Question

Figure 21-11 shows four situations in which five charged particles are evenly spaced along an axis. The charge values are indicated except for the central particle, which has the same charge in all four situations. Rank the situations according to the magnitude of the net electrostatic force on the central particle, greatest first.

Short answer

The rank of the situations according to the magnitude of the electrostatic force is$\left|{\mathrm{F}}_3\right|>\left|{\mathrm{F}}_1\right|>\left|{\mathrm{F}}_2\right|>\left|{\mathrm{F}}_4\right|$

Step-by-step solution

Stating the given data

Figure 21-11 represents four different situations with five charge particles that are evenly spaced along the axis. The central particle has the same polarity and value in all the four situations.

Understanding the concept of electrostatic force

The net force acting on a charged particle is given by adding the force values that are in the same direction and canceling out the force values that are in the opposite direction. This is given as per the concept that opposite charges attract each other while like charges repel each other.

Formula:

The magnitude of the electrostatic force due to the two charges, $\left|\mathrm{F}\right|=\frac{\mathrm{k}\left|{\mathrm{q}}_1\right|\left|{\mathrm{q}}_2\right|}{{\mathrm{r}}^2}$ (i)

Calculation of the rank according to the magnitudes of the forces 

Let us consider that a positive charge$+\mathrm{q}$is placed at the central point of all the given situations and the space between each pair of charges be $\mathrm{r}$.

Thus, the magnitude of the net force acting on the central particle due to all the charges in all the situations can be given using equation (i) as follows (considering the right direction to be positive):

Situation-1:

$\begin{array}{c}\left|{\mathrm{F}}_1\right|=\frac{\mathrm{k}|-\mathrm{e}||+\mathrm{q}|}{{\left(2\mathrm{r}\right)}^2}+\frac{\mathrm{k}|-\mathrm{e}||+\mathrm{q}|}{{\left(\mathrm{r}\right)}^2}+\frac{\mathrm{k}|+\mathrm{e}||+\mathrm{q}|}{{\left(\mathrm{r}\right)}^2}-\frac{\mathrm{k}|-\mathrm{e}||+\mathrm{q}|}{{\left(2\mathrm{r}\right)}^2}\\ =\frac{2\mathrm{kqe}}{{\mathrm{r}}^2}\end{array}$

Situation-2:

$\begin{array}{c}\left|{\mathrm{F}}_2\right|=-\frac{\mathrm{k}|+\mathrm{e}||+\mathrm{q}|}{{\left(2\mathrm{r}\right)}^2}-\frac{\mathrm{k}|+\mathrm{e}||+\mathrm{q}|}{{\left(\mathrm{r}\right)}^2}+\frac{\mathrm{k}|+\mathrm{e}||+\mathrm{q}|}{{\left(\mathrm{r}\right)}^2}-\frac{\mathrm{k}|-\mathrm{e}||+\mathrm{q}|}{{\left(2\mathrm{r}\right)}^2}\\ =-\frac{2\mathrm{kqe}}{4{\mathrm{r}}^2}\\ =-\frac{\mathrm{kqe}}{2{\mathrm{r}}^2}\\ =\frac{\mathrm{kqe}}{2{\mathrm{r}}^2}\text{(}\because \text{negativeindicatestheoppositeforcedirection)}\end{array}$

Situation-3:

$\begin{array}{c}\left|{\mathrm{F}}_3\right|=\frac{\mathrm{k}|-\mathrm{e}||+\mathrm{q}|}{{\left(2\mathrm{r}\right)}^2}+\frac{\mathrm{k}|-\mathrm{e}||+\mathrm{q}|}{{\left(\mathrm{r}\right)}^2}+\frac{\mathrm{k}|+\mathrm{e}||+\mathrm{q}|}{{\left(\mathrm{r}\right)}^2}+\frac{\mathrm{k}|+\mathrm{e}||+\mathrm{q}|}{{\left(2\mathrm{r}\right)}^2}\\ =\frac{2\mathrm{kqe}}{4{\mathrm{r}}^2}+\frac{2\mathrm{kqe}}{{\mathrm{r}}^2}\\ =\frac{5\mathrm{kqe}}{2{\mathrm{r}}^2}\end{array}$

Situation-4:

$\begin{array}{c}\left|{\mathrm{F}}_4\right|=\frac{\mathrm{k}|-\mathrm{e}||+\mathrm{q}|}{{\left(2\mathrm{r}\right)}^2}-\frac{\mathrm{k}|+\mathrm{e}||+\mathrm{q}|}{{\left(\mathrm{r}\right)}^2}+\frac{\mathrm{k}|+\mathrm{e}||+\mathrm{q}|}{{\left(\mathrm{r}\right)}^2}-\frac{\mathrm{k}|-\mathrm{e}||+\mathrm{q}|}{{\left(2\mathrm{r}\right)}^2}\\ =0\end{array}$

Hence, the rank of the situations due to magnitude of forces is$\left|{\mathrm{F}}_3\right|>\left|{\mathrm{F}}_1\right|>\left|{\mathrm{F}}_2\right|>\left|{\mathrm{F}}_4\right|$ .