Question

if 10 cm long rod carries a charge of +50?c distributed uniformly along its length. Find the magnitude of the electric field at a point 10 cm from both the end of the rod.

Answer 1

see diagram.

AB is the rod of length 2L = 10 cm.  L = 5 cm. P is the point on perpendicular bisector such that AP = BP = 10 cm.  So  OP = d = 5√3 cm.

TanФ₀ = OP/AO = √3.  So Ф₀ = 60⁰.

charge density of the rod = ρ = +50 μC /0.10 m = 500 μC/m

Take two elements of length dx and charge ρ dx on either side of O at a distance x.
      x = d cot Ф,   dx = - d cosec² Ф dФ = - (x²+d²)/d * dФ
      
Field due to each dx = dE = ρ dx / (x²+d²)
      dE = - (ρ/d) dФ
      Component of dE along OP = dE cos(90-Ф) = dE sinФ 
             = - (ρ/d) sinΦ dΦ
The components of dE along the line parallel to AB and perpendicular to OP , due to the two elements of dx are cancelled. The are in the opposite directions. The components along OP are added.

    E = - 2 (ρ/d) Integral SinΦ dΦ,     Φ = Φ₀ to 90⁰.
       =  2 (ρ/d) [ cos Φ]   Φ=Φ₀ to 90°
       =  2 (ρ/d) Cos Φ₀
       = 2 * 500*μC * cos 60  / 5√3    V/m
       = 100/√3 * 10⁻⁶  V/m

The direction is along the perpendicular bisector away from the rod.