Electric field due to uniformly charged thin wire :
Consider a long thin wire carrying charge Q distributed uniformly over the wire. Let the linear charge density of the wire is λ. Let a point p where we have to find the electric field making an angle α with the centre. Then the electric field will be -----
As we know,
λ = dQ/dy
dQ = λdy .............................(1)
In the above triangle,
tanα = y/x
y= x tanα
dy = x sec2αdα ..................(2)
In the above triangle,
cosα = x/r
r = x cosα .......................(3)
Electric field at point due to dQ ----
dE= dQ/4πЄ₀r2
dE= λdy/4πЄ₀r2 {since, dQ=λdl from eq. 1} .................(4)
putting the value of dy and r from equation (2) and (3) in eq. (4), we get----
dE = λ x sec2α dα/4πЄ₀ x2sec2α
dE = λdα/4πЄ₀x
On x-axis the electric field will be-----
dEx = λdα cosα/4πЄ₀x
ഽdEx = ഽ-π/2π/2λ cosαdα/4πЄ₀x
ഽdEx = λ /4πЄ₀xഽ-π/2π/2 cosαdα
Ex = λ/4πЄ₀ x {[sinα]-π/2π/2}
Ex = λ/4πЄ₀ x {Sinπ/2 + Sinπ/2}
Ex = 2λ/4πЄ₀ x
so the electric field in x-axis will be-
Ex = λ/2πЄ₀ x.
On the y-axis electric field will be -----
dEy = λdα sinα/4πЄ₀x
ഽdEy = ഽ-π/2π/2λ sinαdα/4πЄ₀x
ഽdEy = λ /4πЄ₀xഽ-π/2π/2 sinαdα
Ey = λ/4πЄ₀ x {[-cosα]-π/2π/2}
Ey = λ/4πЄ₀ x {cosπ/2 + cosπ/2}
Ey = 0.
so the net electric field on the y-axis will be ---
Ey = 0.
