Question

Figure 21-17 shows four arrangements of charged particles. Rank the arrangements according to the magnitude of the net electrostatic force on the particle with charge$\mathbf{+}\mathbf{Q}$, greatest first.

Short answer

The rank of the arrangements according to the net electrostatic force on the particle with charge $+\mathrm{Q}$is $\left|{\mathrm{F}}_1\right|=\left|{\mathrm{F}}_4\right|>\left|{\mathrm{F}}_2\right|=\left|{\mathrm{F}}_3\right|$.

Step-by-step solution

Stating the given data

Figure 21-17 shows four arrangements of charged particles with charge particle$+\mathrm{Q}$ in the arrangement.

Understanding the concept of wave

The transverse speed of the wave is the displacement of the wave in the given period of its oscillation, and the period is given by the reverse of the frequency of the wave. Thus, using the given formulae with the given data, the required values can be calculated.

Formula:

The magnitude of two rectangular forces making angle$\mathrm{\theta }$with each other

$\left|\overrightarrow{{\mathrm{F}}_{\mathrm{net}}}\right|=\sqrt{{\mathrm{F}}_1^2+{\mathrm{F}}_1^2+2\left|{\mathrm{F}}_1\right|\left|{\mathrm{F}}_2\right|\mathrm{cos\theta }}$ (i)

Calculation of the rank according to magnitude of net electrostatic force

In the situation (a):

The forces from the two protons in the system are acting on the positive charge $+\mathrm{Q}$.

Thus, the net force at the charge$+\mathrm{Q}$will be given using the above values in equation (i) as follows:

$\begin{array}{c}\left|\overrightarrow{{\mathrm{F}}_{\left(\mathrm{a}\right)\mathrm{net}}}\right|=\sqrt{{\mathrm{F}}_{\mathrm{p}}^2+{\mathrm{F}}_{\mathrm{p}}^2+2\left|{\mathrm{F}}_{\mathrm{p}}\right|\left|{\mathrm{F}}_{\mathrm{p}}\right|\mathrm{cos\theta }}\\ ={\mathrm{F}}_{\mathrm{p}}\sqrt{2+2\mathrm{cos\theta }}\end{array}$

For situation (b):

The forces from a proton and an electron in the system are acting on the positive charge $+\mathrm{Q}$.

Thus, the net force at the charge $+\mathrm{Q}$is calculated using the above values in equation (i) as follows: role="math" localid="1661780121404" $\left(\because \left|{\mathrm{F}}_{\mathrm{p}}\right|=\left|{\mathrm{F}}_{\mathrm{e}}\right|\right)$

$\begin{array}{c}\left|\overrightarrow{{\mathrm{F}}_{\left(\mathrm{b}\right)\mathrm{net}}}\right|=\sqrt{{\mathrm{F}}_{\mathrm{p}}^2+{\mathrm{F}}_{\mathrm{p}}^2+2\left|{\mathrm{F}}_{\mathrm{p}}\right|\left|{\mathrm{F}}_{\mathrm{p}}\right|\cos (180-\mathrm{\theta })}\\ ={\mathrm{F}}_{\mathrm{p}}\sqrt{2-2\mathrm{cos\theta }}\end{array}$

For situation (c):

The forces from an electron and a proton in the system are acting on the positive charge $+\mathrm{Q}$.

Thus, the net force at the charge $+\mathrm{Q}$is calculated using the above values in equation (i) as follows: role="math" localid="1661780140799" $\left(\because \left|{\mathrm{F}}_{\mathrm{p}}\right|=\left|{\mathrm{F}}_{\mathrm{e}}\right|\right)$

$\begin{array}{c}\left|\overrightarrow{{\mathrm{F}}_{\left(\mathrm{c}\right)\mathrm{net}}}\right|=\sqrt{{\mathrm{F}}_{\mathrm{p}}^2+{\mathrm{F}}_{\mathrm{p}}^2+2\left|{\mathrm{F}}_{\mathrm{p}}\right|\left|{\mathrm{F}}_{\mathrm{p}}\right|\cos (180+\mathrm{\theta })}\\ ={\mathrm{F}}_{\mathrm{p}}\sqrt{2-2\mathrm{cos\theta }}\end{array}$

For situation (d):

The forces from a proton and an electron in the system are acting on the positive charge $+\mathrm{Q}$.

Thus, the net force at the charge $+\mathrm{Q}$is calculated using the above values in equation (i) as follows: $\left(\because \left|{\mathrm{F}}_{\mathrm{p}}\right|=\left|{\mathrm{F}}_{\mathrm{e}}\right|\right)$

$\begin{array}{c}\left|\overrightarrow{{\mathrm{F}}_{\left(\mathrm{d}\right)\mathrm{net}}}\right|=\sqrt{{\mathrm{F}}_{\mathrm{p}}^2+{\mathrm{F}}_{\mathrm{p}}^2+2\left|{\mathrm{F}}_{\mathrm{p}}\right|\left|{\mathrm{F}}_{\mathrm{p}}\right|\cos \left(\mathrm{\theta }\right)}\\ ={\mathrm{F}}_{\mathrm{p}}\sqrt{2+2\mathrm{cos\theta }}\end{array}$

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