Question

Obtain the formula for the electric field due to a long thin wire ofuniform linear charge density E without using Gauss’s law. [Hint:Use Coulomb’s law directly and evaluate the necessary integral.]

Answer 1

Consider a long thin wire XY of uniform linear charge density λ.

Consider a point A at a perpendicular distance l from the mid-point O of the wire.

Let E be the electric field due to the wire XY at point A. Consider a small length of element dx on the wire section with OZ=x.

The charge on this element is given as,

q=λdx

The electric field due to the small element is given as,

dE= 1 4π ε 0 λdx ( AZ ) 2 (1)

Here,

AZ= l 2 + x 2

Substituting the value of AZ in the equation (1), we get

dE= 1 4π ε 0 λdx ( l 2 + x 2 ) 2

The electric field is resolved into two angular components. The perpendicular component is dEcosθ and the parallel component is dEsinθ

When the complete wire is considered the parallel component of electric field is cancelled. Only the perpendicular component will affect the point A.

The effective electric field due to the element dx is given by ,

d E 1 = 1 4π ε 0 λdx⋅cosθ ( l 2 + x 2 ) (2)

In ΔAZO,

tanθ= x l x=ltanθ (3)

Differentiate the equation (3).

dx dθ =l sec 2 θ dx=l sec 2 θdθ (4)

The value of l 2 + x 2 is obtained by using equation (3).

l 2 + x 2 = l 2 + l 2 tan 2 θ = l 2 ( 1+ tan 2 θ ) = l 2 sec 2 θ (5)

Substitute the values of equation (4) and equation (5) in equation (2).

d E 1 = 1 4π ε 0 λ⋅l sec 2 θdθ⋅cosθ l 2 sec 2 θ = 1 4π ε 0 λcosθdθ l (6)

The wire is so long that the θ ranges − π 2 <θ< π 2 .

Integrating the equation (6), we get

∫ − π 2 π 2 d E 1 = ∫ − π 2 π 2 1 4π ε 0 λcosθdθ l E 1 = 1 4π ε 0 λ l ∫ − π 2 π 2 cosθdθ = 1 4π ε 0 λ l [ sinθ ] − π 2 π 2 =2× 1 4π ε 0 λ l

Further simplifying the above expression,

E 1 = 1 2π ε 0 λ l = λ 2π ε 0 l

Thus, the electric field due to a long wire is λ 2π ε 0 l .