Figure 23-26 shows four situations in which four very long rods extend into and out of the page (we see only their cross sections). The value below each cross-section gives that particular rod’s uniform charge density in micro-coulombs per meter. The rods are separated by either $\mathbf{d}$ or role="math" localid="1661874332860" $\mathbf{2}\mathbf{d}$as drawn, and a central point is shown midway between the inner rods. Rank the situations according to the magnitude of the net electric field at that central point, greatest first.

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,

$\mathrm{E}=\frac{\mathrm{\lambda }}{2{\mathrm{\pi \epsilon }}_0\mathrm{r}}$ ….. (i)

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

$\begin{array}{c}{\mathrm{E}}_1=\frac{+3}{2{\mathrm{\pi \epsilon }}_0\left(2\mathrm{d}\right)}+\frac{+2}{2{\mathrm{\pi \epsilon }}_0\mathrm{d}}+\frac{-2}{2{\mathrm{\pi \epsilon }}_0\mathrm{d}}+\frac{-3}{2{\mathrm{\pi \epsilon }}_0\left(2\mathrm{d}\right)}\\ =0\end{array}$

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

$\begin{array}{c}{\mathrm{E}}_2=\frac{+2}{2{\mathrm{\pi \epsilon }}_0\left(2\mathrm{d}\right)}+\frac{-4}{2{\mathrm{\pi \epsilon }}_0\mathrm{d}}+\frac{-4}{2{\mathrm{\pi \epsilon }}_0\mathrm{d}}+\frac{+2}{2{\mathrm{\pi \epsilon }}_0\left(2\mathrm{d}\right)}\\ =\frac{+1}{{\mathrm{\pi \epsilon }}_0\left(\mathrm{d}\right)}\end{array}$

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

$\begin{array}{c}{\mathrm{E}}_3=\frac{+8}{2{\mathrm{\pi \epsilon }}_0\left(3\mathrm{d}\right)}+\frac{-2}{2{\mathrm{\pi \epsilon }}_0\mathrm{d}}+\frac{+2}{2{\mathrm{\pi \epsilon }}_0\mathrm{d}}+\frac{+8}{2{\mathrm{\pi \epsilon }}_0\left(3\mathrm{d}\right)}\\ =\frac{+8}{3{\mathrm{\pi \epsilon }}_0\mathrm{d}}\end{array}$

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

$\begin{array}{c}{\mathrm{E}}_4=\frac{-6}{2{\mathrm{\pi \epsilon }}_0\left(3\mathrm{d}\right)}+\frac{+5}{2{\mathrm{\pi \epsilon }}_0\mathrm{d}}+\frac{+5}{2{\mathrm{\pi \epsilon }}_0\mathrm{d}}+\frac{-6}{2{\mathrm{\pi \epsilon }}_0\left(3\mathrm{d}\right)}\\ =\frac{-1}{3{\mathrm{\pi \epsilon }}_0\mathrm{d}}\end{array}$

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