Question

Electric intensity outside a charged cylinder having the charge per unit length lambda at a distance r from its axis is .........

Answer 1

It's answer is E= ^/2π£•kr

By Gauss theorem
TNEI over closed surface=net charge enclosed by the surface
£E×2πrl= ^l

Therefore, E=^/2𣕠kr,where k is relative permittivity of medium

Answer 2

Answer : E = λ/2πε₀r
Charge per unit length is λ
Let the length of charged cylinder = L
Then, charge on cylinder is Q = λL

Use Gaussian theorem,
Let r is the radius of Gaussian surface as shown in figure.
Then area of Gaussian surface { cylinder } , dS = 2πrL
Now, Φ = E.dS = Q/ε₀ [ by Gauss Theory ]
Here E is electric field intensity and dS is area of Gaussian surface , Q is net charge inclosed in Gaussian surface .
E.2πrL = λL/ε₀
E = λ/2πε₀r

Hence, answer is E = λ/2πε₀r