intensity of electric field E at a point at a perpendicular distance r from infinite line charge having linear charge density lamda is goven by
Attachments:
Answers
Answered by
3
Answer:
Consider a Gaussian surface (cylinder in this case) of radius r and length l.
Then according to the Gauss's Law of electrostatics:
∫
E
.
dA
=
ϵ
0
q
enclosed
From the symmetry of the surface, we can easily say that the electric field would only be along the radial direction. And the amount of charge enclosed by this cylinder is q
enclosed
=λ×l and the surface area of the cylinder A=2πrl.
So, E×2πrl=
ϵ
0
λl
E=(
2πϵ
0
1
)
r
λ
Explanation:
Similar questions