91Ó°ÊÓ

Figure 23-26 showing four situations in which four very long rods extend into and out of the page are given.

The separation of the rods is either$\mathrm{d}$or $2d$and the central point is shown as the midway between the inner rods.

The electric field due to a charged rod at any given point is given by its line charge element and the distance of separation. Thus, using the concept of the electric field at a point for the cylindrical rod, you can get the net electric field at the given midway point.

Formula:

The electric field at any point due to an infinitely long rod,

$\mathrm{E}=\frac{\mathrm{λ}}{2{\mathrm{Ï€Î\mu }}_0\mathrm{r}}$ ….. (i)

For situation 1, the net electric field at the central point can be given using equation (i) as follows:

$\begin{array}{c}{\mathrm{E}}_1=\frac{+3}{2{\mathrm{Ï€Î\mu }}_0\left(2\mathrm{d}\right)}+\frac{+2}{2{\mathrm{Ï€Î\mu }}_0\mathrm{d}}+\frac{−2}{2{\mathrm{Ï€Î\mu }}_0\mathrm{d}}+\frac{−3}{2{\mathrm{Ï€Î\mu }}_0\left(2\mathrm{d}\right)}\\ =0\end{array}$

For situation 2, the net electric field at the central point can be given using equation (i) as follows:

$\begin{array}{c}{\mathrm{E}}_2=\frac{+2}{2{\mathrm{Ï€Î\mu }}_0\left(2\mathrm{d}\right)}+\frac{−4}{2{\mathrm{Ï€Î\mu }}_0\mathrm{d}}+\frac{−4}{2{\mathrm{Ï€Î\mu }}_0\mathrm{d}}+\frac{+2}{2{\mathrm{Ï€Î\mu }}_0\left(2\mathrm{d}\right)}\\ =\frac{+1}{{\mathrm{Ï€Î\mu }}_0\left(\mathrm{d}\right)}\end{array}$

For situation 3, the net electric field at the central point can be given using equation (i) as follows:

$\begin{array}{c}{\mathrm{E}}_3=\frac{+8}{2{\mathrm{Ï€Î\mu }}_0\left(3\mathrm{d}\right)}+\frac{−2}{2{\mathrm{Ï€Î\mu }}_0\mathrm{d}}+\frac{+2}{2{\mathrm{Ï€Î\mu }}_0\mathrm{d}}+\frac{+8}{2{\mathrm{Ï€Î\mu }}_0\left(3\mathrm{d}\right)}\\ =\frac{+8}{3{\mathrm{Ï€Î\mu }}_0\mathrm{d}}\end{array}$

For situation 4, the net electric field at the central point can be given using equation (i) as follows:

$\begin{array}{c}{\mathrm{E}}_4=\frac{−6}{2{\mathrm{Ï€Î\mu }}_0\left(3\mathrm{d}\right)}+\frac{+5}{2{\mathrm{Ï€Î\mu }}_0\mathrm{d}}+\frac{+5}{2{\mathrm{Ï€Î\mu }}_0\mathrm{d}}+\frac{−6}{2{\mathrm{Ï€Î\mu }}_0\left(3\mathrm{d}\right)}\\ =\frac{−1}{3{\mathrm{Ï€Î\mu }}_0\mathrm{d}}\end{array}$

Hence, the rank of the situations according to the electric fields is ${\mathrm{E}}_3>{\mathrm{E}}_2>{\mathrm{E}}_1>{\mathrm{E}}_4$.