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