Question

Question: Figure 21-31 shows an arrangement of four charged particles, with angle $\theta =30.0°$and distance= 2.00 cm. Particle 2 has charge$q_2=+8.00\times {10}^{-19}C$; particles 3 and 4 have charges$q_3=q_4=-1.60\times {10}^{19}C$. (a) What is distanceDbetween the origin and particle 2 if the net electrostatic force on particle 1 due to the other particles is zero? (b) If particles 3 and 4 were moved closer to thex-axis but maintained their symmetry about that axis, would the required value ofDbe greater than, less than, or the same as in part (a)?

Short answer

• a) The distance D between the origin and particle 2 if the net electrostatic force on particle 1 due to other particles is zero is 1.92 cm
• b) If particles 3 and 4 were moved closer to the x-axis but maintaining their symmetry, the value D is less than that in part (a).

Step-by-step solution

The given data

Charges of the particles,$q_2=+8.00\times {10}^{-19}C,and{q}_3=q_4=-1.60\times {10}^{-19}C$, and

Four particles are maintained at an angle $\theta =30.0°$and distance, d = 2.00 cm

Understanding the concept of Coulomb’s law 

Using the same concept of Coulomb's law, we can get an equation for the condition of the net force on particle 1 as zero. Thus, this will determine the distance of the particle from its origin. Again, using the same concept, we can compare the value of the distance.

Formula:

The magnitude of the electrostatic force between any two particles,

$F=K\frac{\left|q_1\right|\left|q_2\right|\cos \theta }{r^2}$

(1)

a) Calculation of the distance D of the second particle from the origin

We note that $\cos (30°)=\frac{\sqrt{3}}{2}$so that the dashed line distance in figure 21-31, the separation between the particles 1 and 3 or particles 1 and 4, $r=\frac{2d}{\sqrt{3}}$The net force on the charge $q_1$due to charges $q_{3}and{q}_4(with{q}_3=q_4=-1.60\times {10}^{-19}C)$onthe y -axis has magnitude using the equation as:

$$\begin{gathered} F_1=F_3+F_4 \\ =2\frac{\left|q_1{q}_3\right|}{4\pi {\epsilon }_0}\cos (30°) \\ =\frac{3\sqrt{3}\left|q_1{q}_3\right|}{16\pi {\epsilon }_0{d}^2} \end{gathered}$$

This must be set equal to the magnitude of the force exerted on $q_1$by $q_2=+8.00\times {10}^{-19}C$so that its net force is zero. Thus, using the same equation (1), it is given as:

$$\begin{gathered} \frac{3\sqrt{3}\left|q_1{q}_3\right|}{16\pi {\epsilon }_0{d}^2}=\frac{\left|q_1{q}_2\right|}{4\pi {\epsilon }_0{\left(D+d\right)}^2} \\ D=d\left(2\sqrt{\frac{5}{3\sqrt{3}}}-1\right) \\ D=0.96\times 2cm \\ D=1.92cm \end{gathered}$$

.

Hence, the value of the distance, D is 1.92 cm.

b) Calculation of the distance D when particles 3 and 4 are moved but maintaining the symmetry 

As the angle decreases, its cosine increases, resulting in a larger contribution from the charges on the y axis. To offset this, the force exerted by $q_2$must be made
stronger, so that it must be brought closer to$q_1$ (keep in mind that Coulomb’s law is inversely proportional to distance-squared). Thus, D must be decreased.