Question

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.

Short answer

The rank of the situations according to their magnitude of the net electric field at that central point is${\mathrm{E}}_3>{\mathrm{E}}_2>{\mathrm{E}}_1>{\mathrm{E}}_4$.

Step-by-step solution

The given data: 

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.

Understanding the concept of electric field: 

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)

Calculation of rank of the situations according to electric fields: 

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 ${\mathrm{E}}_3>{\mathrm{E}}_2>{\mathrm{E}}_1>{\mathrm{E}}_4$.