
[tex]E=\frac{\lambda}{2\pi r\epsilon_0}[/tex]
