Question 30

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.]

Answer

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= 1/4πε₀xλdx. 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 sec² 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