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 figure) of uniform linear charge density `lambda` .
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.
`therefore q=lambda dx`
Electric field due to the piece,
The electric field is resolved into two rectangular components. `dEcostheta` is the perpendicular component and `dEsintheta` is the parallel component.
When the whole wire is considered, the component `dEsintheta` is cancelled.
Only the perpendicular component `dEcostheta` affects point A.
Hence, effective electric field at point A due to the element dx is dE1.
`therefore dE_1=(lambdadx costheta)/(4piin_0(x^2+l^2))` ...(1)
`x=l tantheta` ...(2)
On differentiating equation (2), we obtain `(dx)/(d theta)=lsec^2theta`
From equation (2),
`x^2+l^2=l^2+l^2 tan theta`
`therefore l^2(1+tan^2 theta)=l^2sec^2theta`
Putting equations (3) and (4) in equation (1), we obtain
`dE_1=(lambda l sec^2 d theta)/(4piin_0l^2sec^2 theta)xxcostheta`
`dE_1=(lambda cos theta d theta)/(4 pi in_0 l)` .......(5)
The wire is so long that `theta` tends from `-pi/2`to `+pi/2`.
By integrating equation (5), we obtain the value of field E1 as,
`int_(-pi/2)^(pi/2) dE_1=int_(-pi/2)^(pi/2)lambda/(4piin_0l) cos theta d theta`
`=lambda/(4 pi in_0l)xx2`
Therefore, the electric field due to long wire is`lambda/(2piin_0l)`.