Electric field intensity due to uniformly charged wire of infinite length
Derivation of electric field intensity due to the uniformly charged wire of infinite length:
Let us consider a uniformly-charged (positively charged) wire of infinite length having a constant linear charge density (that is, a charge per unit length) λ coulomb/meter. Let P is a point at a distance r from the wire at which electric field →E has to find.
Let us draw a coaxial Gaussian cylindrical surface of length l through point P. By symmetry, the magnitude E of the electric field will be the same at all points on this surface and directed radially outward.
Thus, Now take small area elements dA1, dA2 and dA3 on the Gaussian surface as shown in figure below. Therefore, the total electric flux passing through area elements is:
ϕE=∮→E⋅→dA1+∮→E⋅→dA2+∮→E⋅→dA3
ϕE=∮EdA1cosθ1+∮EdA2cosθ2+∮EdA3cosθ3
The angle between the area element dA1, dA2 and dA3 with electric field are θ1=0∘ , θ2=90∘, and θ3=90∘ respectively. So total electric flux
ϕE=∮EdA1cos0∘+∮EdA2cos90∘+∮EdA3cos90∘
Here cos0∘=1 and cos90∘=0
Hence the above equation can be written as:
ϕE=∮EdA1
The total electric flux passing through the Gaussian surface is
ϕE=∮EdA1
ϕE=E∮dA1
ϕE=E(2πrl){∵∮dA1=2πrl}
ϕE=E(2πrl)
But, By Gaussian's law, The total flux ϕE must be equal to qϵ0, where q is the total charge enclosed with the Gaussian surface. so that
qϵ0=E(2πrl)
E=q2πrlϵ0
For linear charge distribution →
q=λl
So substitute this value in the above equation which can be written as
E=λl2πrlϵ0
E=λ2πϵ0r
The vector form of the above equation :
→E=λ2πϵ0rˆr
Where ˆr is a unit vector in the direction of r. The direction of →E is radially outwards(for positively charged
wire).
Thus, the electric field (E) due to the linear charge is inversely proportional to the distance (r) from the linear charge and its direction
is outward perpendicular to the linear charge.
Special Note:
A charged cylindrical conductor behaves for external points as the whole charge is distributed along its axis.