Obtain the formula for the electric field due to a long thin wire of uniform linear charge density `lambda` without using Gauss's law. [Hint.use Coulomb's law directly and evaluate the necessary integral ].

Explanation

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

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 =lambdadx`
Electric field due to the piece,
`dE=(1)/(4piepsilon_(0))(lamdadx)/((AZ)^(2))`
`"However", dZ=sqrt((l^(3)+x^(3)))`
`therefore dE=(lambdadx)/(4piepsilon_(0)(l^(2)+x^(2)))`
The electric field is resolved into two rectangular components. `dE cos theta` is the perpendicular component and
`thereforedE_(1)=(lamdadxcostheta)/(4piin_(0)(x^(2)+l^(2)))" "...(1)`
In`DeltaAZO`,
`tantheta=(x)/(l)`
`x=l"tan"theta" "...(2)`
On differrentiating equation (2), we obtain
dxd`theta`=lsec2`theta`dx=lsec2`theta`d`theta`
From equation (2),
`x^(2)+l^(2)=l^(2)+l^(2)tan^(2)theta`
`thereforel^(2)(1+tan^(2)theta)=l^(2)sec^(2)theta`
`x^(2)+l^(2)=l^(2)sin^(2)theta" "...(4)`
Putting equations (3) and (4) in equation (1), we obtain
`thereforedE_(1)=(lamdalsec^(2)d""theta)/(4piin_(0)l^(2)sec^(2)theta)xxcostheta`
`thereforedE_(1)=(lamdacosthetad""theta)/(4piin_(0)l)" "...(5)`
The wire is so long that `theta` tends from `-(pi)/(2)" to "+(pi)/(2)`
By intergrating equation (5), we obtain the value of field `E_(1)` as,
`underset(-(pi)/(2))overset((pi)/(2))intdE_(1)=underset(-(pi)/(2))overset((pi)/(2))int(lamda)/(4piin_(0)l)costhetad""theta`
`E_(1)=(lamda)/(4piin_(0)l)[sintheta]_(-(pi)/(2))^((pi)/(2))`
`=(lamda)/(4piin_(0)l)xx2`
`E_(1)=(lamda)/(2piin_(0)l`
Therefore, the electric field due to long wire is `(lamda)/(2piin_(0)l)`.