Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ 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 l 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.
∴ q = λdx
Electric field due to the piece,
However, 
The electric field is resolved into two rectangular components. dEcosθ is the perpendicular component and dEsinθ is the parallel component. When the whole wire is considered, the component dEsinθ is cancelled. Only the perpendicular component dEcosθ affects point A.
Hence, effective electric field at point A due to the element dx is dE1 .
∴ d E1 = 1/4πε0 x λdx. cos θ/(l2 + x2) ...(1)
In ∆AZO, tanθ = x/l ⇒ x = l.tanθ ...(2)
On differentiating equation (2), we obtain
dx/dθ = l x sec 2 θ ⇒ dx = l x sec 2 θdθ ...(3)
From equation (2), we have
x2 + l2 = l2tan2θ + l2 = l2 (tan2 θ + 1) = l2 sec2 θ ...(4)
Putting equations (3) and (4) in equation (1), we obtain
The wire is so long that θ tends from − π/2 to π/2.
By integrating equation (5), we obtain the value of field E1 as,
Therefore, the electric field due to long wire is λ/2πε0 l .
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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 l from the mid-point O of the wire, as shown in the following figur...
Step-by-step explanation:
• 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 l from the mid-point O of the wire, as shown in the following figure
•
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