Figure 21-21 shows four situations in which a central proton is partially surrounded by protons or electrons fixed in place along a half-circle. The angles$\theta$are identical; the anglesare$\varphi$ also. (a) In each situation, what is the direction of the net force on the central proton due to the other particles? (b) Rank the four situations according to the magnitude of that net force on the central proton, greatest first.

Figure 21-21 shows four situations in which a central proton is partially surrounded by protons or electrons fixed in place along a half-circle with angles $\theta$and role="math" localid="1662711519218" $\varphi$ being identical.

Using the concept of similar charges repelling each other and opposite charges attracting, we can get the directions and magnitude of the forces. Then further distributing the forces according to their x and y-components, we can get the net after canceling all the similar and opposite directed forces.

Formula:

The force due to x-component of a vector, $F_x=F\cos \theta$

For situation (1):

The net field is due to the contribution of the x-components of the force vectors due to four protons on the arc and that is towards negative x-axis as similar charge direction is away from the central particle. That is given using equation (i) as:

$$\begin{gathered} F_{\left(1x\right)}=-F_p\cos (90-\theta )-F_p\cos \varphi -F_p\cos (90-\theta )-F_p\cos \varphi \\ =-\left(2F_p{\cos }^2\right(90-\theta )+2F_p\cos \varphi \\ |F_{\left(1x\right)}|=2F_p\sqrt{{\cos }^2(90-\theta )+{\cos }^2\varphi } \end{gathered}$$

For situation (2):

The net field is due to the contribution of the x-components of the force vectors due to four protons on the arc and that is towards negative x-axis as similar charge direction is away from the central particle.

role="math" localid="1662712173618" $$\begin{gathered} F_{\left(2x\right)}=-F_p\cos (90-\theta )-F_p\cos \varphi -F_p\cos (90-\theta )-F_p\cos \varphi \\ =-\left(2F_p{\cos }^2\right(90-\theta )+2F_p\cos \varphi \\ |F_{\left(2x\right)}|=2F_p\sqrt{{\cos }^2(90-\theta )+{\cos }^2\varphi } \end{gathered}$$

For situation (3):

The net field is due to the contribution of the x-components of the force vectors due to two protons and two electrons on the arc and that is towards positive x-axis as the net field is depends on the angle made by the vectors of electrons and protons where from figure $\varphi >\theta$.

role="math" localid="1662712513650" $$\begin{gathered} F_{(3,x)}=+F_e\cos (90+\theta )-F_p\cos \varphi +F_e\cos (90+\theta )-F_p\cos \varphi \\ =-\left(2F_p{\cos }^2\right(90+\theta )+2F_p\cos \varphi |F_p|=|F_e\left| \\ \right|F_{\left(3\right),x}|=2F_p\sqrt{{\cos }^2(90+\theta )-{\cos }^2\varphi } \end{gathered}$$

For situation (4):

The net field is due to the contribution of the x-components of the force vectors due to two protons and two electrons on the arc and that is towards positive x-axis as the net field is depends on the angle made by the vectors of electrons and protons where from figure$\varphi >\theta$.

$$\begin{gathered} F_{(4,x)}=+F_p\cos (90-\theta )-F_e\cos \varphi -F_p\cos (90-\theta )+F_e\cos \varphi \\ =-\left(2F_p{\cos }^2\right(90-\theta )-2F_p\cos \varphi |F_p|=|F_e\left| \\ \right|F_{\left(4\right),x}|=2F_p\sqrt{{\cos }^2(90-\theta )-{\cos }^2\varphi } \end{gathered}$$

Since,$\varphi >\theta$

Thus,$\cos (90-\theta )>\cos \varphi$

The directions of the net force in situations (1) and (2) are in positive x-axis while the direction of force in situation (3) is in positive x-axis and in situation (4) in negative x direction.

From part (a) calculations, the rank of the situations according to the magnitude of the net forces is $\left|F_{\left(1x\right)}\right|=\left|F_{\left(2\right)x}\right|>\left|F_{\left(3\right)x}\right|=\left|F_{\left(4\right)x}\right|.$