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


 

Take a long thin wire XY (as shown in the following figure) of uniform linear charge density λ

Consider a point A at a perpendicular distance I from the mid-point
O of the wire, as shown in the following figure.
Let E be the electric field at point A due to the wire, XY.
Consider a small length element dx on the wire section with OZ X
Let q be the charge on this piece.

Electric field due to the piece,    q=λdx
dE=14πϵ0λdx(AZ)2
HoweverAZ=l2+x2
 dE =14πϵ0λdx(l2+x2)

The electric field is resolved into two rectangular components. dE cos θ is the perpendicular component and dE sin G is the parallel component. When the whole wire is considered, the component dE sin G is cancelled. Only the perpendicular component dEcos6 affects point A.

Hence, effective electric field at point A due to the element dx is dEj.

In AZO,    dE1=14πϵ0λdxcosθ(l2+x2)(1)tanθ=xlx=l.tanθ.(2)

On differentiating equation (2), we obtain

dxdθ=lsec2θdx=lsec2θdθ (3) 

From equation (2), we have

x2+l2=l2tan2θ+l2=l2(tan2θ+1)=l2sec2θ

Putting equations (3) and (4) in equation (1), we obtain

dE1=14πϵ0λsec2θdθcosθl2sec2θ=14πϵ0λcosθdθl(5)

The wire is so long that 6 tends from π2 to π2

By integrating equation (5), we obtain the value of field E as,

π2π2dE1=ππ214πϵ0λlcosθdθE1=14πϵ0λl[sinθ]π2π2E1=14πϵ0λl×2E1=λ2πϵ0l

Therefore, the electric field due to long wire is λ2πϵ0l.